MCQ Questions for Class 11 Engineering Maths Sequences And Series Quiz 7

The value of the sum

$1.2.3+2.3.4+3.4.5+.......$ upto n terms=

0%
$\frac{1}{6}n^{2}(2n^{2}+1)$
0%
$\frac{1}{6}n^{2}(n^{2}-1)(2n-1)(2n+3)$
0%
$\frac{1}{8}(n^{2}+1)(n^{2}+5)$
0%
$\frac{1}{4}(n)(n+1)(n+2)(n+3)$

Explanation

Sum of

$1.2.3+2.3.4+3.4.5+.......$ upto

$n$ terms

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

$S_{n}=$ Sum of

$T_{n}$

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

$=\sum{n}^{3}+{3}{n}^{2}+2n$

$\sum{n}^{3}+3\sum{n}^{2}+2\sum{n}$

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

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

Hence, it is the correct answer.

If the coefficients of

$x^9,x^{10},x^{11}$ in the expansion of

$(1+x)^n$ are in arithmetic progression then

$n^2-41n=$

0%
$398$
0%
$298$
0%
$-398$
0%
$198$

Explanation

$^nC_9,^nC_{10},^nC_{11}$ are in A.P

$\Rightarrow 2\ ^{n}C_{10}=^nC_9+^nC_{11} \Rightarrow=\dfrac{^nC_9}{^nC_{10}}+\dfrac{^nC_{11}}{^nC_{10}}$

$\Rightarrow 2=\dfrac{10}{n-9}+\dfrac{n-10}{11} \Rightarrow 2=\dfrac{110+n^2-19n+90}{11n-99}$

$\Rightarrow n^2-41n=-398$

The value of

$\cfrac { 1 }{ \left( 2n-1 \right) !0! } +\cfrac { 1 }{ \left( 2n-3 \right) !2! } +\cfrac { 1 }{ \left( 2n-5 \right) !4! } +....+\cfrac { 1 }{ 1!\left( 2n-2 \right) ! }$ equal to

0%
${ 2 }^{ 2n-1 }$
0%
${ 2 }^{ 2n-2 }$
0%
${ 2 }^{ 2n-3 }$
0%
$\cfrac { { 2 }^{ 2n-2 } }{ \left( 2n-1 \right) ! }$

Explanation

We know that,

$^nC_0+^nC_1+^nC_2..........^nC_n=2^n$

$^nC_0+^nC_2+^nC_4..........^nC_{n-1}=$

$^nC_1+^nC_3+^nC_5..........^nC_n=2^{n-1}$

Using the above relation,

$^{2n-1}C_0+^{2n-1}C_1+^{2n-1}C_2..........^{2n-1}C_n=2^{(2n-1)-1}$

Expanding the coefficients we get the desired result,

$\dfrac{(2n-1)!}{(2n-1)!0!}+$

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

$\dfrac{(2n-1)!}{(2n-5)!4!}+$

$...............+\dfrac{(2n-1)!}{1!(2n-2!)}=2^{2n-2}$

$\dfrac{1}{(2n-1)!0!}+$

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

$\dfrac{1}{(2n-5)!4!}+$

$...............+\dfrac{1}{1!(2n-2!)}=\dfrac{2^{2n-2}}{(2n-1)!}$

$x\in R$ , find the minimum value of the expression

$3^x+3^{1-x}$ .

0%
$3 \sqrt 2$
0%
$2 \sqrt 3$
0%
$3 \sqrt 3$
0%
none of these

Explanation

We know,

$AM > GM$

Therefore,

$\dfrac{3^x+3^{1-x}}{2} \geq \sqrt{3^x \times 3^{1-x}}$ for all

$x\in R$ .

$\dfrac{3^x+3^{1-x}}{2} \geq \sqrt{3}$ for all

$x\in R$ .

$3^x+3^{1-x} \geq 2\sqrt{3}$ for all

$x\in R$

Hence, the required value is

$2\sqrt{3}$ .

Sum of the series

${ \left( ^{ 100 }C_{ 1 } \right) }^{ 2 }+2{ \left( ^{ 100 }C_{ 2 } \right) }^{ 2 }+3{ \left( ^{ 100 }C_{ 3 } \right) }^{ 2 }+.....100{ \left( ^{ 100 }C_{ 100 } \right) }^{ 2 }$ equals:

0%
$\dfrac { { 2 }^{ 99 }\left[ 1.3.5.....\left( 199 \right) \right] }{ \left( 99 \right) ! }$
0%
$100.^{ 100 }C_{ 100 }$
0%
$50.^{ 200 }C_{ 100 }$
0%
$100.^{ 199 }C_{ 99 }$

Explanation

Sum of the series

$(\ ^{100}C_{1})^{2}+2(\ ^{100}C_{2})^{2}+3(\ ^{100}C_{3})^{2}+......100(\ ^{100}C_{100})^{2}$

equal

Solution:

$(\ ^{100}C_{1})^{2}+2(\ ^{100}C_{2})^{2}+3(\ ^{100}C_{3})^{2}+.........+100(\ ^{100}C_{100})^{2}$

$\Rightarrow \ ^{100}C_{99}\ ^{100}C_{1}+2.\ ^{100}C_{98}\ ^{100}C_{2}+3.\ ^{100}C_{97}\ ^{100}C_{3}+......+ 100\ ^{100}C_{0}.\ ^{100}C_{100}$

$(1+x)^{100}=\ ^{100}C_{0}+\ ^{100}C_{1}x+\ ^{100}C_{2}x^{2}+\ ^{100}C_{3}x^{3}+ .......+\ ^{100}C_{100}x^{100}$

$100(1+x)^{99}=1.\ ^{100}C_{1}+2x\ ^{100}C_{2}+3x^{2}\ ^{100}C_{3}+......+99 x^{99}C_{99+100}x^{99}\ ^{100}C_{100}\Rightarrow (1)$

$(1+x)^{100}=\ ^{100}C_{0}+\ ^{100}C_{1}x+\ ^{100}C_{2}x^{2}+\ ^{100}C_{3}x^{3}+.....+\ ^{100}C_{100}x^{100}$

from

$(1)$ &

$(2)$

$\Rightarrow 100(1+x)^{99}(1+x)^{100}$

coefficient of

$x^{99}$ is

$\Rightarrow 100(1+x)^{199}$

$\Rightarrow T_{r+1}=100\ ^{199}C_{r}x^{r}$

Here

$r=99$

$\Rightarrow T_{100}=100.\ ^{199}C_{99}x^{99}$

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%
$\displaystyle\frac{3n(n + 1)}{2}$
0%
$\cfrac { 8{ n }^{ 3 }+6{ n }^{ 2 }+n }{ 3 }$
0%
$\displaystyle\frac{n(n + 1)^2}{4}$
0%
$\displaystyle\left [ \frac{n(n + 1)}{4} \right ]^2$

Explanation

${ 1 }^{ 2 }+2.{ 2 }^{ 2 }+{ 3 }^{ 2 }+2.{ 4 }^{ 2 }+{ 5 }^{ 2 }+2.{ 6 }^{ 2 }+.......\\ ({ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.....{ n }^{ 2 })+2({ 2 }^{ 2 }+{ 4 }^{ 2 }+{ 6 }^{ 2 }+......)\\ =({ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.....{ n }^{ 2 })+2.2({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+......)\\ =({ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.....{ n }^{ 2 })+{ 2 }^{ 2 }({ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+......)\\ =\cfrac { n(4{ n }^{ 2 }-1) }{ 3 } +4[\cfrac { n(n+1)(2n+1) }{ 6 } ]\\ =\cfrac { n(4{ n }^{ 2 }-1) }{ 3 } +\cfrac { 2({ n }^{ 2 }+n)(2n+1) }{ 3 } \\ =\cfrac { 4{ n }^{ 3 }-n+(4n+2)({ n }^{ 2 }+n) }{ 3 } \\ =\cfrac { 4{ n }^{ 3 }-n+4{ n }^{ 3 }+4{ n }^{ 2 }+2{ n }^{ 2 }+2n }{ 3 } \\ =\cfrac { 8{ n }^{ 3 }+6{ n }^{ 2 }+n }{ 3 }$

Let

$A$ be the sum of the first

$20$ terms and

$B$ be the sum of the first

$40$ terms of the series

$1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2.6^{2} + .....$ If

$B - 2A = 100\lambda$ , then

$\lambda$ is equal to

0%
$464$
0%
$496$
0%
$232$
0%
$248$

Explanation

$B=1^2+2.2^2+3^2+2.4^2+....+2.40^2$

$A=1^2+2.2^2+3^2+2.4^2+.... +2.20^2$

$B = 1^2+3^3+5^2+...+39^2+2 [2^2+4^2+...+40^2]$

$A = 1^2+3^2+5^2+... + 19^2+2[2^2+4^2+...+20^2]$

Sum of square of first n odd natural number

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

sum of square of first n even natural number

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

$B=\left[\dfrac{n(2n+1)(2n-1)}{3}\right]_{n=20} + 2\left[\dfrac{2n(n+1)(2n+1)}{3}\right]_{n=20}$

$A=\left[\dfrac{n(2n+1)(2n-1)}{3}\right]_{n=10}+2\left[\dfrac{2n(n+1)2n+1)}{3}\right]_{n=10}$

$B = \dfrac{20(41)(39)}{3} + 2\left[\dfrac{(400(21)(41)}{3}\right]$

$B = \dfrac{100860}{3}$

$A=\dfrac{10(21)(19)}{3}+2\left[\dfrac{20(11)(21)}{3}\right]$

$A=\dfrac{13230}{3}$

$B-2A = \left[\dfrac{100860-26460}{3}\right] = \dfrac{74400}{3}$

$B-2A = 24800= 100\lambda$

$\lambda = 248$

$z^2-2z+4=0$ then the value of

$\left(\displaystyle\frac{z}{2}+\frac{2}{z}\right)^2+\left(\displaystyle\frac{z^2}{4}+\frac{4}{z^2}\right)^2+\left(\displaystyle\frac{z^3}{8}+\frac{8}{z^3}\right)^2+........+\left(\displaystyle\frac{z^{24}}{2^{24}}+\frac{2^{24}}{Z^{24}}\right)^2$ is equal to.

0%
$24$
0%
$32$
0%
$48$
0%
None of these

Explanation

Given :-

$z^{2}-2z+4=0$

$z^{2}+4 = 2x$

Problem :

$\sum_{n=1}^{24} \left(\dfrac{z^{n}}{2^{n}}+ \dfrac{2^{n}}{z^{n}} \right)^{2}$

$\sum_{n=1}^{24} \left( \dfrac{z^{2n}+ 2^{2n}}{(2z)^{n}} \right)^{2}$

We know

$(z^{2^{(1)}}+ 2^{2^{(2)}})= 2z^{(1)}$

$(z^{2}+2^{2})= (2z)^{2}$

$z^{4}+2^{4}+2.z^{2} 2^{2}= (2z)^{2}$

$z^{4}+2^{4}= 2^{4}z^{2}-8z^{2}$

$8z^{2}$

When

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

$\left( \dfrac{2z}{2z} \right)^{2}$

$=1.$

When

$n=2 \dfrac{(z^{4}+2^{4})^{2}}{(2^{2} z^{2})^{2}}$

$\dfrac{z^{4}+2^{4}+2.z^{2}.2^{2}}{(2^{2}z^{2})^{2}}$

$\dfrac{8z^{2}}{8z^{2}}$

$=1$ .

So on

$\therefore \sum_{n=1}^{24} \left( \dfrac{z^{2n}+ 2^{2n}}{(2z)^{n}} \right)^{2}= \sum \left( \dfrac{(2z)^{n}}{(2z)^{n}} \right)^{2}$

$= \sum_{n=1}^{24} 1$ .

$=24$

The odd natural numbers have been divided in groups as

$(1, 3); (5, 7, 9, 11); (13, 15, 17, 19, 21, 23); .....$ Then the sum of numbers in the

$10^{th}$ group is

0%
$4000$
0%
$4003$
0%
$4007$
0%
$4008$

Explanation

Numbers in first group

$= 2$
Numbers in second group

$= 4$
Numbers in

$9^{th}$ group

$= 18$
Numbers in

$10^{th}$ group

$= 20$
Total numbers in

$1^{st}$ to

$9^{th}$ group

$= 2 + 4 + .....18$

$= 2 (1 + ..... 9) = 2\times \dfrac {9\times 10}{2} = 90$
Total numbers in

$10^{th}$ group =110
Sum of first

$N$ odd numbers

$=n^2$
sum of numbers in

$10^{th}$ group = sum till

$10^{th}$ group - Sum till

$9^{th}$ group

$= 110^{2} - 90^{2}$

$= (110 + 90)(110 - 90)$

$= 200\times 20 = 4000$ .

$9@ 3 = 12, 15 @ 4 = 22, 16 @ 14 = 4$ , then what is the value of

$6 @ 2 = ?$

0%
$26$
0%
$1$
0%
$30$
0%
$8$

Explanation

$(9 - 3) \times 2 = 12, (15 - 4) \times 2$

$= 22, (16 - 14) \times 2 = 4$
Hence,

$(6 - 2) \times 2 = 8$ .

5	2	7
?	3	1
4	5	2
-15	7	13

Select the missing number from the given alternatives.

0%
$1$
0%
$5$
0%
$9$
0%
$7$

Explanation

Columnwise
First Number x Third Number - Second Number = Lowermost Number

First Column

$5 \times 4 - ? = 15 \Rightarrow 20 - ? = 15$

$\therefore ? = 20 - 15 = 5$

Second Column

$2 \times 5 - 3 = 10 - 3 = 7$

Third Column

$7 \times 2 - 1 = 14 - 1 = 13$

Select the missing number from the given alternatives.

$3$	$4$	$2$	$14$
$6$	$5$	$4$	$44$
$5$	$2$	$7$	?

0%
$58$
0%
$14$
0%
$49$
0%
$4$

Explanation

$3\times 2 + 4\times 2 = 6 + 8 = 14$

$6\times 4 + 5\times 4 = 24 + 20 = 44$
Hence,

$5\times 7 + 2 \times 7 = 35 + 14 = 49$ .

$12\times 16 = 188$ and

$14\times 18 = 248$ , then find the value of

$16\times 20 = ?$

0%
$320$
0%
$360$
0%
$316$
0%
$318$

Explanation

$12\times 16 = 192 - 4 = 188$

$14\times 18 = 252 - 4 = 248$

$16\times 20 = 320 - 4 = 316$ .

Select the missing number from the given alternatives.

0%
110
0%
115
0%
121
0%
54

Explanation

$\div$ 2 = 24;
24

$\times$ 3 = 72: 72

$\div$ 2 = 36;
36

$\div$ 3 = 108; 108

$\div$ 2 = 54.

$28\div 11 = 8, 39\div 21 = 7, 45\div 27 = 4$ , then

$95\div 25 = ?$

0%
$11$
0%
$9$
0%
$16$
0%
$5$

Explanation

$28\div 11 \Rightarrow [(2 + 8) \div (1 + 1)] + 3$

$= (10\div 2) + 3 = 5 + 3 = 8$

$39\div 21 \Rightarrow [(3 + 9) \div (2 + 1)] + 3$

$= (12\div 3) + 3 = 4 + 3 = 7$

$45\div 27 \Rightarrow [(4 + 5) \div (2 + 7)] + 3$

$= (9\div 9) + 3 = 1 + 3 = 4$

$\therefore 95\div 25\Rightarrow [(9 + 5) \div (2 + 5)] + 3$

$= (14\div 7) + 3 = 2 + 3 = 5$ .

In a row of girls, Mridula is

$18^{th}$ from the right and Sanjana is

$18^{th}$ from the left. If both of them interchange their position, Sanjana becomes

$25^{th}$ from the left, how many girls are there in the row?

0%
$40$
0%
$41$
0%
$42$
0%
$35$

Explanation

Total number of girls

$=$ (Position of Sanjana from left

$+$ Position of Sanjana from right)

$- 1 = (25 + 18) - 1 = 42$ .

Choose the correct answer from the alternatives given :
The sum of

$\dfrac{1}{\sqrt 2 + 1} \, + \, \dfrac{1}{\sqrt 3 + \sqrt 2} \, + \, \dfrac{1}{\sqrt 4 + \sqrt 3} \, + \, ..... \, + \dfrac{1}{\sqrt {100} + \sqrt {99}}$ is

0%
$9$
0%
$10$
0%
$11$
0%
None of these

Explanation

$\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{4} + \sqrt{3}} + .......+\frac{1}{\sqrt{100} + \sqrt{99}}$
Rationalising the denominator,

$\displaystyle\, = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} + \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$\displaystyle\, + \frac{1}{\sqrt{4} + \sqrt{3}} \times \frac{\sqrt{4} - \sqrt{3}}{\sqrt{4} - \sqrt{3}} + ......+ \frac{1}{\sqrt{100} + \sqrt{99}} \times \frac{\sqrt{100} - \sqrt{99}}{\sqrt{100} - \sqrt{99}}$

$\displaystyle\, \sqrt{2} 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + .+ \sqrt{100} - \sqrt{99}$
=

$-1 + 10 = 9$

Choose the correct answer alternatives given.
Select the missing number from the given alternatives.

0%
18
0%
21
0%
5
0%
17

Explanation

80 = 8

$\times$ (8 + 2)
143 = 11

$\times$ (11 + 2)
323 = ?

$\times$ (? + 2)

$\therefore$ ? = 17

$1^{3} + 2^{3} + .... + 10^{3} = 3025$ , then the value of

$2^{3} + 4^{3} + ..... + 20^{3}$ is

0%
$7590$
0%
$5060$
0%
$24200$
0%
$12100$

Explanation

Given that,

$1^{3} + 2^{3} + .... + 10^{3} = 3025$
Now,

$2^{3} + 4^{3} + .... + 20^{3}$

$= 2^{3} (1 + 2^{3} + .... + 10^{3}) = 8\times 3025 = 24200$ .

In our number system the base is ten. If the base were changed to four you would count as follows:

$1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30,....$ The twentieth number would be:

0%
$20$
0%
$38$
0%
$44$
0%
$104$
0%
$110$

Explanation

If we change the base from 10 to 4 then the number 20 will be written as

$20$ =

$1\times 4^2$ +

$1\times4^1$ +

$0\times4^0$ .

so the required number is 110.

we can also find by dividing the number by 4

$\dfrac{20}{4}$ =

$5$ . now

$\frac{5}{4}$ =1 as dividend, now

$\frac{1}{4}$ =0 as dividend. so the required number is 110.

so option

$E$ is correct option.

The value of the product

${ 6 }^{ \frac { 1 }{ 2 } }\times { 6 }^{ \frac { 1 }{ 4 } }\times { 6 }^{ \frac { 1 }{ 8 } }\times { 6 }^{ \frac { 1 }{ 16 } }\times ..$ up to infinite terms is

0%
$6$
0%
$36$
0%
$216$
0%
$512$

Explanation

$6^{\tfrac12}\times 6^{\tfrac14}\times 6^{\tfrac18}........$

$=6^{\left(\tfrac12+\tfrac14+\tfrac18.......\right)}$

$=6^{\left(\tfrac{\tfrac12}{1-\tfrac12}\right)}$

$=6^1=6$

Hence, the answer is

$6$ .

If a and b are two unequal positive numbers, the:

0%
$\frac{2ab}{a+b}>\sqrt{ab}>\frac{a+b}{2}$
0%
$\sqrt{ab}>\frac{2ab}{a+b}>\frac{a+b}{2}$
0%
$\frac{2ab}{a+b}>\frac{a+b}{2}>\sqrt{ab}$
0%
$\frac{a+b}{2}>\frac{2ab}{a+b}>\sqrt{ab}$
0%
$\frac{a+b}{2}>\sqrt{ab}>\frac{2ab}{a+b}$

Explanation

a and b are two unequal positive numbers.

$\therefore A.M>G.M>H.M\\ \cfrac { a+b }{ 2 } >\sqrt { ab } >\cfrac { 2ab }{ a+b }$

$y = x+x^2+x^3+...$ up to infinite terms, where

$x<1$ , then which one of the following is correct?

0%
$x=\dfrac { y }{ 1+y }$
0%
$x=\dfrac { y }{ 1-y }$
0%
$x=\dfrac {1+ y }{ y }$
0%
$x=\dfrac { 1-y }{ y }$

Explanation

Given

$y=x+x^2+x^3+...\infty$

$y=x(1+x+x^2+x^3+.....\infty)$

We know that

$x+x^2+x^3+x^4....\infty =y$

$y=x(1+y)$

$x=\dfrac{y}{1+y}$

$1.3 + 2.3^2 + 3.3^3 +...+ n.3^n$ =

$\dfrac { (2n-1)3^ a+b }{ 4 }$ then

$a$ and

$b$ are respectively

0%
$n,2$
0%
$n,3$
0%
$n +1,2$
0%
$n + 1,3$

Explanation

$1.3 + 2.3^2 + 3.3^3 +...+ n.3^n$

this can be wriiten as

$0.3^0+1.3 + 2.3^2 + 3.3^3 +...+ n.3^n$

Sum of

$n$ terms in AGP is

$S_n=\dfrac{a-[a+(n-1)d]r^n}{1-r}+\dfrac{dr(1-r^{n-1})}{(1-r)^2}$

Here

$a=0,d=1,n=n ,r=3$

Substituting in the given formula we get,

$S_n=\dfrac{(n-1)3^n}{2}+\dfrac{3(1-3^{n-1})}{4}$

$S_n=\dfrac{2(n-1)3^n+3-3^n}{4}$

$S_n=\dfrac{(2n-1)3^{n+1}+3}{4}$

Thus,

$a$ and

$b$ are

$n+1$ and

$3$ respectively.

The value of

$\dfrac { 1 }{ \log _{ 3 }{ e } } +\dfrac { 1 }{ \log _{ 3 }{ e^2 } } +\dfrac { 1 }{ \log _{ 3 }{ e^4 } } +...$ up to infinite terms is

0%
$\log _{ e }{ 9 }$
0%
$0$
0%
$1$
0%
$\log _{ e }{ 3 }$

Explanation

We can write

$\log_{3} e^2=2\log_{3} e$

Similarly,

$\log_{3} e^4=4\log_{3} e$ and so on

Let

$t=\dfrac{1}{\log_{ \ 3}e}$

So,

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

$=t+\dfrac{t}{2}+\dfrac{t}{4}....$

$=t\left(1+\dfrac12+\dfrac14+\dfrac18+.....\right)$

$=t\times \dfrac{1}{1-\dfrac12}=2t$

Now,

$2t=2\dfrac{1}{\log_{ \ 3}e} =2\dfrac{1}{\dfrac{\log e}{\log3}}=2\dfrac{\log 3}{\log e}=2\log_{e}3$

$=\log_{e}9$

Therefore, the answer is

$\log_{e}9$ .

$x =1-y+{ y }^{ 2 }-{ y }^{ 3 }+...$ upto infinite terms, where |y| < 1, then which one of the following is correct?

0%
$x =\frac { 1 }{ 1+y }$
0%
$x =\frac { 1 }{ 1-y }$
0%
$x = \frac { y }{ 1+y }$
0%
$x=\frac { y }{ 1-y }$

$\displaystyle \sum _{ k=1 }^{ 6 }{ \left[ sin\frac { 2k\pi }{ 7 } -i\quad cos\quad \frac { 2k\pi }{ 7 } \right] } =$

0%
$-1$
0%
$0$
0%
$-i$
0%
$i$

Explanation

$\displaystyle-i\sum _{ k=1 }^{ 6 }{ \left( cos\frac { 2k\pi }{ 7 } +i\quad sin\frac { 2k\pi }{ 7 } \right) } =$

$-i\quad \left[ 1+\alpha +{ \alpha }^{ 2 }+....+{ \alpha }^{ 6 }-1 \right] =\quad -i\left[ 0-1 \right] =i$

Arrange these numbers in ascending order.

$756, 567, 657, 676$

0%
$657, 567, 756, 676$
0%
$567, 657, 676, 756$
0%
$756, 676, 657, 567$
0%
$676, 756, 567, 657$

Explanation

$\Rightarrow$ Numbers are said to be in ascending order when they are arranged from the smallest to the largest number.

$\Rightarrow$ The numbers which we have to arrange in ascending order are

$756,\,567,\,657$ and

$676$

$\Rightarrow$

$567<657<676<756$

$\therefore$ Ascending order

$=567,\,657,\,676,\,756$

For some natural

$N$ , the number of positive integral

$x$ satisfying the equation,

$1!+2!+3!+......+(x)!=(N)^2$ is :

0%
None
0%
One
0%
Two
0%
Infinite

Explanation

$1!+2!+3!+......+(x)!=(N)^2$ is true for

$x=1, x=3$ in that case

$N=1,N=3$ respectively.

For,

$x=4$ , we have

$1!+2!+3!+4!=33$ which is not a perfect square.

Again for

$x\geq 5$

$1!+2!+....+(x)!$ is of the form

$10k+3$ for

$k$ is some natural number. In these cases, the given sum is not going to be a perfect square.

So, two values of

$x$ satisfies the given equation.

State True or False.

$1 \, + \, \dfrac{1}{5} \, + \, \dfrac{3}{5^2} \, + \, \dfrac{5}{5^3} \, + \, ..... \infty$ =

$\dfrac {13}{8}$

0%
True
0%
False

Explanation

Let

$S=1 \, + \, \dfrac{1}{5} \, + \, \dfrac{3}{5^2} \, + \, \dfrac{5}{5^3} \, + \, ..... \infty$

$...........(1)$

$\dfrac{S}{5}=\dfrac{1}{5} \, + \, \dfrac{1}{5^2} \, + \, \dfrac{3}{5^3} \, + \, \dfrac{5}{5^4} \, + \, ..... \infty$

$............(2)$

From

$(1)-(2)$ , we get

$S-\dfrac{S}{5}=1+ \, \dfrac{2}{5^2} \, + \, \dfrac{2}{5^3} \, + \, \dfrac{2}{5^4} \, + \, ..... \infty$

$............(2)$

$\dfrac{4S}{5}=1+ \, \dfrac{2}{5^2} \, + \, \dfrac{2}{5^3} \, + \, \dfrac{2}{5^4} \, + \, ..... \infty$

$\dfrac{4S}{5}-1=\dfrac{2}{5^2} \, + \, \dfrac{2}{5^3} \, + \, \dfrac{2}{5^4} \, + \, ..... \infty$

$\dfrac{4S}{5}-1=2\left(\dfrac{1}{5^2} \, + \, \dfrac{1}{5^3} \, + \, \dfrac{1}{5^4} \, + \, ..... \infty \right)$

$\dfrac{4S}{5}-1=2\left(\dfrac{\dfrac{1}{5^2}}{1-\dfrac{1}{5}}\right)$

$\dfrac{4S}{5}-1=2\left(\dfrac{1}{20}\right)$

$\dfrac{4S}{5}-1=\left(\dfrac{1}{10}\right)$

$\dfrac{4S}{5}=\dfrac{11}{10}$

$4S=\dfrac{11}{2}$

$S=\dfrac{11}{8}$

Hence, this is the answer. Hence the answer is false.

State true or false.

$199.1 + 197.3 + 195.5 +.....3.197 + 1.397 = 666700$

0%
True
0%
False

Explanation

We have,

$199.1+197.3+195.5+.....+3.197+1.397$

Since,

$a=199.1, d=-1.8, l=1.397$

Therefore, the numbers of terms

$1.397=199.1+(n-1)\times -1.8$

$-197.703=(n-1)\times -1.8$

$n-1=109.835$

$n=110.835 \approx 111$

Sum of the given series

$=\dfrac{111}{2}(2\times 199.1+(111-1)\times -1.8)$

$=\dfrac{111}{2}(398.2-198)$

$=\dfrac{111}{2}(200.2)$

$=11127.58$

Hence, this is the answer.

The sum of the given series

$1+3+5+7+9+......+53$ is

$729$ .

0%
True
0%
False

Explanation

We have,

$1+3+5+7+9+......+53$

$a=1, d=2, l=53$

We know that

$l=a+(n-1)d$

$53=1+(n-1)\times2$

$n=27$

We know that

$S_{27}=\dfrac{27}{2}[2+(27-1)\times 2]$

$S_{27}=\dfrac{27}{2}[2+26\times 2]$

$S_{27}=\dfrac{27}{2}[2+52]$

$S_{27}=\dfrac{27}{2}[54]$

$S_{27}=729$

Hence, this is the answer.

$S_n=\sum\limits_{r=1}^n \dfrac{2r+1}{r^4+2r^3+r^2}$ then

$S_{20}$ =

0%
$\dfrac{220}{221}$
0%
$\dfrac{420}{441}$
0%
$\dfrac{439}{221}$
0%
$\dfrac{440}{441}$

Explanation

$\dfrac{2r+1}{r^4+2r^3+r^2}=\dfrac{2r+1}{r^2(r+1)^2}=\dfrac{(r+1)^2-r^2}{r^2(r+1)^2}$
=

$\dfrac{1}{r^2}-\dfrac{1}{(r+1)^2} \therefore \sum\limits_{r=1}^{20}\Big[\dfrac{1}{r^2}-\dfrac{1}{(r+1)^2}\Big]$
Put r=1, 2, 3, .........20 and terms will cancel diagonally.

$\therefore \sum \dfrac{1}{1}-\dfrac{1}{(20+1)^2}=1-\dfrac{1}{441}=\dfrac{440}{441}$ .

The sum to

$50$ terms of the series

$\dfrac {1}{2} + \dfrac {3}{4} + \dfrac {7}{8} + \dfrac {15}{16} + ....$ is equal to

0%
$2^{50} - 51$
0%
$1 - 2^{-50}$
0%
$2^{-50}+49$
0%
$2^{50} - 1$

Explanation

$T_{n} = \dfrac {2^{n} - 1}{2^{n}} = 1 - \left (\dfrac {1}{2}\right )^{n}$

$\therefore S_{n} = n - \displaystyle \sum_{n = 1}^{n} \left (\dfrac {1}{2}\right )^{n} = n - \dfrac {\dfrac {1}{2} . [1 - (1/2)^{n}]}{1 - 1/2}$

$= n + 2^{-n} - 1$ .

$\sum\limits_{r=1}^{50}\Big[ \dfrac{1}{49+r}-\dfrac{1}{2r(2r-1)}\Big]=$

0%
$\dfrac{1}{50}$
0%
$\dfrac{1}{99}$
0%
$\dfrac{1}{100}$
0%
$\dfrac{1}{101}$

Explanation

$\sum\limits_{r=1}^{50} \cfrac{1}{2r(2r-1)}=\sum\limits_{r=1}^{50}\Big(\cfrac{1}{2r-1}-\cfrac{1}{2r}\Big)$

$=\Big( 1-\cfrac{1}{2}+\cfrac{1}{3}-\cfrac{1}{4}+........+\cfrac{1}{99}-\cfrac{1}{100}\Big)$

$= \Big(1+\cfrac{1}{2}+\cfrac{1}{3}+\cfrac{1}{4}+.........+\cfrac{1}{100}\Big)-2\Big(\cfrac{1}{2}+\cfrac{1}{4}+............+\cfrac{1}{100}\Big)$

$=\sum\limits{r=1}^{100}\cfrac{1}{r}-\Big(1+\cfrac{1}{2}+.....+\cfrac{1}{50}\Big)$ , i.e.,

$\sum\limits_{r=1}^{50}\cfrac{1}{r}=\sum\limits_{r=51}^{100}\cfrac{1}{r}$

$\therefore E=\sum\limits_{r=1}^{50}\cfrac{1}{49+r}-\sum\limits_{r=51}^{100}\cfrac{1}{r}$

$=\Big[\cfrac{1}{50}+\cfrac{1}{51}+...........+\cfrac{1}{99}\Big] -[\cfrac{1}{51}+\cfrac{1}{52}+............+\cfrac{1}{100}\Big]$

$=\cfrac{1}{50}-\cfrac{1}{100}=\cfrac{2}{100}-\cfrac{1}{100}=\cfrac{1}{100}\Rightarrow (C)$

The value of

$x$ satisfying the equation

$\dfrac{5050-\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+......+\dfrac{5049}{5050}\right)}{1+\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{5050}}=\dfrac{x}{5050}$ is

0%
$1$
0%
$5049$
0%
$5050$
0%
$5051$

Explanation

First we solve,

$\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.......+\dfrac{5049}{5050}\right)$

We can conclude general term or

$n^{th}$ term,

$=\sum_{n=1}^{5049}\dfrac{n}{n+1}$

$=\sum_{n=1}^{5049}\dfrac{n+1-1}{n+1}$

$=\sum_{n=1}^{5049}\left ( 1-\dfrac{1}{n+1} \right )$

$=(1-\dfrac{1}{1+1})+(1-\dfrac{1}{2+1})+.......+(1-\dfrac{1}{5049+1})$

$=(1-\dfrac{1}{2})+(1-\dfrac{1}{3})+.......+(1-\dfrac{1}{5050})$

$=5049-(\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{5050})$

Substituting this value in

$LHS$ of the given expression,

$\dfrac{5050-(\dfrac{1}{2}+\dfrac{2}{3}+.......+\dfrac{5049}{5050})}{1+\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{5050}}$

$=\dfrac{5050-5049+(\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{5050})}{1+\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{5050}}$

$=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{5050}}{1+\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{5050}}$

$=1$

Now,

$\dfrac{x}{5050}=1$

$x=5050$

This is the required value of

$x$ .

$1 + {x^2} = \sqrt {3}x$ , then

$\displaystyle \prod\limits_{n = 1}^{24} {\left( {x^n} + \dfrac {1} {x^n} \right)}$ is equal to

0%
$48$
0%
$-48$
0%
$\pm 48\left( {\omega - {\omega ^2}} \right)$
0%
none of these

Explanation

$1+{x}^{2}=\sqrt{3}x$

$\Rightarrow\,\dfrac{1}{x}+x=\sqrt{3}$

$\Rightarrow\,x+\dfrac{1}{x}=\sqrt{3}$

$\Rightarrow\,{\left(x+\dfrac{1}{x}\right)}^{2}={\left(\sqrt{3}\right)}^{2}$

$\Rightarrow\,{x}^{2}+\dfrac{1}{{x}^{2}}+2x\times \dfrac{1}{x}=3$

$\Rightarrow\,{x}^{2}+\dfrac{1}{{x}^{2}}+2=3$

$\Rightarrow\,{x}^{2}+\dfrac{1}{{x}^{2}}=3-2=1$

$\Rightarrow\,{\left(x+\dfrac{1}{x}\right)}^{3}={\left(\sqrt{3}\right)}^{3}$

$\Rightarrow\,{x}^{3}+\dfrac{1}{{x}^{3}}+3x\times \dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=3\sqrt{3}$

$\Rightarrow\,{x}^{3}+\dfrac{1}{{x}^{3}}+3\left(\sqrt{3}\right)=3\sqrt{3}$ since

$x+\dfrac{1}{x}=\sqrt{3}$

$\Rightarrow\,{x}^{3}+\dfrac{1}{{x}^{3}}=3\sqrt{3}-3\sqrt{3}=0$

$\displaystyle\prod_{n=1}^{24}{{x}^{n}+\dfrac{1}{{x}^{n}}}=\left(x+\dfrac{1}{x}\right)\left({x}^{2}+\dfrac{1}{{x}^{2}}\right)\left({x}^{3}+\dfrac{1}{{x}^{3}}\right)...\left({x}^{24}+\dfrac{1}{{x}^{24}}\right)=0$ since

${x}^{3}+\dfrac{1}{{x}^{3}}=0$

$\left| a \right| < 1$ and

$\left| b \right| < 1$ then

$s = 1 + \left( {1 + a} \right)b + \left( {1 + a + {a^2}} \right){b^2} + \left( {1 + a + {a^2} + {a^3}} \right){b^3} + ...is - .$

0%
${\dfrac 1 {\left( {1 - b} \right)\left( {1 - ab} \right)}}$
0%
${\dfrac 1 {\left( {1 + b} \right)\left( {1 - ab} \right)}}$
0%
${\dfrac 1 {\left( {1 - b} \right)\left( {1 + ab} \right)}}$
0%
${\dfrac 1 {\left( {1 + b} \right)\left( {1 + ab} \right)}}$

Explanation

$S=(1+b+b^2+b^3+....)+ab(1+b+b^2+....)+a^2b^2(1+b+b^2+...)+...$

Since using sum to infinity for G.P we get

$S=\cfrac{1}{1-b}+\cfrac{ab}{1-b}+\cfrac{a^2b^2}{1-b}+....\\S=\cfrac{1}{1-b}(1+ab+a^2b^2+....)$

Using G.P for sum to infinity, we get

$S=\cfrac{1}{1-b}\times\cfrac{1}{1-ab}=\cfrac{1}{(1-b)(1-ab)}$

The sum of the series

$\dfrac{1}{(1\times 2)}+\dfrac{1}{(2\times 3)}+\dfrac{1}{(3\times 4)}+.......+\dfrac{1}{(100\times 101)}$ is equal to

0%
$\dfrac{200}{101}$
0%
$\dfrac{100}{101}$
0%
$\dfrac{50}{101}$
0%
$\dfrac{25}{101}$

Explanation

$\dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\cdots +\dfrac{1}{100\times 101}=\dfrac{2-1}{1\times 2}+\dfrac{3-2}{2\times 3}+\cdots+\dfrac{101-100}{100\times 101}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\cdots +\dfrac{1}{100}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}$

$\left| a \right| < 1$ and

$\left| b \right| < 1$ then

$\eqalign{ & \cr & S = 1 + \left( {1 + a} \right)b + \left( {1 + a + {a^2} } \right){b^2} + ...}$ =

0%
${1 \over {\left( {1 - b} \right)\left( {1 - ab} \right)}}$
0%
${1 \over {\left( {1 + b} \right)\left( {1 - ab} \right)}}$
0%
${1 \over {\left( {1 - b} \right)\left( {1 + ab} \right)}}$
0%
${1 \over {\left( {1 + b} \right)\left( {1 + ab} \right)}}$

Explanation

Let

$S = 1 + (a + a)b + (1-a+a^2)b^2 + (a+a+a^2+a^3)b^3 + ....$

then

$bs = b + (1+a)b^2 + (1+a+a^2)b^3 + (1+a+a^2 + a^3)b^4+.......$

Apply

$(S-bS)$ :

$S-bS = 1 + b+ab + b^2 + ab^2+a^2-b^2+b^3+ab^3+a^2b^3 + a^3b^2 + .....$

$-b-b^2 - ab^2 - b^3 - ab^3-a^2b^3-b^4-ab^4 - a^2b^4-a^3b^4$ ......

$S-bS = 1 + ab + a^2b^2 + a^3b^3 +.....$

the above series is in G.P.

$S(1-b) = \dfrac{1}{(1-ab)}$

$S = \dfrac{1}{(1-ab)(1-b)}$

$\therefore$ option A is correct.

$(1+3+5+....+p)+(1+3+5+....+q)=(1+3+5+.....+r),$ where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of

$p+q+r$ , (where

$P>6$ ) is :

0%
$12$
0%
$21$
0%
$27$
0%
$24$

Explanation

$1+3+5+...+2n-1 = n^{2}$

no. of terms

$= \dfrac{n+1}{2}$

According to question we can write

$\left (\dfrac{p+1}{2}\right)^{2}+\left(\dfrac{q+1}{2}\right)^{2}\pm \left(\dfrac{r+1}{2}\right)^{2}$

It forms Pythagoras triplet Thus for min value

take 3, 4 and 5 as our Pythagoras triplet

$\therefore \dfrac{p+1}{2} = 4$

$\dfrac{q+1}{2} = 3$

$\dfrac{r+1}{2} = 5$

$p = 7$

$q = 5$

$r = 9$

$p + q + r = 21$

1185644_1037392_ans_677b4b406e3a4098b697394f648d0144.jpg

Let

$T_{n}=\displaystyle \sum _{ r=1 }^{ n }{ \dfrac { n }{ { r }^{ 2 }-2r.n+2{ n }^{ 2 } } ,{ S }_{ n }=\displaystyle \sum _{ r=0 }^{ n-1 }{ \dfrac { n }{ { r }^{ 2 }-2r.n+{ 2n }^{ 2 } } } }$ , then

0%
$T_{n} > S_{n} \forall n \epsilon N$
0%
$T_{n} > \dfrac{\pi}{4}$
0%
$S_{n} < \dfrac{\pi}{4}$
0%
$\displaystyle \lim _{ n-\infty }{ { S }_{ n }=\dfrac { \pi }{ 4 } }$

Explanation

Sol

$T_{n}=\displaystyle \sum _{ r=1 }^{ n }{ \dfrac { n }{ { r }^{ 2 }-2r.n+2{ n }^{ 2 } } ,{ S }_{ n }=\displaystyle \sum _{ r=0 }^{ n-1 }{ \dfrac { n }{ { r }^{ 2 }-2r.n+{ 2n }^{ 2 } } } }$

On expanding

$T_{n}$ and

$S_{2}$ we get

$T_{n} = \dfrac{n}{2n^{2}-2n+1}+\dfrac{n}{2n^{2}-4n+4}+\dfrac{n}{2n^{2}-6n+9}+...+\dfrac{n}{n^{2}} ... (1)$

and

$S_{n} = \dfrac{n}{2n^{2}}+\dfrac{n}{2n^{2}-2n+1}+\dfrac{n}{2n^{2}-4n+4}+...+\dfrac{n}{2n^{1}-2n^{2}+2n+(n-1)^{2}}$

$S_{n} = \dfrac{n}{2n^{2}}+\dfrac{n}{2n^{2}-2n+1}+\dfrac{n}{2n^{2}-2n+4}+...+\dfrac{n}{n^{2}+1}...(2)$

from eq (1)

on comparing

$T_{n}$ &

$S_{2}$ we can clearly

see that

$T_{n}\Rightarrow S_{n}$

$T_{n} = P+\dfrac{1}{n}$

and

$S_{n} = P+\dfrac{1}{2n}$

$P =$ common part in eq (1) & (2)

$[T_{n}> S_{n} \forall\ n \epsilon N]$

Hence from equation (1) & (2)

$[T_{n} = S_{n}+\dfrac{1}{2n}]$

1172444_1037627_ans_f470c571a66742de956a729cf3918f59.jpg

$S_{n}=\dfrac {3}{2}+\dfrac {33}{2^{2}}+\dfrac {333}{2^{2}}+....$ upto

$n$ terms

$=\dfrac {a^{n+1}+b^{b-n}-c}{d}$ (where

$a,b,c,d \epsilon\ N$ ), then ?

0%
$a+b+c+d=28$
0%
$b+d=a+c$
0%
$a-c=b+d$
0%
$d-c=a-b$

Sum of the series

$1+2.2+3.2^{2}+4.2^{3}+..+100.2^{99}$ is

0%
$100.2^{100}+1$
0%
$99.2^{100}+1$
0%
$99.2^{100}-1$
0%
$100.2^{100}-1$

Explanation

It is been clear that the series is in Arithmetic-Geometric progression with

$A.P=\quad 1+2+3+.......+100\\ and,\quad G.P=\quad 1+2+{ 2 }^{ 2 }+.....+{ 2 }^{ 99 }\\ Let,\quad S=1+2.2+3.{ 2 }^{ 2 }+4.{ 2 }^{ 3 }+.....+100.{ 2 }^{ 99 }\quad \quad \quad \quad \quad \longrightarrow \quad (1)\\ \text{Multiply the above eauation by} 2\\ 2S=2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+.....+99.{ 2 }^{ 99 }+100.{ 2 }^{ 100 }\quad \quad \quad \quad \quad \quad \longrightarrow \quad (2)\\ now\quad subtract\quad (2)\quad from\quad (1)\quad we\quad get,\\ \quad \quad \quad \quad S-2S=(1+2+{ 2 }^{ 2 }+....+{ 2 }^{ 99 })-100.{ 2 }^{ 100 }\\ \Rightarrow \quad -S=\frac { ({ 2 }^{ 100 }-1) }{ 2-1 } -100.{ 2 }^{ 100 }\quad \quad (since\quad the\quad above\quad series\quad in\quad bracket\quad is\quad in\quad G.P\quad so\quad { S }_{ n }=\frac { a({ r }^{ n }-1) }{ r-1 } \quad and\quad this\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad is\quad a\quad G.P\quad with\quad 100\quad terms\quad and\quad a=1;\quad r=2)\\ \Rightarrow \quad S=\quad -{ 2 }^{ 100 }+1+100.{ 2 }^{ 100 }\\ \therefore \quad S=\quad 99.{ 2 }^{ 100 }+1\\$

The value of sum

$\sum _{ n=1 }^{n=\infty }{ \frac { 2 }{ { 3 }^{ n } } is\ equal\ to: }$

0%
1
0%
3
0%
15/4
0%
7/3

Pam likes the numbers 1689 and 6891. Knowing this, which pair of numbers will she like among the ones below?

0%
1981 and 1891
0%
19 and 91
0%
190 and 160
0%
1198911 and 1168611

Let

${S_n} = \frac{3}{{{1^2}}} + \frac{5}{{{1^2} + {2^3}}} + \frac{7}{{{1^2} + {2^3} + {3^3}}} + \frac{9}{{{1^2} + {2^3} + {3^3} + {4^3}}} + .......{\text{ up to n terms, then }}\mathop {\lim }\limits_{n \to \infty } {S_n}{\text{ }}$ is equal to

0%
2
0%
3
0%
6
0%
1

$A = (2 + 1)({2^2} + 1)({2^4} + 1).....\left( {{2^{2016}} + 1} \right)$ . The value of

${\left( {A + 1} \right)^{\frac{1}{{2016}}}}$ is

0%
$4$
0%
$2016$
0%
$^{{2^{4032}}}$
0%
$2$

Explanation

$A=(2+1)(2^2+1)(2^4+1)(2^8+1).....(2^{2016}+1)$

$A=(2-1) (2+1) (2^2+1)(2^4+1)(2^8+1)....(2^{2016}+1)$

$A=(2^2-1)(2^2+1)(2^4+1)......(2^{2016}+1)$

Further simplification gives

$A=2^{4032}-1$

$\Rightarrow \ A+1=2^{4032}$

$\Rightarrow \ (A+1)^{1/2016} =(2^{4032})^{1/2016}$

$=2^2$

$=4$

Let

${n}^{th}$ term of sequence

$1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,.....$ is given by

${t}_{n}$ , then ?

0%
${t}_{100}=14$
0%
${t}_{200}=20$
0%
${t}_{300}=24$
0%
${t}_{400}=28$

Explanation

$no.\>of\>terms\>with\>same\>values\>are\>in\>AP(let's\>call\>them\>as\>a\>group)\\100=(\frac{n(n+1)}{2})\\n(n+1)=200\\for\>n\>=14\\n(n+1)=210\>>\>200\\hence\>all\>terms\>with\>values\>14,\>will\>cover\>group\>including\>100th\>term$

given relation is $\frac{1}{n+1} + \frac{1}{n+2}+.....+ \frac{1}{2n} > \frac{13}{24}$ < for all natural numbers $n>1$

0%
True
0%
False

CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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