MCQ Questions for Class 11 Engineering Maths Sequences And Series Quiz 4

The sum of the series

$1^2 - 2^2 + 3^2 - 4^2 + .......... - 1000^2 + 1001^2$ is

0%
$-501\times1001$
0%
$501\times1001$
0%
$499\times1001$
0%
$-499\times1001$

Explanation

We would like to regroup the series as follows:

$1^2+(3^2-2^2)+(5^2-4^2)+....(1001^2-1000^2)$

$a^2-b^2=(a+b)(a-b)$ , in our case

$a-b=1$
Hence,

$1+5+9+13+... +2001$
There are

$501$ terms, then
Sum

$=501\times1001$

$\displaystyle \frac{1.2^{2}+2.3^{2}+3.4^{2}+.\cdot.\cdot.\cdot.\cdot+n(n+1)^{2}}{1^{2}.2+2^{2}.3+3^{2}.4++n^{2}(n+1)}=$

0%
$\displaystyle \frac{3n+1}{3n+5}$
0%
$\displaystyle \frac{3n+5}{3n+1}$
0%
$(3n+1)(3n+5)$
0%
$\displaystyle \frac{3n+5}{3n+7}$

Explanation

Numerator :

$\sum_{k=1}^n k(k+1)^2$

$= \sum_{k=1}^n (k^3+2k^2+k)$
Denominator :

$\sum_{k=1}^n k^2(k+1)$ =

$\sum_{k=1}^n (k^3 + k^2)$
On evaluating each sum we get
Numerator =

$n(n+1)(\frac{n(n+1)}{4} + \frac{2n+1}{3} + \frac{1}{2})$
Denominator =

$n(n+1)(\frac{n(n+1)}{4}+\frac{2n+1}{6})$
Now,

$Answer=\frac{numerator}{denominator}=\frac{n(n+1)}{n(n+1)}\times\frac{(n+2)(3n+5)}{(n+2)(3n+1)}=\frac{3n+5}{3n+1}$
Hence, option 'B' is correct.

$1^{2}+1+2^{2}+2+3^{2}+3+\ldots..+n^{2}+n=$

0%
$\displaystyle \frac{(n+1)(n+2)}{3}$
0%
$\displaystyle \frac{n(n+1)(n+2)}{3}$
0%
$\displaystyle \frac{n(n+2)(n+3)}{6}$
0%
$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}$

Explanation

$S=\Sigma n^2 + \Sigma n$
Therefore,

$S=\dfrac{n(n+1)(2n+1)}6+\dfrac{n(n+1)}2$
Hence,

$S=\dfrac{n(n+1)(n+2)}3$

The sum of

$7$ terms of the series

$1^2-2^2+3^2-4^2+5^2-6^2+...$ is

0%
$-21$
0%
$15$
0%
$28$
0%
$-35$

Explanation

We can see that first 6 terms taken in pair and is equal to

$-3+(-7)+(-11)$
So the sum will be

$-21+7^2$

$=28$

$1^3+1^2+1+2^{3}+2^{2}+2+3^{3}+3^{2}+3+ ... 3\mathrm{n}$ terms

$=$

0%
$\displaystyle \frac{n(n+1)(n^{2}+12n+5)}{12}$
0%
$\displaystyle \frac{n(n+1)(3n^{2}+7n+8)}{12}$
0%
$\displaystyle \frac{n(n+1)(n+2)(n^{2}+5n+6)}{12}$
0%
$\displaystyle \frac{(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}$

Explanation

$S=\Sigma r^3 +\Sigma r^2 +\Sigma r$

$S=\dfrac{n^2(n+1)^2}4 +\dfrac{n(n+1)(2n+1)}6 +\dfrac{n(n+1)}2$

$S=\dfrac{n(n+1)(3n^2 +7n+8)}{12}$

Hence Option B

The sum

$\mathrm{V}_{1}+\mathrm{V}_{2}+\ldots+\mathrm{V}_{\mathrm{n}}$ is

0%
$\displaystyle \frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}-\mathrm{n}+1)$
0%
$\displaystyle \frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}+\mathrm{n}+2)$
0%
$\displaystyle \frac{1}{2}\mathrm{n}(2\mathrm{n}^{2}-\mathrm{n}+1)$
0%
$\displaystyle \frac{1}{3}(2\mathrm{n}^{3}-2\mathrm{n}+3)$

Explanation

$\displaystyle \mathrm{V}_{\mathrm{r}}=\frac{\mathrm{r}}{2}[2\mathrm{r}+(\mathrm{r}-1)(2\mathrm{r}-1)]=\frac{1}{2}(2\mathrm{r}^{3}-\mathrm{r}^{2}+\mathrm{r})$

$\therefore \displaystyle \sum \mathrm{V}_{\mathrm{r}}=\frac{1}{2}\left(2 \sum r^3-\sum r^2+\sum r\right)=\frac{1}{12}\mathrm{n}(\mathrm{n}+1)(3\mathrm{n}^{2}+\mathrm{n}+2)$ .

The sum of the infinite series,

$\dfrac{1}{1.3}+\dfrac{2}{3.7}+\dfrac{3}{7.13}+\dfrac{4}{13.21}+ ..........$ is

0%
$\dfrac{1}{2}$
0%
$\dfrac{25}{24}$
0%
$\dfrac{25}{54}$
0%
$\dfrac{125}{252}$

Find

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n-1)!}$

0%
$\displaystyle \frac{1}{2}(e+e^{-1})$
0%
$\displaystyle \frac{1}{2}(e-e^{-1})$
0%
$e+e^{-1}$
0%
$\mathrm{e}-\mathrm{e}^{-1}$

$\displaystyle \frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots+n$ terms

$=$

0%
$\displaystyle \frac{n(n+3)}{4}$
0%
$\displaystyle \frac{n(n+3)}{5}$
0%
$\displaystyle \frac{n(n+2)}{3}$
0%
$\displaystyle \frac{n(n+5)}{6}$

Explanation

We Know,

$\sum_{i=1}^{n} k =1 + 2 + 3 +… + n = \dfrac{n(n+1)}{2}$ (sum of first n natural numbers)

$\sum_{i=1}^{n} k^2 =1^2 + 2^2 + 3^2 +… + n^2=\dfrac{n(n+1)(2n+1)}{6}$ (sum of squares of the first n natural numbers)

$\therefore t_{r}=\dfrac {\Sigma k^2}{\Sigma k}=\dfrac {\dfrac {n(n+1)(2n+1)}{6}}{\dfrac {n(n+1)}{2}}=\dfrac {2n+1}{3}$

$S=\dfrac{2}{3} \Sigma n +\dfrac{1}{3} \Sigma 1$

$S=\dfrac{2}{3} {\dfrac {n(n+1)}{2}} +\dfrac{1}{3} n$

$S=\dfrac {n(n+2)}{3}$

Evaluate

$1.6+2.9+3.12+...+\mathrm{n}(3\mathrm{n}+3)=$

0%
$\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)$
0%
$(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)$
0%
$(\mathrm{n}+2)(\mathrm{n}+3)(\mathrm{n}+4)$
0%
$(n-1) n(n+1)$

Explanation

$\sum _{ k=1 }^{ n }{k(3k+3)}=3\sum _{ k=1 }^{ n }{k^2} +3\sum _{ k=1 }^{ n }{k}$

$=3\times \dfrac{n(n+1)(2n+1)}{6}+3\times \dfrac{n(n+1)}{2}$

$=\dfrac{n(n+1)}{2}(2n+1+3)$

$=n(n+1)(n+2)$

$2.4+4.7+6.10+...(n-1)$ terms

0%
$2\mathrm{n}^{3}-2\mathrm{n}^{2}$
0%
$\displaystyle \frac{n^{3}+3n^{2}+1}{6}$
0%
$2n^{3}+2n$
0%
$2\mathrm{n}^{3}-\mathrm{n}^{2}$

Explanation

General term is given by

$(2k)(1+3k)$ where

$k$ varies from

$1$ to

$n-1$

$\displaystyle \sum_{k=1}^{n-1} (2k)(1+3k) = 2 \sum_{k=1}^{n-1} (k+3k^2)$ .
Substituting the values in the formula (Note that

$n = n-1$ ),
We get

$(n)(n-1)(1+(2(n-1)+1))$

$=2n^3-2n^2$

The sum of the first

$n$ terms of the series

$1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+.....$ is

$\dfrac {n(n+1)^2}{2}$ , when

$n$ is even,when

$n$ is odd, the sum is

0%
$\dfrac {n^2(n+1)}{2}$
0%
$\left[ \dfrac {n(n+1)}{2}\right]^{2}$
0%
$\dfrac {3n(n+1)}{2}$
0%
$\dfrac {n(n+1)^{2}}{4}$

Explanation

For odd, let

$n=2m+1$

$\Rightarrow { S }_{ 2m+1 }={ S }_{ 2m }+{ (2m+1) }^{ th } term$

$...(1)$
Also, given that

${ S }_{ 2m }=\dfrac { m{ (m+1) }^{ 2 } }{ 2 }$ .
Since,

$n=2m+1$

$\Rightarrow 2m=n-1$
So,

${ S }_{ 2m }=\dfrac { (n-1){ (n-1+1) }^{ 2 } }{ 2 }$

$\Rightarrow { S }_{ 2m }=\dfrac { (n-1){ n }^{ 2 } }{ 2 }$

$...(2)$
Also, the odd terms of the given series are

${ 3 }^{ 2 },{ 5 }^{ 2 },...etc$ .

$\Rightarrow { (2m+1) }^{ th } term={ (2m+1) }^{ 2 }$

$\Rightarrow { (2m+1) }^{ th } term={ n }^{ 2 }$

$...(3)$
Now, substitute

$(2)$ and

$(3)$ in

$(1)$ ; we get

${ S }_{ 2m+1 }=\dfrac { (n-1){ n }^{ 2 } }{ 2 } +{ n }^{ 2 }$

$\Rightarrow { S }_{ 2m+1 }=\dfrac { { n }^{ 3 }-{ n }^{ 2 }+2{ n }^{ 2 } }{ 2 }$

$\Rightarrow { S }_{ 2m+1 }=\dfrac { { n }^{ 3 }+{ n }^{ 2 } }{ 2 }$

$\Rightarrow { S }_{ 2m+1 }=\dfrac { { n }^{ 2 }(n+1) }{ 2 }$
If

$n = 4l+3$ , then

${S}_{2m+1}=\dfrac { ( 4l+3 )^{ 2 }(4l+4) }{ 2 }$ , which is even.
Hence, options 'A' and 'D' are correct.

$x = (-1)^{a^1} + (-1)^{a^2} + .....(-1)^{a^{1006}}$

$y = (-1)^{a^{1007}} + (-1)^{a^{1008}} + ...+(-1)^{a^{2013}}$
Then which of the following is true?

0%
$(-1)^{x} = 1; (-1)^{y} =1$
0%
$(-1)^{x} = 1; (-1)^{y} = -1$
0%
$(-1)^{x} = -1; (-1)^{y} =1$
0%
$(-1)^{x} = -1; (-1)^{y} = -1$

Explanation

$x=(-1)^{ { a }^{ 1 } }+(-1)^{ { a }^{ 2 } }+..............+(-1)^{ { a }^{ 1006 } }\\ \because \quad a\quad is\quad not\quad distinct\quad and\quad it\quad is\quad an\quad integer.\\ Let\quad us\quad take\quad a\quad as\quad even.\\ \therefore \quad { a }^{ n }\quad will\quad be\quad even\quad for\quad all\quad values\quad of\quad n.\\ \therefore \quad (-1)^{ { a }^{ n } }\quad will\quad be\quad 1\\ Total\quad number\quad of\quad terms\quad is\quad 1006.\\ \therefore \quad x=\sum _{ n=1 }^{ 1006 }{ (-1)^{ n } } =1006\\ \therefore \quad x=1006\longrightarrow an\quad even\quad number.\\ Again,\quad let\quad a\quad be\quad odd.\quad Then\quad { a }^{ n }\quad will\quad be\quad odd\quad for\quad all\quad \\ values\quad of\quad n.\\ \therefore \quad (-1)^{ { a }^{ n } }=-1\\ \therefore \quad \sum _{ n=1 }^{ 1006 }{ (-1)^{ n } } =\quad -1006\rightarrow an\quad even\quad negative\quad number.\\ Again\quad y=(-1)^{ { a }^{ 1007 } }+(-1)^{ { a }^{ 1008 } }+...................+(-1)^{ { a }^{ 2013 } }\\ As\quad per\quad the\quad previous\quad discussion\\ (-1)^{ { a }^{ n } }=1\quad in\quad case\quad a\quad is\quad even\quad and\quad \\ (-1)^{ { a }^{ n } }=\quad -1\quad in\quad case\quad a\quad is\quad odd\\ Total\quad number\quad of\quad terms=(2013-1007)+1=1007\\ Then\quad y=1007\quad in\quad case\quad a\quad is\quad even\\ \quad \quad \quad \quad \quad =-1007\quad in\quad case\quad a\quad is\quad odd\\ y\quad is\quad an\quad odd\quad number,\quad positive\quad and\quad negative\quad respectively.\\ Let\quad us\quad investigate\quad the\quad options.\\ In\quad case\quad a\quad is\quad odd\longrightarrow \left\{ x\quad is\quad even,\quad negative\\ y\quad is\quad odd,\quad negative \right\} ----(1)\\ In\quad case\quad a\quad is\quad even\longrightarrow \left\{ x\quad is\quad even,\quad positive\\ y\quad is\quad odd,\quad positive \right\} ----(2) Option\quad A\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\longrightarrow a\quad odd.\\ a\quad even\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\\ Option\quad A\quad is\quad not\quad correct.\\ Option\quad B\longrightarrow a\quad is\quad odd\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad a\quad is\quad even\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\\ Option\quad B\quad is\quad correct.\\ Obviously\quad option\quad C\quad and\quad D\quad are\quad not\quad correct.\\ Answer-\quad Option\quad B\quad is\quad correct.$

What number corresponds to the fifth missing planet ?

0%
28
0%
78
0%
32
0%
36

Explanation

After each term starting with

$p_3$ , the number in from of 4 is twice the number in front of 4 in the preceding term.
Thus the number for the unknown plant is

$2 \times 12 +4 =28$

State true or false:

Given sequence

$15, 30, 60, 120, 240$ is in G.P.

0%
True
0%
False

Explanation

Given series is

$15,30,60,120,240$

Ratio between first two terms

$=$

$\dfrac{30}{15}$

$=2$
Ratio between second and third terms

$=$

$\dfrac{60}{30}$

$=2$
Ratio between third and fourth terms

$=$

$\dfrac{120}{60}$

$=2$
Ratio between fourth and fifth terms

$=$

$\dfrac{240}{120}$

$=2$
Since, the ratio between the terms is the same. The series forms a G.P.

Let a sequence

$\left \{a_n\right \}$ be defined by

$a_n=\dfrac {1}{n+1}+\dfrac {1}{n+2}+\dfrac {1}{n+3}+.....+\dfrac {1}{3n}$ , then

0%
$a_2=\dfrac {7}{12}$
0%
$a_2=\dfrac {19}{20}$
0%
$a_{n+1}-a_n=\dfrac {9n+5}{(3n+1)(3n+2)(3n+3)}$
0%
$a_{n+1}-a_n=\dfrac {-2}{3(n+1)}$

Explanation

$\displaystyle a_n$ is sum of reciprocals of natural numbers starting from

$n + 1$ and ending at

$3n$ .

$\displaystyle \therefore a_2=\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}=\frac {19}{20}$

$\displaystyle \therefore a_{n+1}=\frac {1}{n+2}+\frac {1}{n+3}+ ....+\frac {1}{3n}+\frac {1}{3n+1}+\frac {1}{3n+2}+\frac {1}{3n+3}$

$\displaystyle \Rightarrow a_{n+1}-a_n=\frac {1}{3n+1}+\frac {1}{3n+2}+\frac {1}{3n+3}-\frac {1}{n+1}$

$\displaystyle =\frac {1}{3n+1}+\frac {1}{3n+2}-\frac {2}{3n+3}$

$\displaystyle =\frac {(3n+2)(3n+3)}{(3n+1)(3n+2)(3n+3)}+\frac {(3n+1)(3n+3)}{(3n+1)(3n+2)(3n+3)}-\frac {2(3n+1)(3n+2)}{(3n+1)(3n+2)(3n+3)}$

$\displaystyle =\frac {9n+5}{(3n+1)(3n+2)(3n+3)}$
Hence, options 'B' and 'C' are correct.

In the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, .......... starting with the third terms, each term is the sum of the two preceding terms :

$2=1+1, 3=1+2, 5=2+3$ , and so on.
Choose the correct options -

0%
Eighth term in Fibonacci sequence is 21
0%
Tenth term in Fibonacci sequence is 55
0%
Twelfth term in Fibonacci sequence is 144
0%
Twelfth term in Fibonacci sequence is 112

Explanation

From the sequence give we can find general formula

$a_n = a_{n-2}+ a_{n-1}$

$a_8 = a_7+a_6 =13 +8=21$

$a_9 = a_8+a_7 = 21+13 = 34$

$a_{10}= a_9+a_8 = 34+21=55$

$a_{11} = a_{10}+a_9 = 55 + 34 = 89$

$a_{12} = a_{10}+ a_{11}=55+89=144$

State true or false:

A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.

Parts of red pigment	1	4	7	12	20
Parts of base	8	...	...	...	...

Answer:

Parts of red pigment	1	4	7	12	20
Parts of base	8	32	56	96	150

0%
True
0%
False

The value of

$\alpha$ for which

$\sum_{n=1}^{\infty}A_n=\frac {8}{3}$ is-

0%
1/3, 2/3
0%
1/4, 3/4
0%
1/5, 4/5
0%
1/2

Explanation

$a_1=1, a_{n+1}^{2}=(2\alpha^2-2\alpha+1)^n$

$\therefore \sum_{n=1}^{\infty}A_n=\dfrac {1}{2\alpha-2\alpha^2}=\dfrac {8}{3}$

$\Rightarrow 2\alpha-2\alpha^2=3/8$

$\Rightarrow 16\alpha^2-16\alpha+3=0$

$\Rightarrow \alpha=1/4, 3/4$

Out of nine cells of a square, one cell is left blank and in the rest of the cells, numbers are written follow some rule. Get the rule and find out the proper option for the blank cell

2	72	56
?	0	42
12	20	30

0%
4
0%
6
0%
8
0%
10

Explanation

The pattern of numbers are as followed

$1^2-1=0$

$2^2-1=3$

$3^2-3=6$

$4^2-4=12$

$5^2-5=20$

$6^2-6=30$

$7^2-7=42$

$8^2-8=56$

$9^2-9=72$

Which of the following is not a G.P.?

0%
$2, 4, 6, 8....$
0%
$5, 25, 125, 625....$
0%
$1.5, 3.0, 6.0, 12.0....$
0%
$8, 16, 24, 32, ....$

Explanation

In series

$2,4,6,8,....$ difference is same i.e.

$2$

$8,16,24,32,......$ difference again is same

$8$

$\therefore$ Both the series (a) and (b) are in AP as the difference between their consecutive terms is the same.

The value of

$\sum_{n=1}^{10}\int_{0}^{n}xdx$ is

0%
an even integer
0%
an odd integer
0%
a rational number
0%
an irrational number

Explanation

$\displaystyle \sum_{n=1}^{10} \int_{0}^{n} x dx = \sum_{n=1}^{10} \frac{n^{2}}{2}$

$=\displaystyle \frac{1^{2}+3^{2}+...+9^{2}}{2} + \frac{2^{2}+4^{2}+...+10^{2}}{2}$

$= A +B$

Now,

$B$ is even divided by

$2 \Rightarrow B$ is an integer

$A$ contains

$5$ odd terms not divided by

$2$

$\therefore A$ is a fraction . Thus,

$A+B$ is rational

Hence, option 'C' is correct.

If n is an odd integer greater than or equal to 1 then the value of

$n^{3}-\left ( n-1 \right )^{3}+\left ( n-2 \right )^{3}-...+\left ( -1 \right )^{n-1}.1^{3}$ is

0%
$\dfrac{\left ( n+1 \right )^{2}.\left ( 2n-1 \right )}{4}$
0%
$\dfrac{\left ( n-1 \right )^{2}.\left ( 2n-1 \right )}{4}$
0%
$\dfrac{\left ( n+1 \right )^{2}.\left ( 2n+1 \right )}{4}$
0%
none of these

Explanation

The sequence can be written as :

${ S }_{ n }={ n }^{ 3 }+{ \left( n-2 \right) }^{ 3 }+{ \left( n-4 \right) }^{ 3 }+......+{ 1 }^{ 3 }-\left[ { \left( n-1 \right) }^{ 3 }+{ \left( n-3 \right) }^{ 3 }+{ \left( n-5 \right) }^{ 3 }+....+{ 2 }^{ 3 } \right] \\ ={ n }^{ 3 }+{ \left( n-2 \right) }^{ 3 }+{ \left( n-4 \right) }^{ 3 }+......+{ 1 }^{ 3 }+\left[ { \left( n-1 \right) }^{ 3 }+{ \left( n-3 \right) }^{ 3 }+{ \left( n-5 \right) }^{ 3 }+....+{ 2 }^{ 3 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -2\left[ { \left( n-1 \right) }^{ 3 }+{ \left( n-3 \right) }^{ 3 }+{ \left( n-5 \right) }^{ 3 }+....+{ 2 }^{ 3 } \right] \\ =\sum _{ r=1 }^{ n }{ r^{ 3 } } -2\times { 2 }^{ 3 }\left[ { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+............+{ \left( \dfrac { n-1 }{ 2 } \right) }^{ 3 } \right]$
we have taken 8 common from the 2nd part of the sequence.

$=\sum _{ r=1 }^{ n }{ r^{ 3 } } -16\sum _{ 1 }^{ \dfrac { n-1 }{ 2 } }{ r^{ 3 } } \\ =\left[ { \dfrac { n\left( n+1 \right) }{ 2 } }^{ 2 } \right] -16\left[ { \dfrac { \dfrac { n-1 }{ 2 } \left( \dfrac { n-1 }{ 2 } +1 \right) }{ 2 } }^{ 2 } \right] \\ ={ \dfrac { n\left( n+1 \right) }{ 2 } }^{ 2 }-16\left[ { \dfrac { \left( n-1 \right) \left( n+1 \right) }{ 8 } }^{ 2 } \right] \\ =\dfrac { { \left( n+1 \right) }^{ 2 } }{ 4 } \left[ { n }^{ 2 }-{ \left( n-1 \right) }^{ 2 } \right] \\ =\dfrac { { \left( n+1 \right) }^{ 2 }\left( 2n-1 \right) }{ 4 }$

Find the sum of first

$n$ terms of the series.

$1^{3}+3\times2^{2}+3^{3}+3\times4^{2}+5^{3}+3\times6^{2}+....$ If

$n$ is even.

0%
$\displaystyle \frac{n}{8}(n^{3}+4n^{2}+10n+6)$
0%
$\displaystyle \frac{n}{8}(n^{2}+4n^{2}+10n+6)$
0%
$\displaystyle \frac{n}{8}(n^{3}+4n^{2}+10n+8)$
0%
$\displaystyle \frac{n}{8}(n^{2}+4n^{2}+10n+8)$

Explanation

$n$ is even,
Let

$n=2m$

$\therefore S=1^{3}+3.2^{2}+3^{3}+3.4^{2}+5^{3}+3.6^{2}+....+(2m-1)^{3}+3(2m)^{2}$

$=\left \{ 1^{3}+3^{3}+5^{3}+...+(2m-1)^{3} \right \}+3\left \{ 2^{2}+4^{2}+6^{2}+....+(2m)^{2} \right \}$

$=\sum_{r=1}^{m}(2r-1)^{3}+3.4\sum_{r=1}^{m}r^{2}$

$=\sum_{r=1}^{m}\left \{ 8r^{3}-12r^{2}+6r-1 \right \}+12\sum_{r=1}^{m}r^{2}$

$=8\sum_{r=1}^{m}r^{3}-12\sum_{r=1}^{m}r^{2}+6\sum_{r=1}^{m}r-\sum_{r=1}^{m}1+12 \sum_{r=1}^{m}r^{2}$

$=8\sum_{r=1}^{m}r^{3}+6\sum_{r=1}^{m}r-\sum_{r=1}^{m}1$

$=8.\displaystyle \frac{m^{2}(m+1)^{2}}{4}+6\displaystyle \frac{m(m+1)}{2}-m$

$=2m^{2}(m+1)^{2}+3m(m+1)-m$

$=m[2m^{3}+4m^{2}+5m+2]$

$=\displaystyle \frac{n}{2}\left [ 2\left ( \displaystyle \frac{n}{2} \right )^{3}+4\left ( \displaystyle \frac{n}{2} \right )^{2}+5\left ( \displaystyle \frac{n}{2}\right )+2 \right ]$

$(\because m=\displaystyle \frac {n}{2})$

$=\displaystyle \frac{n}{8}(n^{3}+4n^{2}+10n+8)$

Let

$S_{n}$ denote the sum of cubes of the first

$n$ natural numbers and

$s_{n}$ denote the sum of the first

$n$ natural numbers. Then

$\sum_{r=1}^{n}\dfrac{S_{r}}{S_{r}}$ is equal to

0%
$\dfrac{n\left ( n+1 \right )\left ( n+2 \right )}{6}$
0%
$\dfrac{n\left ( n+1 \right )}{2}$
0%
$\dfrac{n^{2}+3n+2}{6}$
0%
none of these

Explanation

Since,

$S_{ n }$ denote the sum of cubes of the first

$n$ natural numbers.
Therefore,

$S_{ n }=\dfrac { { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } }{ 4 }$
Since,

$s_{ n }$ denote the sum of the first n natural numbers.
Therefore,

$s_{ n }=\dfrac { { n }{ \left( n+1 \right) } }{ 2 }$

Now,

${ T }_{ n }=\sum _{ r=1 }^{ n } \dfrac { S_{ r } }{ s_{ r } } =\sum _{ r=1 }^{ n } \dfrac { r\left( r+1 \right) }{ 2 }$

$\Rightarrow { T }_{ n }=\dfrac { 1 }{ 2 } \left\{ \sum _{ r=1 }^{ n }{ { r }^{ 2 } } +\sum _{ r=1 }^{ n } r \right\} =\dfrac { { n }{ \left( n+1 \right) }{ \left( 2n+1 \right) } }{ 12 } +\dfrac { { n }{ \left( n+1 \right) } }{ 4 }$

$\Rightarrow { T }_{ n }=\dfrac { n\left( n+1 \right) \left( n+2 \right) }{ 6 }$

Let a sequence be defined by

$a_{1}= 0$ and

$a_{n+1}= a_{n}+4n+3$ for all

$n\geq 1 (n\epsilon N)$

The value of

$a_{k}$ in terms of k is

$(k\in N)$

0%
$(k-1)(2k+3)$
0%
$(k-1)(3k+1)$
0%
$(k-1)(k+3)$
0%
$(k-1)(2k+5)$

Explanation

${ a }_{ n+1 }={ a }_{ n }+4n+3$

${ a }_{ 2 }={ a }_{ 1 }+4.1+3=7=\left( 2-1 \right) \left( 2 \cdot 2+3 \right)$

${ a }_{ 3 }={ a }_{ 2 }+4.2+3=18=\left( 3-1 \right) \left( 3 \cdot 2+3 \right)$

${ a }_{ k }=\left( k-1 \right) \left( 2k+3 \right)$

Let

$\displaystyle V_{r}$ denote the sum of the first r terms of an arithmetic progression (AP) whose first term is r and the common difference is

$\displaystyle (2r-1).$ Let

$\displaystyle T_{r}=V_{r+1}-V_{r}-2$ and

$\displaystyle Q_{r}=T_{r+1}-T_{r}$ for

$r=1, 2, ...$

The sum

$\displaystyle V_{1}+V_{2}+... +V_{n}$ is

0%
$\displaystyle \frac{1}{12}n\left ( n+1 \right )\left ( 3n^{2}-n+1 \right )$
0%
$\displaystyle \frac{1}{12}n\left ( n+1 \right )\left ( 3n^{2}+n+2 \right )$
0%
$\displaystyle \frac{1}{2}n\left ( 2n^{2}-n+1 \right )$
0%
$\displaystyle \frac{1}{3}\left ( 2n^{3}-2n+3 \right )$

Explanation

$V_r = \dfrac{r}{2}(r+(r-1)(2r-1))$

$=\dfrac{1}{2}(2r^3 - r^2 +r)$

$V_1 +V_2 +V_3 +....... = \sum V_r$

$\sum V_r =\dfrac{1}{2}( {2 \sum r^3 - \sum r^2 + \sum r})$

$=\dfrac{1}{2} (2\times \dfrac{r^2 (r+1)^2}{4} - \dfrac{r (r+a)(2r+1)}{6} + \dfrac{r(r+1)}{2})$

$= \dfrac{1}{12} {r (r+1)(3r^2+r+2)}$

for

$r=n$ (i.e. sum upto n terms)

$= \dfrac{1}{12}{n(n+1)(3n^2 +n+2)}$

Find the sum of the products of every pair of the first

$n$ natural numbers.

0%
$S=\displaystyle \frac{(n-1)^2(3n+2)}{24}$
0%
$S=\displaystyle \frac{n(n+1)(n-1)(3n-2)}{24}$
0%
$S=\displaystyle \frac{n(n-1)^2(3n+2)}{24}$
0%
$S=\displaystyle \frac{n(n+1)(n-1)(3n+2)}{24}$

Explanation

We find that

$S=1.2+1.3+1.4+....+2.3+2.4+.......+3.4+3.5+.....+(n-1)n \cdots (1)$

$\because (1+2+3+...+(n-1)+n)^{2}=1^{2}+2^{2}+3^{2}+...(n-1)^{2}+n^{2}$

$+2(1.2+1.3+1.4+....+2.3+2.4+.......+3.4+3.5+.....+(n-1)n )$ from

$(1)$

$(\sum n)^{2}=\sum n^{2}+2S$

$\Rightarrow S=\displaystyle \frac{(\sum n)^{2}-\sum n^{2}}{2}$

$=\displaystyle \frac{\left \{ \displaystyle \frac{n(n+1)}{2} \right \}^{2}-\displaystyle \frac{n(n+1)(2n+1)}{6}}{2}$

$=\displaystyle \frac { \displaystyle \frac{n^{2}(n+1)}{4}^{2}-\displaystyle \frac{n(n+1)(2n+1)}{6}}{2}$

$=\displaystyle \frac{n(n+1)}{24}\left \{ 3n(n+1)+2(2n+1) \right \}$

$=\displaystyle \frac{n(n+1)(3n^{2}-n-2)}{24}$

$=\displaystyle \frac{(n+1)n(n-1)(3n+2)}{24}$

The value of

$\sum_{r=1}^{n}\left \{ \left ( 2r-1 \right )a+\dfrac{1}{b^{r}} \right \}$ is equal to

0%
$an^{2}+\dfrac{b^{n-1}-1}{b^{n-1}\left ( b-1 \right )}$
0%
$an^{2}+\dfrac{b^{n}-1}{b^{n}\left ( b-1 \right )}$
0%
$an^{3}+\dfrac{b^{n-1}-1}{b^{n}\left ( b-1 \right )}$
0%
none of these

Explanation

Let

$S=\sum _{ r=1 }^{ n } \left\{ \left( 2r-1 \right) a+\dfrac { 1 }{ b^{ r } } \right\}$

$\Rightarrow S=2a\sum _{ r=1 }^{ n } r-a\sum _{ r=1 }^{ n } +\sum _{ r=1 }^{ n } \left( \dfrac { 1 }{ b^{ r } } \right)$

$\Rightarrow S=an\left( n+1 \right) -an+\dfrac { 1 }{ b } \left[ \dfrac { { \left( 1/b \right) }^{ n }-1 }{ \left( 1/b \right) -1 } \right]$

$\Rightarrow S=a{ n }^{ 2 }+\dfrac { { b }^{ n }-1 }{ { b }^{ n }\left( b-1 \right) }$

Ans: B

Sum of the series

$11+23+45+87 ...$ up to

$n$ terms

0%
$10(2^n-1)+n^2$
0%
$10(2^n+1)+n^2$
0%
$20(2^n-n)+n^2$
0%
None of these

Explanation

Let

$S=11+23+45+87+...$ up to

$n$ terms

$\Rightarrow \ S=\left( 5.2+1 \right) +\left( 5.{ 2 }^{ 2 }+3 \right) +\left( 5.{ 2 }^{ 3 }+5 \right) +\left( 5.{ 2 }^{ 4 }+7 \right) +...$ up to

$n$ terms

Now,

${ a }_{ n }=5{ .2 }^{ n }+2n-1$
Therefore,

$S=\sum _{ n=1 }^{ n }{ { a }_{ n } } =\sum _{ n=1 }^{ n }{ \left( 5{ .2 }^{ n }+2n-1 \right) }$

$\Rightarrow S=5\sum _{ n=1 }^{ n }{ { 2 }^{ n } } +2\sum _{ n=1 }^{ n }{ n } -\sum _{ n=1 }^{ n }{ 1 } =\dfrac { 5.2\left( { 2 }^{ n }-1 \right) }{ 2-1 } +\dfrac { 2n\left( n+1 \right) }{ 2 } -n$

$\Rightarrow S=10\left( { 2 }^{ n }-1 \right) +{ n }^{ 2 }$

$\displaystyle 1^{2}+\left ( 1^{2}+2^{2} \right )+\left (1^{2}+2^{2}+3^{2} \right )+...$ to

$n$ terms.

0%
$\dfrac{1}{12}n(n+1)(2n^3+3n^2+n)$
0%
$\dfrac{1}{12}n(n+1)(3n^3+2n^2+n)$
0%
$\dfrac{1}{6}n(n+1)(2n+1)$
0%
$\dfrac{1}{4}n^2(n+1)^2$

Explanation

nth term

$T_n = 1^2+2^2+3^2+.....+n^2=\sum n^2=\cfrac{1}{6}n(n+1)(2n+1)=\cfrac{1}{6}(2n^3+3n^2+n)$

Thus required summation is

$\displaystyle =\sum T_n =\frac{1}{6}\left(\sum 2n^3+\sum 3n^2+ \sum n\right)$

$\displaystyle = \frac{1}{6}\left(2\frac{n^2(n+1)^2}{4}+3\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right) =\frac{1}{12}n(n+1)(2n^3+3n^2+n)$

$\displaystyle 1^{3}-2^{3}+3^{3}-4^{3}+...+9^{3}$ equals

0%
425
0%
400
0%
405
0%
395

Explanation

Let,

$S=1^{ 3 }-2^{ 3 }+3^{ 3 }-4^{ 3 }+...+9^{ 3 }$

$\Rightarrow S=\left( 1^{ 3 }+3^{ 3 }+5^{ 3 }+...+9^{ 3 } \right) -\left( 2^{ 3 }+4^{ 3 }+6^{ 3 }+8^{ 3 } \right)$

$\Rightarrow S=\left( 1^{ 3 }+3^{ 3 }+5^{ 3 }+...+9^{ 3 } \right) -2^{ 3 }\left( 1^{ 3 }+2^{ 3 }+3^{ 3 }+4^{ 3 } \right)$

$\Rightarrow S=\sum _{ n=1 }^{ 5 }{ { \left( 2n-1 \right) }^{ 3 } } -2^{ 3 }\sum _{ n=1 }^{ 4 }{ { n }^{ 3 } }$

$\Rightarrow S=8\sum _{ n=1 }^{ 5 }{ { n }^{ 3 } } -12\sum _{ n=1 }^{ 5 }{ { n }^{ 2 } } +6\sum _{ n=1 }^{ 5 }{ { n } } -\sum _{ n=1 }^{ 5 }{ 1 } -2^{ 3 }\sum _{ n=1 }^{ 4 }{ { n }^{ 3 } }$

$\Rightarrow S=8\left[ \dfrac{n^2(n+1)^2}{4}\right] -12\left[\dfrac{n(n+1)(2n+1)}{6} \right] +6\left[ \dfrac{n(n+1)}{2} \right] -5-2^{ 3 }\left[ \dfrac{n^2(n+1)^2}{4} \right]$

$\Rightarrow S=8\left[ \dfrac { { 5 }^{ 2 }{ \left( 5+1 \right) }^{ 2 } }{ 4 } \right] -12\left[ \dfrac { 5\left( 5+1 \right) \left( 2.5+1 \right) }{ 6 } \right] +6\left[ \dfrac { 5\left( 5+1 \right) }{ 2 } \right] -5-2^{ 3 }\left[ \dfrac { 4^{ 2 }{ \left( 4+1 \right) }^{ 2 } }{ 4 } \right]$

$\Rightarrow S=1800-660+90-5-800=425$

Ans: A

The sum of all possible product of

$1^{st}$

$n$ natural numbers taken two at a time is

0%
$\displaystyle \frac{n\left ( n+1 \right )\left ( n+2 \right )\left ( 2n+3 \right )}{24}$
0%
$\displaystyle \frac{n\left ( n^{2}-1 \right )\left ( 3n+2 \right )}{24}$
0%
$\displaystyle \frac{n\left ( n^{2}+1 \right )\left ( 3n+2 \right )}{6}$
0%
$\displaystyle \frac{n\left ( n^{2}-1 \right )\left ( 3n+2 \right )}{6}$

Explanation

Consider

$\displaystyle \left ( b_{1}+b_{2}+b_{3}+...+b_{n} \right )^{2}=b_{1}^{2}+b_{2}^{2}+...+b_{n}^{2}+2\sum_{i= j}b_ib_j$
taking

$\displaystyle b_{1}=1,b_{2}=2,...,b_{n}=n$

$\displaystyle \therefore \left ( 1+2+3+...+n \right )^{2}=1^{2}+2^{2}+3^{2}+...+n^{2}+2\displaystyle \sum$ (Product of number taken two at a time)

$\Rightarrow 2 \sum{b_ib_j} \displaystyle =\left ( 1+2+3+...+n \right )^{2}-\sum_{n=1}^{n}n^{2}\displaystyle=\frac{n^{2}\left ( n+1 \right )^{2}}{4}-\frac{n\left ( n+1 \right )\left ( 2n+1 \right )}{6}=\frac{n\left ( n^{2}-1 \right )\left ( 3n+2 \right )}{12}$

$\Rightarrow \sum{b_ib_j} =\dfrac{n\left ( n^{2}-1 \right )\left ( 3n+2 \right )}{24}$

Sum of the series

$2.3.1+3.4.4+4.5.7+...$ up to

$n$ terms is

0%
$\displaystyle \frac{n}{12}\left [ 9n^{3}+46n^{2}+51n-34 \right ]$
0%
$\displaystyle \frac{n}{6}\left [ 9n^{3}+46n^{2}+51n-34 \right ]$
0%
$\displaystyle \frac{\left ( n+1 \right )}{6}\left [ 9n^{3}+46n^{2}-34 \right ]$
0%
$\displaystyle \frac{n}{12}\left [ 9n^{3}+46n^{2}-34 \right ]$

Explanation

$S=2.3.1+3.4.4+4.5.7\\ S=\sum _{ r=1 }^{ n }{ \left( r+1 \right) \left( r+2 \right) \left( 3r-2 \right) } =\sum _{ r=1 }^{ n }{ \left( r+1 \right) \left( { 3r }^{ 2 }+4r-4 \right) }$

$\displaystyle =\sum _{ r=1 }^{ n }{ \left( { 3r }^{ 3 }+{ 7r }^{ 2 }-4 \right) } =3{ \left( \frac { n\left( n+1 \right) }{ 2 } \right) }^{ 2 }+7\left( \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } \right) -4$

$\displaystyle =\frac { n }{ 12 } \left[ { 9n }^{ 3 }+{ 46n }^{ 2 }+51n-34 \right]$

Let

$\displaystyle \left ( a_{n} \right )n\geq 1$ be an increasing sequence of positive integers such that 1.

$\displaystyle a_{2n}=a_{n}+n$ for all

$\displaystyle n\geq 1$ 2.if

$\displaystyle a_{n}$ is prime, then n is a prime. Prove that

$\displaystyle a_{n}=n,$ for all

$\displaystyle n\geq 1.$

0%
$\displaystyle n\geq 1.$
0%
$\displaystyle n\ne 1.$
0%
$\displaystyle n= 1.$
0%
None of the above

Explanation

Let $a_1=c$ . Then $a_2=a_1+1=c+1$ , $a_4=a_2+2=c+3$ . Since the sequence is increasing,it follows that $a_3=c+2$ . We Prove that $a_n=c+n1$ for all $n1$ . Indeed, if $n=2k$ for some integer $k$ this follows by induction on $k$ . Suppose that $a_{2^k}=c+2^k1$ . Then $a_{2^{k+1}}=a_{2.2^k}=a_{2^k}+2^k=c+2^{k+1}1$ . If $2^k<n<2^{k+1}$ , then $c+2^{k}1=a_{2^k}<a_{2^k+1}<...<a_n<...<a_{2^k+1}=c+2^{k+1}1$ add this is possible only if $a_n=c+n1$ .

Next we prove that $c=1$ . Suppose that $c2$ and let $p<q$ be two consecutive prime numbers greater than $c$ . We have $a_{qc+1}=c+qc=q$ , hence $qc+1$ is a prime and clearly $qc+1p$ . It follows that for any consecutive prime numbers $p<q$ we have $qc+1$ . The numbers $(c+1)!+2,(c+1)!+3...,(c+1)!+c+1$ are all composite, hence if $p$ and $q$ are the consecutive primes such that $p<(c+1)!+2<(c+1)!+c+1<q$ then $qp>c1$ , a contradiction. It follows that $c=0$ and $a_n=n$ for all $n1$ .

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

$\displaystyle \sum_{r=1}^{n}r(n-r)^{3}= \sum_{r=1}^{n}(n-r)r^{3}$

$=\displaystyle \sum_{r=1}^{n}\left ( nr^{3}-r^{4} \right )$

$=\displaystyle n \sum_{r=1}^{n} r^{3}-\sum_{r=1}^{n} r^{4}=n \left \{ \frac{n(n+1)}{2} \right \}^{2}-f(n)$

Ans: A

$\displaystyle \frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + .....$ upto n terms is

0%
$\displaystyle \frac{1}P{3} (2n + 1)$
0%
$\displaystyle \frac{1}{3} n^{2}$
0%
$\displaystyle \frac{1}{3} (n + 2)$
0%
$\displaystyle \frac{1}{3} n(n + 2)$

Explanation

General term of the given series is,

$\displaystyle T_n =\frac{\sum_{k=1}^n k^2}{\sum_{k=1}^n k}=\frac{\dfrac{1}{6}n(n+1)(2n+1)}{\dfrac{1}{2}n(n+1)}=\frac{1}{3}(2n+1)$
The required summation is,

$\displaystyle \sum_{k=1}^nT_k=\frac{1}{3}\left(2\sum_{k=1}^n k+\sum_{k=1}^n 1 \right)=\frac{1}{3}[n(n+1)+n]=\frac{1}{3}n(n+2)$
Hence option 'D' is correct choice.

The sum of n terms of

$1^{2}\, +\, (1^{2}\, +\, 2^{2})\, +\, (1^{2}\, +\, 2^{2}\, +\, 3^{2})\, +\, ....$ .

0%
$\displaystyle \frac{n(n\, +\,a)\, (2n\, +\, 1)}{6}$
0%
$\displaystyle \frac{n(n\, +\, 1)\, (2n\, -\, 1)}{6}$
0%
$\displaystyle \frac{1}{12}\, n(n\, +\, 1)^{2}\, (n\, +\, 2)$
0%
$\displaystyle \frac{1}{12}n^{2}(n\, +\, 1)^{2}$

Explanation

$S={ 1 }^{ 2 }+\left( { 1 }^{ 2 }+{ 2 }^{ 2 } \right) +\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } \right) +...$

$\displaystyle=\sum _{ r=1 }^{ n }{ \left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ r }^{ 2 } \right) }$

$\displaystyle=\sum _{ r=1 }^{ n }{ \left( \frac { r\left( r+1 \right) \left( 2r+1 \right) }{ 6 } \right) }$

$\displaystyle=\frac { 1 }{ 6 } \sum _{ r=1 }^{ n }{ \left( r\left( { 2r }^{ 2 }+3r+1 \right) \right) }$

$\displaystyle=\frac { 1 }{ 6 } \sum _{ r=1 }^{ n }{ \left( { 2r }^{ 3 }+3{ r }^{ 2 }+r \right) } =\frac { 2 }{ 6 } \sum _{ r=1 }^{ n }{ { r }^{ 3 } } +\frac { 3 }{ 6 } \sum _{ r=1 }^{ n }{ { r }^{ 2 } } +\frac { 1 }{ 6 } \sum _{ r=1 }^{ n }{ { r } }$

$\displaystyle=\frac { 1 }{ 3 } { \left( \frac { n\left( n+1 \right) }{ 2 } \right) }^{ 2 }+\frac { 1 }{ 2 } \left( \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } \right) +\frac { 1 }{ 6 } \left( \frac { n\left( n+1 \right) }{ 2 } \right)$

$=\displaystyle{ \left( \frac { { n }^{ 2 }\left( n+1 \right) }{ 12 } \right) }^{ 2 }+\frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 12 } +\frac { n\left( n+1 \right) }{ 12 }$

$=\displaystyle\frac { 1 }{ 12 } \left( n{ \left( n+1 \right) }^{ 2 }\left( n+2 \right) \right)$

1 + 3 + 6 + 10 + .... upto n terms is equal to

0%
$\displaystyle \frac{1}{3} n(n + 1)(n + 2)$
0%
$\displaystyle \frac{1}{6} n(n + 1)(n + 2)$
0%
$\displaystyle \frac{1}{12} n(n + 2)(n + 3)$
0%
$\displaystyle \frac{1}{12} n(n + 1)(n + 2)$

Explanation

Let

$S =1+3+6+10+..............+T_n$ ........(1)
Also

$S = 1+3 + 6+ 10+..........+T_{n-1} +T_n$ ........(2)
Now by

$(1)-(2)$

$0=1+2+3+.........+n-T_n$

$\Rightarrow T_n = 1+2+3......+n=\cfrac{1}{2}n(n+1)$
Hence required summation is,

$\displaystyle \sum_{k=1}^nT_k=\frac{1}{2}\left(\sum_{k=1}^nk^2+\sum_{k=1}^n k \right)$

$\displaystyle =\frac{1}{2}\left(\frac{1}{6}n(n+1)(2n+1)+\frac{1}{2}n(n+1) \right)$

$\quad =\displaystyle \frac{1}{6}n(n+1)(n+2)$

Determine the fourth degree expression in n which is equal to

$\displaystyle \sum_{r=1}^{n}r\left ( r+1 \right )\left ( 2r+3 \right ).$

0%
$\displaystyle \frac{1}{6}\left ( 3n^{4}+16n^{3}+27n^{2}+14n \right ).$
0%
$\displaystyle \frac{1}{6}\left ( 3n^{4}+9n^{3}+27n^{2}+36n \right ).$
0%
$\displaystyle \frac{1}{6}\left ( 3n^{4}+36n^{3}+27n^{2}+9n \right ).$
0%
$\displaystyle \frac{1}{6}\left ( 3n^{4}+14n^{3}+27n^{2}+16n \right ).$

Explanation

Sum

$=\displaystyle \sum_{r=1}^{n}r\left ( r+1 \right )\left ( 2r+3 \right )=\sum_{r=1}^{n}(2r^3+5r^2+3r)$

$= 2\sum_{r=1}^{n}r^3+5\sum_{r=1}^{n}r^2+3\sum_{r=1}^{n}r$

$= 2(\cfrac{n(n+1)}{2})^2+5\cfrac{n(n+1)(2n+1)}{6}+3\cfrac{n(n+1)}{2}$

$=\displaystyle \frac{1}{6}\left ( 3n^{4}+16n^{3}+27n^{2}+14n \right ).$

The sum of n terms of the series

$1 + (1 + a) + (1 + a + a^{2}) + (1 + a + a^{2} + a^{3}) + .....$ , is

0%
$\displaystyle \frac{n}{1 - a} - \frac{a(1 - a^{n})}{(1 - a)^{2}}$
0%
$\displaystyle \frac{n}{1 - a} + \frac{a(1 - a^{n})}{(1 - a)^{2}}$
0%
$\displaystyle \frac{n}{1 - a} - \frac{a(1 + a^{n})}{(1 - a)^{2}}$
0%
$\displaystyle - \frac {n}{1 - a} + \frac{a(1 - a^{n})}{(1 - a)^{2}}$

Explanation

$T_n = 1 + a + a^{2} + ..... + a^{n - 1} = \displaystyle \frac{1 - a^{n}}{1 - a}$

$S_n = \Sigma T_n =\displaystyle \frac{1}{(1 -a)} [\Sigma 1 - \Sigma a^{n}]$

$= \displaystyle \frac{1}{(1 - a)} [n - (a + a^{2} + a^{3} + ..... a^{n})]$

$= \displaystyle \frac{1}{(1 - a)} \left[n - \frac{a(1 - a^{n})}{(1 - a)} \right] = \frac{n}{1 - a} - \frac{a(1 - a^{n})}{(1 - a)^{2}}$

Observe the given multiples of 37.

${37\times3=111}$

${37\times 6 =222}$

${37\times9=333}$

${37\times12=444}$ -------------------------------
Find the product of

${37\times27}$

0%
$999$
0%
Greatest $3$ digit number
0%
Either A or B
0%
Smallest $3$ digit number

Explanation

${37\times3=37\times(3\times1)=111}$

${37\times 6=37\times(3\times2)=222}$

${37\times9=37\times(3\times3)=333}$

${37\times27=37\times(3\times9)=999}$

Sum to

$20$ terms of the series

$1.{ 3 }^{ 2 }+2.{ 5 }^{ 2 }+3.{ 7 }^{ 2 }+...$ is

0%
$178090$
0%
$168090$
0%
$188090$
0%
None of these

Explanation

We have,

${ t }_{ n }=[nth$ term of

$1,2,3,...]\times [nth$ term of

$3,5,7,...{ ] }^{ 2 }$

$=n{ \left( 2n+1 \right) }^{ 2 }=4{ n }^{ 3 }+4{ n }^{ 2 }+n$

$\displaystyle \therefore { S }_{ n }=\sum { { t }_{ n } } =4\sum { { n }^{ 3 } } +4\sum { { n }^{ 2 } } +\sum { n }$

$\displaystyle =4.{ \left\{ \frac { n\left( n+1 \right) }{ 2 } \right\} }^{ 2 }+4.\frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } +\frac { n\left( n+1 \right) }{ 2 }$

$\displaystyle ={ n }^{ 2 }{ \left( n+1 \right) }^{ 2 }+\frac { 2 }{ 3 } n\left( n+1 \right) \left( 2n+1 \right) +\frac { 1 }{ 2 } n\left( n+1 \right)$

$\displaystyle \therefore { S }_{ 20 }={ 20 }^{ 2 }.{ 21 }^{ 2 }+\frac { 2 }{ 3 } .20.21.41+\frac { 1 }{ 2 } .20.21=188090$

Find the value of

$\displaystyle \left ( 1-\frac{1}{2^{2}} \right )\left ( 1-\frac{1}{3^{3}} \right )\left (1 -\frac{1}{4^{2}} \right )\left ( 1-\frac{1}{5^{2}} \right )........\left ( 1-\frac{1}{9^{2}} \right )\left ( 1-\frac{1}{10^{2}} \right )$

0%
$\displaystyle \frac{5}{12}$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{11}{20}$
0%
$\displaystyle \frac{7}{10}$

Explanation

Given expression can be expanded as

$\displaystyle \left ( 1-\frac{1}{4} \right )\left ( 1-\frac{1}{9} \right )\left ( 1-\frac{1}{16} \right )\left ( 1-\frac{1}{25} \right )$ ............

$\displaystyle \left ( 1-\frac{1}{81} \right )\left ( 1-\frac{1}{100} \right )$

$\displaystyle =\frac{3^{1}}{4_{1}}\times \frac{8^{2^{1}}}{9_{8_{1}}}\times \frac{15^{5^{1}}}{25_{5_{1}}}\times \frac{24^{3^{1}}}{36_{12_{1}}}\times \frac{48^{4^{1}}}{49_{7_{1}}}$

$\displaystyle =\frac{63^{7^{1}}}{64_{16^{1}}}\times \frac{80^{5^{1}}}{81_{9_{1}}}\times \frac{99^{11}}{100_{20}}=\frac{11}{20}$

$\displaystyle \sum_{s=1}^{n}\, \left \{ \displaystyle \sum_{r=1}^{s}r \right \}\, =\, an^3\, +\, bn^2\, +\, cn$ , then find the value of a + b + c.

0%
1
0%
0
0%
2
0%
3

Explanation

$\displaystyle \sum_{s=1}^{n}\, \left \{ \displaystyle \sum_{r=1}^{s}r \right \}\, =\, \displaystyle \sum_{s=1}^{n}\, \displaystyle \frac {s(s+1)}{2}$

$=\, \displaystyle \frac {1}{2}\, \left [ \displaystyle \frac {n(n+1)(2n+1)}{6}\, +\, \displaystyle \frac {n(n+1)}{2} \right ]$

$=\, \displaystyle \frac {n(n+1)}{4}\, \left [ \displaystyle \frac {2n+1+3}{3} \right ]\, =\, \displaystyle \frac {1}{12}\, n(n+1)(2n+4)$

$\Rightarrow\, a\, =\, \displaystyle \frac {1}{6},\, b\, =\, \displaystyle \frac {1}{2},\, c\, =\, \displaystyle \frac {1}{3}$

$\Rightarrow a+b+c=1$

The sum of the series

$4 + 8 + 16 + 32 + .......$ . till

$10$ terms is

0%
$6\, \times\, 2^{10}\, -\, 1$
0%
$4\, \times\, 2^{10}\, +\, 1$
0%
$4(2^{10}\, -\, 1)$
0%
$4(2^{10}\, +\, 1)$

Explanation

Let

$S_{10}=4 + 8 + 16 + 32 + ........$ upto

$10$ terms

$=4(1+2+4+8+.........$ .upto

$10$ terms
Clearly,

$1+2+2^2+2^3+...$ is an GP with first term

$=1$ and common ratio

$=2$ .

$\therefore S_{10}=4[\displaystyle \frac{1(2^{10}-1)}{2-1}]=4(2^{10}-1)$

Thus the sum of the given series is

$4(2^{10}-1)$

$\displaystyle \sum _{ p=1 }^{ 32 }{ \left( 3p+2 \right) { \left[ \sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi }{ 11 } } -i\cos { \frac { 2q\pi }{ 11 } } \right) } \right] }^{ p } } =$

0%
$8\left( 1-i \right)$
0%
$16\left( 1-i \right)$
0%
$48\left( 1-i \right)$
0%
None of these

Explanation

$\displaystyle\sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi }{ 11 } -i } \sin { \frac { 2q\pi }{ 11 } } \right) }$

$\displaystyle=\left\{ \sin { \frac { 2\pi }{ 11 } } +\sin { \frac { 4\pi }{ 11 } } +...+10 \right\}- i\left\{ \cos { \frac { 2\pi }{ 11 } } +\cos { \frac { 4\pi }{ 11 } } +...+10 \right\}$

$\displaystyle=\frac { \sin { \left( \frac { 2\pi }{ 11 } +\frac { 9\pi }{ 11 } \right) } \sin { \frac { 10\pi }{ 11 } } }{ \sin { \frac { \pi }{ 11 } } } -i\frac { \cos { \left( \frac { 2\pi }{ 11 } +\frac { 9\pi }{ 11 } \right) \sin { \frac { 10\pi }{ 11 } } } }{ \sin { \frac { \pi }{ 11 } } }$

$=0-i\left( -1 \right) =i$ ...(1)

$\displaystyle\therefore S=\sum _{ p=1 }^{ 32 }{ \left( 3q+2 \right) { \left[ \sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi }{ 11 } -i } \cos { \frac { 2q\pi }{ 11 } } \right) } \right] }^{ p } }$

$\displaystyle=\sum _{ p=1 }^{ 32 }{ \left( 3p+2 \right) { i }^{ p } } =3\sum _{ p=1 }^{ 32 }{ p } { i }^{ p }+2\sum _{ p=1 }^{ 32 }{ { i }^{ p } } =3A+2B$ ...(2)

Now,

$\displaystyle A=i+2{ i }^{ 2 }+3{ i }^{ 3 }+...+32{ i }^{ 32 }$

$\displaystyle \Rightarrow Ai={ i }^{ 2 }+2{ i }^{ 3 }+...+31{ i }^{ 32 }+32{ i }^{ 33 }$

$\displaystyle\Rightarrow A\left( 1-i \right) =i+{ i }^{ 2 }+...+{ i }^{ 32 }-32{ i }^{ 33 }$

$\displaystyle =\frac { { i }^{ 32 }-1 }{ i-1 } -32i=-32i\quad \left[ \because { i }^{ 32 }=1 \right]$

$\displaystyle\therefore A=\frac { -32i }{ 1-i } =\frac { 32\left( 1+i \right) }{ 2 } =16\left( 1-i \right)$ ...(3)

and,

$\displaystyle B=i+{ i }^{ 2 }+...+{ i }^{ 32 }=0$ ...(4)

Hence,

$\displaystyle S=3\times 16\left( 1-i \right) =48\left( 1-i \right) .$

Find the sum the infinite G.P.:

$1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}$

0%
$\displaystyle \frac{3}{5}$
0%
$\displaystyle \frac{3}{2}$
0%
$\displaystyle \frac{49}{27}$
0%
$\displaystyle \frac{8}{5}$

Explanation

Formula used: Sum of infinite GP

$S_\infty=a+ar+ar^2+ar^3+...=\dfrac{a}{1-r}$ , where

$|r|<1$
Given,

$1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}$
Clearly, it is a GP with first term

$a=1$ and common difference

$d=\frac13$

$\therefore1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}=\frac{1}{1-\frac13}=\frac32$
Hence, option B is correct.

The sum of series

$\displaystyle \frac {1^2}{1}+\frac {1^2+2^2}{1+2}+\frac {1^2+2^2+3^2}{1+2+3}+....$ upto n terms is

0%
$\displaystyle\frac {1}{3}(2n+1)$
0%
$\displaystyle\frac {1}{3}n^2$
0%
$\displaystyle\frac {1}{3}(n+2)$
0%
$\displaystyle\frac {1}{3}n(n+2)$

Explanation

$\displaystyle S=\frac { { 1 }^{ 2 } }{ 1 } +\frac { { 1 }^{ 2 }+{ 2 }^{ 2 } }{ 1+2 } +\frac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 1+2+3 } +...$

$\Rightarrow\displaystyle S=\sum _{ r=1 }^{ n }{ \frac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ r }^{ 2 } }{ 1+2+3+...+r } }$

$\displaystyle=\sum _{ r=1 }^{ n }{ \frac { \displaystyle\frac { r\left( r+1 \right) \left( 2r+1 \right) }{ 6 } }{ \displaystyle\frac { r\left( r+1 \right) }{ 2 } } } =\sum _{ r=1 }^{ n }{ \left( \displaystyle\frac { 2r+1 }{ 3 } \right) }$

$\displaystyle=\frac { 2 }{ 3 } \sum _{ r=1 }^{ n }{ r } +\frac { 1 }{ 3 } \sum _{ r=1 }^{ n }{ 1 } =\frac { 2 }{ 3 } \frac { n(n+1) }{ 2 } +\frac { 1 }{ 3 } n$

$\displaystyle =\frac { n }{ 3 } \left( n+1+1 \right) =\frac { n }{ 3 } \left( n+2 \right)$

$\dfrac {1^3+2^3+3^3+4^3+........12^3}{1^2+2^2+3^2+4^2+......12^2}=$

0%
$\dfrac {234}{25}$
0%
$\dfrac {243}{35}$
0%
$\dfrac {263}{27}$
0%
None of these

Explanation

Clearly required ratio is

$=\cfrac{\sum n^3}{\sum n^2}=\cfrac{\cfrac{n^2(n+1)^2}{4}}{\cfrac{ n(n+1) (2n+1)}{6}}=\cfrac{3n(n+1)}{2(2n+1)}$
Substituting

$n =12$ we get required value

$=\cfrac{234}{25}$
Hence, option 'A' is correct.

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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