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CBSE Questions for Class 11 Engineering Maths Sequences And Series Quiz 9 - MCQExams.com
CBSE
Class 11 Engineering Maths
Sequences And Series
Quiz 9
The value of
∞
∑
n
=
2
(
1
−
1
n
2
)
equals
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0%
−
l
n
3
0%
0
0%
−
l
n
2
0%
−
l
n
5
solve that :-
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0%
14
0%
18
0%
11
0%
26
Explanation
Given,
Image
A
4
+
2
+
5
+
3
=
14
×
2
=
28
Image
B
7
+
5
+
4
+
3
=
19
×
2
=
38
∴
Image
C
2+1+3+7=13 \times 2=26
The numbers
{\log _{180}}12,{\log _{2160}}12,{\log _{25920}}12
are in
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0%
AP
0%
GP
0%
HP
0%
None of these
If
[x]
denotes the greates integer
\le x
, then
\left[\dfrac{2}{3} \right] + \left[\dfrac{2}{3} + \dfrac{1}{99} \right] + \left[\dfrac{2}{3} + \dfrac{2}{99} \right] + .... + \left[\dfrac{2}{3} + \dfrac{98}{99} \right] =
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0%
99
0%
98
0%
66
0%
65
Explanation
\left [ \cfrac{2}{3} \right ]=0
\left [ \cfrac{2}{3}+\cfrac{1}{99} \right ]=0
\left [ \cfrac{2}{3}+\cfrac{2}{99} \right ]=0
...
...
\left [ \cfrac{2}{3}+\cfrac{33}{99} \right ]=\left [ \cfrac{2}{3}+\cfrac{1}{3} \right ]=\left [ 1 \right ]=1
\left [ \cfrac{2}{3}+\cfrac{34}{99} \right ]=\left [ 1+\cfrac{1}{99} \right ]=1
\left [ \cfrac{2}{3}+\cfrac{98}{99} \right ]=\left [ 1+\cfrac{65}{99} \right ]=1
So, the sum is
1+1+1+......... 66
times
=66
If
(1+ax)^{n }=1+8x+24x^{2}+...;
then
\cfrac {a-n}{a+n}
is equal to
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0%
3
0%
\dfrac {-1}{3}
0%
-3
0%
\dfrac {1}{3}
Explanation
(1+ax)^{ n }=1+n\left( ax \right) +^{ n }{ C }_{ 2 }{ \left( ax \right) }^{ 2 }+^{ n }{ C }_{ 3 }{ \left( ax \right) }^{ 3 }....
According to question
(1+ax)^{ n }=1+8x+24x^{ 2 }+...
\Rightarrow an=8
{ a }^{ 2 }=\cfrac { (n)(n-1) }{ 2 } =24
\Rightarrow { a }^{ 2 }\left( n \right) \left( n-1 \right) =48
\Rightarrow a\left( n-1 \right) =6
\Rightarrow \cfrac { n }{ n-1 } =\cfrac { 8 }{ 6 } =\cfrac { 4 }{ 3 }
\Rightarrow n=4,\Rightarrow a=2
\therefore \cfrac { a-n }{ a+n } =\cfrac { 2-4 }{ 2+4 } =\cfrac { -2 }{ 6 } =\cfrac { -1 }{ 3 }
If
S={1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}....
upto
n
terms and
n
is even, then
S
equals _____
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0%
\cfrac{n(n+1)}{2}
0%
\cfrac{n(n-1)}{2}
0%
\cfrac{-n(n+1)}{2}
0%
\cfrac{-n(n-1)}{2}
Explanation
S=1^{2}-2^{2}+3^{2}-4^{2}.........
upto
n
term
=(1-2)(1+2)+(3-4)(3+4)+......+(n-1-n)(n-1+n)
=-(1+2+3+4+.....+n)
S=\dfrac{-n(n+1)}{2}
The sum
\sum _{ i=0 }^{ m }{ \left( \begin{matrix} 10 \\ i \end{matrix} \right) \left( \begin{matrix} 20 \\ m-i \end{matrix} \right) }
(where
\left( \begin{matrix} p \\ q \end{matrix} \right)=0
if
p<q
) is maximum where
m
is
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0%
5
0%
10
0%
15
0%
20
The series
{1}^{2}-{2}^{2}+{3}^{2}-{4}^{2}+.....+{99}^{2}-{100}^{2}=
_____
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0%
-5050
0%
5050
0%
11000
0%
-11000
Explanation
The sum of nth term of the series is given as
S=(-1)^{n+1}\dfrac{n(n+1)}{2}
Here
n=100
So,
S=(-1)^{100+1}\dfrac{100(100+1)}{2}
=(-1)\dfrac{10100}{2}
=-5050
If
x
and
y
are the number of possibilities that
A
can assume such that the unit digit of A and
A^3
are same and the unit digit of
A^2
and
A^3
are same respectively ,then the value of
x-y
is (where
A
is a single digit number)
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0%
4
0%
2
0%
3
0%
5
Explanation
Here, according to the question
if
A=1
, then
{ A }^{ 3 }={ 1 }
,
We see that unit digit number of
A
and
{ A }^{ 3 }
are same.
for,
A=2\Rightarrow { A }^{ 3 }=8,
, unit digits are not same
A=3\Rightarrow { A }^{ 3 }=27,
unit digits are not same
A=4\Rightarrow { A }^{ 3 }=64,
unit digits are same
A=5\Rightarrow { A }^{ 3 }=125,
unit digits are same
A=6\Rightarrow { A }^{ 3 }=216,
unit digits are not same
A=7\Rightarrow { A }^{ 3 }=343,
unit digits are not same
A=8\Rightarrow { A }^{ 3 }=512,
unit digits are not same
A=8\Rightarrow { A }^{ 3 }=512,
unit digits are same
So, there are five possible solutions where unit digits of
A
and
{ A }^{ 3 }
are same.
\therefore x=5
Again, if we take
A=1,{ A }^{ 2 }=1,{ A }^{ 3 }={ 1 }
, unit digits are same
A=2\Rightarrow { A }^{ 2 }=4,{ A }^{ 3 }=8,
unit digits are not same
A=3\Rightarrow { A }^{ 2 }=9,{ A }^{ 3 }=27,
unit digits are not same
A=4\Rightarrow { A }^{ 2 }=16,{ A }^{ 3 }=64,
unit digits are not same
A=5\Rightarrow { A }^{ 2 }=25,{ A }^{ 3 }=125,
unit digits are same
A=6\Rightarrow { A }^{ 2 }=36,{ A }^{ 3 }=216,
unit digits are same
A=7\Rightarrow { A }^{ 2 }=49,{ A }^{ 3 }=343,
unit digits are not same
A=8\Rightarrow { A }^{ 2 }=64,{ A }^{ 3 }=512,
unit digits are not same
A=9\Rightarrow { A }^{ 2 }=81,{ A }^{ 3 }=729
unit digits are not same
Here, we can see that there are three possible solutions where unit digits of
{ A }^{ 2 },{ A }^{ 3 }
are same.
So,
y=3
Hence,
x-y=5-3=2.
\begin{bmatrix} 12 & 47 & 21 \\ 10 & 52 & 4 \\ 64 & ? & 24 \end{bmatrix}
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0%
40
0%
83
0%
62
0%
16
Explanation
\begin{bmatrix} 12 & 47 & 21\\ 10 & 52 & 4\\ 64 & ? & 24\end{bmatrix}
We can see that dividing third number in each row with second digit of middle number and multiplying with first digit gives first member.
Similarly, out of
4
options, dividing
24
by
3
and multiplying by
8
gives
64
.
\Rightarrow
Option
(B)
.
In a triangle
ABC
,
acos^2(\frac{C}{2})+c\;cos^2(\frac{A}{2})=\dfrac{3b}{2}
, then the sides
a,b,c
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Satisfy
a+b=c
0%
are in
A.P.
0%
are in
G.P.
0%
are in
H.P.
Explanation
a\cos^{2}\left(\dfrac{C}{2}\right)+C\cos^{2}\left(\dfrac{A}{2}\right)=\dfrac{3b}{2}
a\left[\dfrac{1+\cos C}{2}\right]+C\left[\dfrac{1+\cos A}{2}\right]=\dfrac{3b}{2}
\dfrac{a}{2}+\dfrac{c}{2}+\dfrac{1}{2}\left[a\cos C+C \cos A\right]=\dfrac{3b}{2}
a+c+a\cos C+ c \cos A=3b
2R\sin A+2R\sin C+2R\sin A\cos C+2R\sin C\cos A =3(2R)(\sin B)
\sin A+\sin C+\sin (A+C)=3\sin (B)
A+B+C=\pi\Rightarrow \sin(A+C)=\sin B
\sin A+\sin C+\sin B=3\sin (B)
\sin A+\sin C=2\sin B
2R\sin A+2R\sin C=2R\sin B
a+c=2b
\therefore a,b,c
are an
A.P
The value of
\displaystyle \sum^{n-1}_{r=1}\sin^{2}\dfrac {r \pi}{n}
is equal to
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0%
n
0%
\dfrac {n}{2}
0%
n+1
0%
Zero
The sum of the series
10.^{n}C_{0}+10^{2}.^{n}C_{1}+10^{3}.^{n}C_{2}+...10^{n+1}.^{n}C_{n}
is
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0%
11^{n}
0%
10.11^{n}
0%
11^{n+1}
0%
11^{n}-1
Explanation
\rightarrow 10(^{n}C_{0}+10.^{n}C_{1}+...+10^{n}
^{n}C_{n})
= 10\displaystyle(\sum_{r=1}^{n}
^{n}C_{r}(10)^{r}(1)^{n-r} )
= 10(1+10)^{n} = 10.11^{n} = (B)
1 + \left( {1 + a} \right)x + 1\left( {1 + a + {a^2}} \right){x^2} + \left( {1 + a + {a^2} + {a^3}} \right){x^3} + - - - - - \;{\text{where}}\;0 < a,x < 1,\;{\text{is}} -
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0%
\frac{1}{{\left( {1 - x} \right)\left( {1 - a} \right)}}
0%
\frac{1}{{\left( {1 - x} \right)\left( {1 - ax} \right)}}
0%
\frac{1}{{\left( {1 - a} \right)\left( {1 - ax} \right)}}
0%
None of these
Explanation
We have,
\begin{matrix} 1+\left( { 1+a } \right) x+1\left( { 1+a+{ a^{ 2 } } } \right) { x^{ 2 } }+\left( { 1+a+{ a^{ 2 } }+{ a^{ 3 } } } \right) { x^{ 3 } }+.......... \\ \frac { 1 }{ { 1-a } } \left[ { \left( { 1+a } \right) +\left( { 1-{ a^{ 2 } } } \right) x+\left( { 1-{ a^{ 3 } } } \right) { x^{ 2 } }+\left( { 1-{ a^{ 4 } } } \right) { x^{ 3 } }.......... } \right] \\ \frac { 1 }{ { 1-a } } \left[ { 1+x+{ x^{ 2 } }+{ x^{ 3 } }+.........-\left( { a+{ a^{ 2 } }x+{ a^{ 3 } }{ x^{ 2 } }+{ a^{ 4 } }{ x^{ 3 } }+.... } \right) } \right] \\ \end{matrix}
\begin{matrix} \frac { 1 }{ { 1-a } } \left[ { \frac { 1 }{ { 1-x } } -\frac { a }{ { 1-ax } } } \right] \\ \frac { 1 }{ { 1-a } } \left[ { \frac { { 1-ax-a+ax } }{ { \left( { 1-x } \right) \left( { 1-ax } \right) } } } \right] \\ \frac { 1 }{ { \left( { 1-x } \right) \left( { 1-ax } \right) } } \\ \end{matrix}
Then,
Option
A
is correct answer.
The sum of the series
\displaystyle \sum_{r = 1}^{n} (-1)^{r - 1}.^{n}C_{r} (a - r)
is
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0%
a
0%
0
0%
n.2^{n - 1} + a
0%
None of these
Explanation
\begin{array}{l} \sum _{ r=1 }^{ n }{ { { \left( { -1 } \right) }^{ r-1 } } }\, { \, ^{ n } }{ c_{ r } }\left( { a-r } \right) & \\ Put\, \, n=1 & \\ { \Rightarrow ^{ n } }{ c_{ 1 } }\left( { a-1 } \right) { -^{ n } }{ c_{ 2 } }\left( { a-2 } \right) { +^{ n } }{ c_{ 3 } }\left( { a-1 } \right) +.......{ \left( { -1 } \right) ^{ n-1 } }\, { \, ^{ n } }{ c_{ n } }\left( { a-n } \right) & \\ \Rightarrow a\left[ { ^{ n }{ c_{ 1 } }{ -^{ n } }{ c_{ 2 } }{ +^{ n } }{ c_{ 3 } }{ -^{ n } }{ c_{ 4 } }{ +^{ n } }{ c_{ 5 } }+.......{ -^{ n } }{ c_{ n } } } \right] { -^{ n } }{ c_{ 1 } }+{ 2^{ n } }{ c_{ 2 } }-{ 3^{ n } }{ c_{ 3 } }+{ 4^{ n } }{ c_{ 4 } }......... & \\ =a\left[ { ^{ n }{ c_{ 1 } }{ -^{ n } }{ c_{ 2 } }{ +^{ n } }{ c_{ 3 } }{ -^{ n } }{ c_{ 4 } }{ +^{ n } }{ c_{ 5 } }+...... } \right] -\left[ { ^{ n }{ c_{ 1 } }-{ 2^{ n } }{ c_{ 2 } }+{ 3^{ n } }{ c_{ 3 } }+{ 4^{ n } }{ c_{ 4 } }+....... } \right] & \\ { \left( { 1-x } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ -^{ n } }{ c_{ 1 } }x{ +^{ n } }{ c_{ 2 } }{ x^{ 2 } }{ -^{ n } }{ c_{ 3 } }{ x^{ 3 } }....... & \left[ { According\, \, to\, \, this\, \, formula } \right] \\ \Rightarrow Add\, \, \& \, \, subtract\, { \, ^{ n } }{ c_{ 0 } }\, \, in\, a\, \, have\, \, equation\, \, \& \, \, taking\, \, out\, \, common\, \, \left( { -1 } \right) \, \, also. & \\ \Rightarrow a\left[ { -\left( { ^{ n }{ c_{ 0 } }{ -^{ n } }{ c_{ 1 } }{ +^{ n } }{ c_{ 2 } }{ -^{ n } }{ c_{ 3 } }+...... } \right) { +^{ n } }{ c_{ o } } } \right] -\left[ { ^{ n }{ c_{ 1 } }-{ 2^{ n } }{ c_{ 2 } }+{ 3^{ n } }{ c_{ 3 } }-{ 4^{ n } }{ c_{ 4 } } } \right] & \\ Put\, \, x=1 & \\ { \left( { 1-1 } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ -^{ n } }{ c_{ 1 } }{ +^{ n } }{ c_{ 2 } }{ -^{ n } }{ c_{ 3 } }=0 & \\ So, & \\ a\left[ { -0{ +^{ n } }{ c_{ 0 } } } \right] -\left[ { ^{ n }{ c_{ 1 } }-{ 2^{ n } }{ c_{ 2 } }+{ 3^{ n } }{ c_{ 3 } }-{ 4^{ n } }{ c_{ 4 } }+......... } \right] \to \left( i \right) & \\ { \left( { 1+n } \right) ^{ n } }{ =^{ n } }{ c_{ 0 } }{ +^{ n } }{ c_{ 1 } }x{ +^{ n } }{ c_{ 2 } }{ x^{ 2 } }{ +^{ n } }{ c_{ 3 } }{ x^{ 3 } }{ +^{ n } }{ c_{ 4 } }{ x^{ 4 } }+....... & \left[ { \leftarrow formula } \right] \\ Diff.\, \, w.r.t.\, \, n,\, we\, \, get & \\ n{ \left( { 1+x } \right) ^{ n-1 } }{ =^{ n } }{ c_{ 1 } }+2{ x^{ n } }{ c_{ 2 } }+3{ x^{ n } }{ c_{ 3 } }+4{ x^{ n } }{ c_{ 4 } }+.... & \\ Put\, \, x=-1\, \, So, & \\ x{ \left( { 1-x } \right) ^{ n-1 } }{ =^{ n } }{ c_{ 1 } }-2{ x^{ n } }{ c_{ 2 } }+3{ x^{ 2 } }^{ n }{ c_{ 3 } }+{ 4^{ n } }{ c_{ 4 } }+......... & \\ 0{ =^{ n } }{ c_{ 1 } }-{ 2^{ n } }{ c_{ 2 } }+{ 3^{ n } }{ c_{ 3 } }-{ 4^{ n } }{ c_{ 4 } } & \end{array}
So, put the value in equation in (1)
\begin{array}{l} =a\left[ { ^{ n }{ c_{ 0 } } } \right] -\left[ 0 \right] \Rightarrow a\left( 1 \right) =a \\ \sum _{ r=1 }^{ n }{ { { \left( { -1 } \right) }^{ r-1 } }\, { \, ^{ n } }{ c_{ r } }\left( { a-r } \right) =a } \end{array}
1.3.4+2.5.8+3.6.9+
upto
n
terms is equal to ________ .
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0%
\dfrac {n(n+1)(n+2)}{6}
0%
\dfrac {n(n+1)(3n^{2}+23n+46)}{12}
0%
\dfrac {n(27n^{3}+90n^{2}+45n-5)}{4}
0%
\dfrac {n(n+1)(2n+1)}{6}
0%
None\ of\ these
The sum of the series
1+2.2+3.2^{2}+4.2^{3}+5.2^{4}+...+1000.2^{999}
is
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0%
999.2^{999}-1
0%
999.2^{1000}-1
0%
999.2^{1000}+1
0%
999.2^{999}+1
Explanation
S = 1+2.2+3.2^{2}+4.2^{3}+5.2^{4}+..... +1000.2^{999}
__ (1)
Multiply with '2' on both sides
2S = 1.2+2.2^{2}+3.2^{2}+4.2^{4}+5.2^{5}+... +999.2^{999}+1000.2^{1000}
__ (2)
subtracting equation (1) & (2)
(S-2S) = [ 1+(2.2-1.2)+(3.2^{2}-2.2^{2})+(4.2^{3}-3.2^{3})+(5.2^{4}-4.2^{4})+
...
+(1000.2^{999}-999.2^{999})-1000.2^{1000}
(S-2S) = 1+1.2+1.2^{2}+1.2^{3}+1.2^{4}+.... +1.2^{999}-1000.2^{1000}
-S = 1+(2+2^{2}+2^{3}+2^{4}+...+2^{999})-1000.2^{1000}
\left [ \because a+ar+ar^{2}+...+ar^{n-1} = \dfrac{a.(r^{n}-1)}{r-1} \right ]
a = 2, r = 2, n = 999
-S= \dfrac{1+2.(2^{999}-1)}{2-1} - 1000.2^{1000}
-S = 1+2^{1000}-2-1000.2^{1000}
-S = -1+2^{1000}(1-1000)
-S = -1-999.2^{1000}
S = 999.2^{1000}+1
\displaystyle\sum^n_{r=1}\displaystyle\sum^{r-1}_{p=0}
^{n}C_r\cdot {^{r}C_p}\cdot 2^p
is equal to?
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0%
4^n-3^n+1
0%
4^n-3^n-1
0%
4^n-3^n+2
0%
4^n-3^n
If
| x| < 1
, then the sum of series
1+ 2x + 3x^2 + 4x^3 + .........\infty
will be
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0%
\dfrac{1}{1-x}
0%
\dfrac{1}{1+x}
0%
\dfrac{1}{(1+x)^2}
0%
\dfrac{1}{(1-x)^2}
Explanation
Let
S=1+2x+3{x}^{2}+......\infty
\dots(1)
multiply by
x
on both sides
{x}{S}=x+2{x}^{2}+3{x}^{3}+.....\infty
\dots(2)
Subtract
(1)
from
(2)
S=1+2x+3{x}^{2}+......\infty
{x}{S}=x+2{x}^{2}+3{x}^{3}+.....\infty
- - - -
--------------------------------------------------------
S(1-x)=1+x+{x}^{2}+{x}^{3}+.....\infty
S(1-x)=\cfrac{1}{1-x}
[sum of infinte G.P
S_{\infty}=\dfrac{1}{1-r}]
S=\cfrac{1}{(1-x)^{2}}
\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+..n
terms
=
Report Question
0%
\dfrac{3n}{2(3n+2)}
0%
\dfrac{3n}{3n+2}
0%
\dfrac{n}{2(3n+2)}
0%
\dfrac{n}{3n+2}
Explanation
\cfrac { 1 }{ 2.5 } +\cfrac { 1 }{ 5.8 } +\cfrac { 1 }{ 8.11 } +.....+\cfrac { 1 }{ \left[ 2+3\left( n-1 \right) \right] .\left[ 5+3\left( n-1 \right) \right] }
=\sum _{ n=1 }^{ n }{ \cfrac { 1 }{ \left[ 2+3n-3 \right] .\left[ 5+3n-3 \right] } }
=\sum _{ n=1 }^{ n }{ \cfrac { 1 }{ \left( 3n-1 \right) .\left( 3n+2 \right) } } =\cfrac { 1 }{ 3 } \sum { \cfrac { \left[ 3n+2-\left( 3n-1 \right) \right] }{ \left( 3n-1 \right) \left( 3n+2 \right) } }
=\cfrac { 1 }{ 3 } \sum { \cfrac { 1 }{ 3n-1 } -\cfrac { 1 }{ 3n+2 } }
=\cfrac { 1 }{ 3 } \left[ \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 5 } +\cfrac { 1 }{ 5 } -\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 8 } -\cfrac { 1 }{ 11 } +...+\cfrac { 1 }{ 3n-4 } -\cfrac { 1 }{ 3n-1 } +\cfrac { 1 }{ 3n-1 } -\cfrac { 1 }{ 3n+2 } \right]
=\cfrac { 1 }{ 3 } \left[ \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 3n+2 } \right] =\cfrac { 3n }{ 3\left( 3n+2 \right) .2 } =\cfrac { n }{ 2\left( 3n+2 \right) }
The sum of the series
{5 \over {13}} + {{55} \over {{{13}^2}}} + {{555} \over {{{13}^3}}} + .........\infty
is
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0%
{65\over {36}}
0%
{65\over {32}}
0%
{25\over {36}}
0%
none of these
Explanation
S=\dfrac{5}{13}+\dfrac{55}{13^2}+\dfrac{555}{13^3}+\dfrac{5555}{13^4}+.....
(1)
Now multiply by
\dfrac{1}{13}
, we get
\dfrac{1}{13}s=\dfrac{5}{13^2}+\dfrac{55}{13^3}+\dfrac{555}{13^4}+.....
(2)
Subtracting
(2)
from
(1)
s-\dfrac{1}{13}s\left[\dfrac{5}{13}+\dfrac{55}{13^2}+\dfrac{555}{13^3}+.....\right]-\left[\dfrac{5}{13^2}+\dfrac{55}{13^3}+..…\right]
\dfrac{12s}{13}=\dfrac{5}{13}+\left(\dfrac{55}{13^2}+\dfrac{5}{13^2}\right)+\left(\dfrac{555}{13^3}+\dfrac{55}{13^3}\right)+.....
\dfrac{12s}{13}=\dfrac{5}{13}+\dfrac{50}{13^2}+\dfrac{500}{13^3}+....
=\dfrac{\dfrac{5}{13}}{1-\dfrac{10}{13}}
\dfrac{12s}{13}=\dfrac{5}{3}
\Rightarrow s=\dfrac{65}{36}
.
Sum to n terms the following series :
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5 + 11 + 19 + 29 + 41 + .......
0%
3 + 7 + 14 + 24 + 37 +......
0%
6 + 9 + 16 + 27 + 42 + .....
0%
5 + 7 + 13 + 31 + 85 + ....
\sum _{ k=1 }^{ 2n+1 }{ (-1)^{k-1} }k^2=
Report Question
0%
(n+1)(2n+1)
0%
(n+1)(2n-1)
0%
(n-1)(2n-1)
0%
(n-1)(2n+1)
Explanation
\displaystyle \sum_{k = 1}^{2n + 1} (-1)^{k -1} \, k^2
= 1^2 - 2^2 + 3^2 - 4^2 + 5^2 ...... - (2n)^2 + (2n + 1)^2
= 1^2 + 3^2 + 5^2 + ..... + (2n + 1)^2 - [2^2 + 4^2 + ... + (2n)^2]
= 1^2 + 2^2 + 3^2 + ..... + (2n + 1)^2 - 2 [2^2 + 4^2 + .... (2n)^2]
Adding and subtract even terms to complete the series.
Using identify
1^2 + 2^2 + .... + n^2 = \dfrac{n (n + 1) (2n + 1)}{6}
The addition will become
= 1^2 + 2^2 + 3^2 + .... + (2n + 1) - 2(2^2) [1^2 + 2^2 + ..... + n^2]
= \dfrac{(2n + 1)(2n + 2)(4n + 3)}{6} - \dfrac{8 n (n + 1)(2n + 1)}{6}
= \dfrac{2(2n + 1)(n + 1)(4n + 3)}{6} - \dfrac{8 n(n + 1)(2n + 1)}{6}
= \dfrac{2}{6} (n + 1)(2n + 1) [4n + 3 - 4n]
= \dfrac{1}{3} (n + 1) (2n + 1) (3)
= (n + 1) (2n + 1)
If
S = 1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + \frac{1}{{1 + 2 + 3 + 4}}......,
then
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0%
{S_n} = \frac{{2n}}{{n + 1}}
0%
{S_n} = \frac{{2n}}{{n - 1}}
0%
{S_\infty } = 2
0%
{S_\infty } = 1
Explanation
S_{n} = \dfrac {1}{1} + \dfrac {1}{1 + 2} + \dfrac {1}{1 + 2 + 3} + \dfrac {1}{1 + 2 + 3 + 4} + ......
T_{n} = \dfrac {1}{1 + 2 + 3 + ...n} = \dfrac {1}{n\dfrac {(n + 1)}{2}}
T_{n} =\dfrac {2}{n(n + 1)} = 2\left [\dfrac {(n + 1) - n}{n(n + 1)}\right ]
= 2\left [\dfrac {1}{n} - \dfrac {1}{n + 1}\right ]
T_{1} = 2\left [\dfrac {1}{1} - \dfrac {1}{2}\right ]
T_{2} = 2\left [\dfrac {1}{2} - \dfrac {1}{3}\right ]
.
.
.
.
T_{n} = 2\left [\dfrac {1}{n} - \dfrac {1}{n + 1}\right ]
Add the above terms
S_{n} = 2\left [1 - \dfrac {1}{n + 1}\right ] = 2\left [\dfrac {n + 1 - 1}{n + 1}\right ]
= \dfrac {2n}{n + 1}
S_{\infty} = \displaystyle \lim_{n\rightarrow 0} \dfrac {2n}{n + 1} = \displaystyle \lim_{n\rightarrow \infty} = \dfrac {2}{1 + \dfrac {1}{n}}
S_{\infty} = 2
.
If
S = \tan ^ { - 1 } \left( \frac { 1 } { n ^ { 2 } + n + 1 } \right) + \tan ^ { - 1 } \left( \frac { 1 } { n ^ { 2 } + 3 n + 3 } \right) + \ldots + \tan ^ { - 1 } \left( \frac { 1 } { 1 + ( n + 19 ) ( n + 20 ) } \right)
then
\tan S
is equal to
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0%
\frac { 20 } { 401 + 20 n }
0%
\frac { n } { n ^ { 2 } + 20 n + 1 }
0%
\frac { 20 } { n ^ { 2 } + 20 n + 1 }
0%
\frac { n } { 401 + 20 n }
Explanation
\begin{array}{l} S={ \tan ^{ -1 } }\left( { \frac { 1 }{ { { n^{ 2 } }+n+1 } } } \right) +{ \tan ^{ -1 } }\left( { \frac { 1 }{ { { n^{ 2 } }+3n+3 } } } \right) +....+{ \tan ^{ -1 } }\left( { \frac { 1 }{ { 1+\left( { n+19 } \right) \left( { n+20 } \right) } } } \right) \\ S={ \tan ^{ -1 } }\left( { \frac { { n+1-n } }{ { 1+n\left( { n+1 } \right) } } } \right) +{ \tan ^{ -1 } }\left[ { \frac { { \left( { n+2 } \right) -\left( { n-1 } \right) } }{ { 1+\left( { n+1 } \right) \left( { n+2 } \right) } } } \right] +.....+{ \tan ^{ -1 } }\left[ { \frac { { \left( { n+20 } \right) -\left( { n+19 } \right) } }{ { 1+\left( { n+19 } \right) \left( { n+20 } \right) } } } \right] \\ =\left[ { { { \tan }^{ -1 } }\left( { n+1 } \right) -{ { \tan }^{ -1 } }\left( n \right) } \right] +\left[ { { { \tan }^{ -1 } }\left( { n+2 } \right) -{ { \tan }^{ -1 } }\left( { n+1 } \right) } \right] +......+\left[ { { { \tan }^{ -1 } }\left( { n+20 } \right) -{ { \tan }^{ -1 } }\left( { n+19 } \right) } \right] \\ ={ \tan ^{ -1 } }\left( { n+20 } \right) -ta{ n^{ -1 } }\left( n \right) \\ ={ \tan ^{ -1 } }\left( { \frac { { 20 } }{ { 1+{ n^{ 2 } }+20n } } } \right) \\ { { tans } }=\frac { { 20 } }{ { { n^{ 2 } }+20n+1 } } \\ Hence,\, the\, option\, C\, is\, correct\, answer. \end{array}
7, 11, 23, 51, 103 ?
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0%
186
0%
188
0%
185
0%
187
0%
none of these
Explanation
7,11,23,51,103,?
7+4\times 1= 11
11+4\times 3=23
\leftarrow
difference of 2
23+4\times 7=51
\leftarrow
difference of 4
51+4\times 13= 103
\leftarrow
difference of 6
103+4\times 21= 187
\leftarrow
difference will be of 8
The value of the expression
\sum _{ r=0 }^{ n }{ { (-1) }^{ r } } \left( \dfrac { ^nC_r }{^{r+3}C_r } \right)
is
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0%
\dfrac{n(n+1)}{2}
0%
\dfrac{n+3}{3}
0%
\dfrac{3}{n+3}
0%
\dfrac{n+2}{2}
S=\tan^{-1}\left(\dfrac{1}{n^2+n+1}\right)+\tan^{-1}\left(\dfrac{1}{n^2+3n+3}\right)+.....+\tan^{-1}\left(\dfrac{1}{1+(n+19)(n+20)}\right)
, then
\tan S
is equal to?
Report Question
0%
\dfrac{20}{401+20n}
0%
\dfrac{n}{n^2+20n+1}
0%
\dfrac{20}{n^2+20n+1}
0%
\dfrac{n}{401+20n}
Explanation
S = \sum_{r=0}^{19} tan^{-1} \left ( \frac{1}{1+(n+r)(n+r+1)} \right ) = tan^{-1}\left ( \frac{(n+r+1)-(n+r)}{1+(n+r)(n+r+1)} \right )
\Rightarrow S = \sum_{r=0}^{19} \left ( tan^{-1}(n+r+1)-tan^{-1}(n+r) \right )
=tan^{-1}(n+1) - tan^{-1}(n) = tan^{-1}(n+20)-tan^{-1}(n+19)
+ tan^{-1}(n+2) - tan^{-1}(n+1)
tan^{-1}(n+20) - tan^{-1}(n+19)
\Rightarrow tan \, S = \frac{(n+20)-(n)}{1+n(n+20)} = \frac{20}{n^{2}+20n+1}
\Rightarrow (C)
13, 16, 22, 33, 51 ?
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0%
89
0%
78
0%
102
0%
69
0%
none of these
Explanation
REF.Image
13,16,22,33,51, ?
Evaluate:-
If
\sum\limits_{r - 0}^n {{{\left\{ {\frac{{^n{C_{r - 1}}}}{{^n{C_r}{ + ^n}{C_{r - 1}}}}} \right\}}^3} = \frac{{25}}{{24}}}
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0%
3
0%
6
0%
4
0%
5
655, 439, 314, 250, 223 ?
Report Question
0%
215
0%
210
0%
195
0%
190
0%
none of these
Explanation
Solution
\rightarrow
Given terms are 655,439,314,250,223,?
As
a_{1}=655
a_{2}= 439
a_{3}=314
a_{4}= 250
a_{5}= 223
a_{6}=?
So,
a_{1}-a_{2}=655- 439= 216=6^{3}
a_{3}-a_{2}= 439- 314= 125=5^{3}
a_{3}-a_{4}= -250+314= 64=4^{3}
a_{4}-a_{5}=-223+250= 27=3^{3}
So, we can observe difference b/w terms are cubes
of 6,5,4,3 so, is similar way
a_{5}-a_{6}=2^{3}
223-a_{6}= 8
a_{6} = 215
So, we find next term is 215
If the expansion of
\left( x+a \right) ^{ n }
if the sum of odd terms be P & sum of even terms be Q, prove that
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0%
{ P }^{ 2 }-{ Q }^{ 2 }=({ x }^{ 2 }-{ a }^{ 2 })^{ n }
0%
4PQ=(x+a)^{ 2n }-(x-a)^{ 2n }
0%
P^2-Q^2=(x^2+a^2)^{n}
0%
None\ of\ these
Sum of the series
S=1^{2}-2^{2}+3^{2}-4^{2}+..... -2000^{2}+2003^{2}
is
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0%
2007006
0%
1005004
0%
2000506
0%
None
Find the next term
210,209,205,196,180,?
Report Question
0%
138
0%
77
0%
155
0%
327
Explanation
T_1=210
,T_2=209,
T_3=205,
T_4=196,
T_5=180,
T_6=?
T_1-T_2=1^2\\
T_2-T_3=2^2\\
T_3-T_4=3^2\\
T_4-T_5=4^2\\
T_5-T_6=5^2\\
\implies T_6=T_5 - 5^2=180-25=155
Determine the next term
20, 24, 33, 49, 74, 110, ?
Report Question
0%
133
0%
147
0%
159
0%
163
0%
171
Explanation
20,24,33,49,74,110,...
Let the number be
x
difference
24-20=4,33-24=9,49-33=16,74-49=25,110-74=36,x-110=49
\implies24-20=2^2, 33-24= 3^2, 49-33= 4^2, \,\, 74-49= 5^2\,\, 110-74=6^2,\,\,\, x-110= 7^2
So,
110+49=159
ANSWER IS C
Find the next term of the series
1728, 2744, 4096, 5832, 8000, 10648, ?
Report Question
0%
2167
0%
13824
0%
15625
0%
9261
0%
17576
Explanation
1728,2744,4096,5832,8000,10648.....
12^3\rightarrow 1728
14^3\rightarrow 2744
16^3\rightarrow 4096
18^3\rightarrow 5832
20^3\rightarrow 8000
22^3\rightarrow 10648
24^3\rightarrow 13824
ANSWER IS B
4,6,12,30,90,315,?
Report Question
0%
945
0%
1102
0%
1260
0%
1417.5
0%
None\ of\ these
Explanation
=315\times 4
=1260.
Hence, the answer is
1260.
Find next term
462,552,650,756,870,992,?
Report Question
0%
1040
0%
1122
0%
1132
0%
105
Find the missing number from the given alternatives.
Report Question
0%
39,116
0%
52,156
0%
30,117
0%
31,116
Explanation
13\quad\quad 39\\26\quad\quad 78\\?\quad\quad ?
The pattern satisfies
x\quad\quad 3x
Therefore correct option is
52\quad\quad 156
8, 31, 122, 485, 1936, 7739, ?
Report Question
0%
30950
0%
46430
0%
34650
0%
42850
0%
38540
The sum of infinite series
\begin{vmatrix} 1 & 2 \\ 6 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 2 } & 2 \\ 2 & 4 \end{vmatrix}+\begin{vmatrix} \frac { 1 }{ 4 } & 2 \\ \frac { 2 }{ 3 } & 4 \end{vmatrix}+.........
Report Question
0%
-10
0%
0
0%
10
0%
\infty
If
\dfrac{\pi}{4}-1+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{1}{13}+=0
then value of
\dfrac{1}{1\times3}+\dfrac{1}{5\times7}+\dfrac{1}{9\times11}+\dfrac{1}{13\times15}+..
is
Report Question
0%
\dfrac{\pi}{8}
0%
\dfrac{\pi}{6}
0%
\dfrac{\pi}{4}
0%
\dfrac{\pi}{34}
Let
t_{r}=\frac{r}{1+r^{2}+r^{4}}
then,
\lim_{n\rightarrow \infty }\sum_{r=1}^{n}t_{r}
equals
Report Question
0%
\frac{1}{4}
0%
1
0%
\frac{1}{2}
0%
None of these
Explanation
t_r=\dfrac{r}{1+r^2+r^4}
=\dfrac{r}{r^4+2r^2+1-r^2}
=\dfrac{r}{(r^2+1)^2-r^2}=\dfrac{r}{(r^2+1-r)(r^2+1+r)}
=\dfrac{1}{2}\dfrac{2r}{(r^2+1-r)(r^2+r+1)}
=\dfrac{1}{2}\dfrac{(r^2+r+1)-(r^2-r+1)}{(r^2-r+1)(r^2+r+1)}
=\dfrac{1}{2}\left[\dfrac{1}{r^2-r+1}-\dfrac{1}{r^2+r+1}\right]
=\dfrac{1}{2}\left[\dfrac{1}{r^2-r+1}-\dfrac{1}{(r+1)^2-(r+1)+1}\right]
\displaystyle\sum^n_{r=1}t_r=\dfrac{1}{2}\left[\dfrac{1}{1-1+1}-\dfrac{1}{(n+1)^2-(n+1)+1}\right]
=\dfrac{1}{2}\left[1-\dfrac{1}{n^2+n+1}\right]=\dfrac{1}{2}\left[\dfrac{n^2+n+1-1}{n^2+n+1}\right]
\displaystyle\sum t_n=\dfrac{1}{2}\left[\dfrac{1+\dfrac{1}{n}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}\right]
\underset{n\rightarrow \infty}{lt}\displaystyle\sum t_r=\dfrac{1}{2}\left[\dfrac{1+0}{1+0+0}\right]
=\dfrac{1}{2}
.
The sum to
n
terms of the series
\dfrac {3}{1^{2}}+\dfrac {5}{1^{2}+2^{2}}+\dfrac {7}{1^{2}+2^{2}+3^{2}}+.........
is
Report Question
0%
\dfrac {6n}{n+1}
0%
\dfrac {9n}{n+1}
0%
\dfrac {12n}{n+1}
0%
\dfrac {3n}{n+1}
Explanation
\begin{array}{l} Let \\ { T_{ r } }=\dfrac { { 2r+1 } }{ { { 1^{ 2 } }+{ 2^{ 2 } }+{ r^{ 2 } } } } \\ =\dfrac { { \left( { 2r+1 } \right) } }{ { r\left( { r+1 } \right) \left( { 2r+1 } \right) } } \\ =\dfrac { 6 }{ { r\left( { 2+1 } \right) } } \\ \sum { { T_{ r } } } =\sum { \dfrac { 6 }{ { r\left( { r+1 } \right) } } } \\ =6\sum _{ r=1 }^{ n }{ \dfrac { { \left( { r+1 } \right) -r } }{ { r\left( { r+1 } \right) } } } \\ =6\sum _{ r=1 }^{ n }{ \left[ { \dfrac { 1 }{ r } -\dfrac { 1 }{ { r+1 } } } \right] } \\ =6\left[ { \left( { 1-\dfrac { 1 }{ 2 } } \right) +\left( { \dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } } \right) +.......+\left( { \dfrac { 1 }{ n } -\dfrac { 1 }{ { n+1 } } } \right) } \right] \\ =6\left[ { 1-\dfrac { 1 }{ { n+1 } } } \right] \\ =\dfrac { { 6n } }{ { n+1 } } \\ Hence,\, option\, A\, is\, \, the\, correct\, answer. \end{array}
499, 622, 868, 1237, 1729, 2344, ?
Report Question
0%
3205
0%
3082
0%
2959
0%
3462
0%
2876
Explanation
499, 622, 868, 1237, 1729, 2344,
______
499+1\times 123=499+123=622
622+2\times 123=622+146=868
868+3\times 123=868+369=1237
1237+4\times 123=1237+492=1729
1729+5\times 123=1729+615=2344
2344+6\times 123=2344+738=3082
.
The sum
\dfrac{1}{1+1^{2}+1^{4}}+\dfrac{2}{1+2^{2}+2^{4}}+\dfrac{3}{1+3^{2}+3^{4}}+...+\dfrac{99}{1+99^{2}+99^{4}}
lies between
Report Question
0%
0.46
and
0.47
0%
0.52
and
1.0
0%
0.48
and
0.49
0%
0.49
and
0.50
Explanation
We have,
\dfrac { 1 }{ { 1+{ 1^{ 2 } }+{ 1^{ 4 } } } } +\dfrac { 2 }{ { 1+{ 2^{ 2 } }+{ 2^{ 4 } } } } +\dfrac { 3 }{ { { 1^{ 3 } }+{ 2^{ 2 } }+{ 3^{ 4 } } } } +.........+\dfrac { { 99 } }{ { 1+{ { 99 }^{ 2 } }+{ { 99 }^{ 4 } } } } \\ =\sum _{ n=1 }^{ 99 }{ \dfrac { x }{ { 1+{ x^{ 2 } }+{ x^{ 4 } } } } } \\ =\sum _{ n=1 }^{ 99 }{ \dfrac { x }{ { { { \left( { { x^{ 2 } }+1 } \right) }^{ 2 } }-{ x^{ 2 } } } } } \\ =\sum _{ n=1 }^{ 99 }{ \dfrac { { \left( { { x^{ 2 } }+x+1 } \right) -\left( { { x^{ 2 } }-x+1 } \right) } }{ { 2\left( { { x^{ 2 } }+x+1 } \right) \left( { { x^{ 2 } }-x+1 } \right) } } } \\ =\dfrac { 1 }{ 2 } \sum _{ n=1 }^{ 99 }{ \left( { \dfrac { 1 }{ { { x^{ 2 } }-x+1 } } -\dfrac { 1 }{ { { x^{ 2 } }+x+1 } } } \right) } \\ =\dfrac { 1 }{ 2 } \left( { \dfrac { 1 }{ 1 } -\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 3 } -\dfrac { 1 }{ 7 } +\dfrac { 1 }{ 7 } +.........+\dfrac { 1 }{ { { { 99 }^{ 2 } }+99+1 } } } \right) \\ =\dfrac { 1 }{ 2 } \left( { 1-\dfrac { 1 }{ { 9901 } } } \right) \\ =\dfrac { { 4950 } }{ { 9901 } } \\ \approx \dfrac { 1 }{ 2 }
\\ Hence,\, option\, D\, is\, correct\, answer.
Find :
12,15,21,24,30,33 , ? , ?
Report Question
0%
39,42
0%
37,42
0%
38,47
0%
39,51
Explanation
12,15,21,24,30,33,?,?
12
12 + 3 = 15
15+3\times 2=21
21+3=24
24+3\times 2=30
30+3=33
33+3\times 2=39
39+3=42
Find the missing number :
Report Question
0%
125
0%
90
0%
105
0%
225
Explanation
\textbf{Step 1: Observe the numbers opposite to each other}
\text{We can see that the number opposite to 315 is 21. Also, } \dfrac{315}{21}=15
\text{Hence, we can say the missing number has at least one factor of 15 }
\text{We see the missing number is opposite to 15}
\text{Also, } 15\times15=225
\textbf{Step 2: Verification}
\text{Let the missing number be x.}
\text{ATQ, }
\Rightarrow 21+15+x=261
\Rightarrow x=261-21-15
x=225
\text{Hence, it is verified that x=225}
\textbf{Thus, the missing number is D 225.}
23,29,47,75 , ?
Report Question
0%
87
0%
93
0%
110
0%
117
Explanation
2\underbrace { 3\quad ,\quad 2 } \underbrace { 9\quad ,\quad 4 } \underbrace { 7\quad ,\quad 7 } \underbrace { 5\quad ,\quad ? } \quad =75+35
6
18
28
35
\left( 2\times 3 \right) \uparrow
\left( 2\times 9 \right) \uparrow
\left( 4\times 7 \right) \uparrow
\left( 7\times 5 \right) \uparrow
=110
The value of
\dfrac{1}{6.10}+\dfrac{1}{10.14}+\dfrac{1}{14.18}+....\infty
equals to
Report Question
0%
\dfrac{1}{(24)^2}
0%
\dfrac{1}{6}
0%
\dfrac{1}{24}
0%
\dfrac{1}{(24)^3}
Explanation
S=\frac{1}{6 \cdot 10}, \frac{1}{10 \cdot 14}+\frac{1}{1419}+.....\infty
S=\frac{1}{4}\left(\frac{1}{6}-\frac{1}{10}\right)+\frac{1}{4}\left(\frac{1}{10}-\frac{1}{14}\right)+\frac{1}{4}\left(\frac{1}{14}-\frac{1}{18}\right)+.....\infty
Now take
\frac{1}{4}
common
S=\frac{1}{4}\left[\frac{1}{6}-\frac{1}{16}+\frac{1}{10}-\frac{1}{1 / 4}+\frac{1}{14}-\frac{1}{18}+1,-\infty\right]
S=\frac{1}{4} \times \frac{1}{6}=\frac{1}{24}
option
c
is correct
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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