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CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 12 - MCQExams.com
CBSE
Class 11 Engineering Maths
Binomial Theorem
Quiz 12
If
C
r
=
(
100
C
r
)
,
t
h
e
n
E
=
n
+
4
∑
r
=
0
(
−
1
)
r
c
r
c
r
+
1
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0%
(
100
C
45
)
0%
(
100
C
47
)
0%
(
101
C
50
)
0%
(
100
C
51
)
Explanation
(
1
+
x
)
101
=
C
0
+
C
1
x
+
C
2
x
2
+
.
.
.
.
.
(
1
−
x
)
101
=
C
0
x
101
−
C
1
x
100
+
C
2
x
99
−
.
.
.
.
Multiplying both sides,
(
x
2
−
1
)
101
=
(
C
0
+
C
1
x
+
C
2
x
2
+
.
.
.
.
)
(
C
0
x
101
−
C
1
x
100
+
C
2
x
99
−
.
.
.
.
)
....(1)
Now E =
C
0
C
1
−
C
−
1
C
2
+
C
2
C
3
−
.
.
.
.
= -[Coefficient of
x
100
in the product of R.H.S. of (1)]
or = -[Coefficient of
(
x
2
)
50
in the product of R.H.S. of (1)] ....(2)
Now
(
x
2
−
1
)
101
=
(
−
1
)
101
(
1
−
x
2
)
101
=
−
(
1
−
x
2
)
101
∴
Coefficient of
(x^2)^{50}
= -
(- 1)^{50} \, \, ^{101}C_{50} \, = \, - ^{101}C_{50}
...(3)
\therefore \, \, \, \, E \, = \, (-) \, (-) \, ^{101}C_{50} \, = \, ^{101}C_{50} \, by \, (2) \, and \, (3)
If the
rth
term in the expansion of
\left(\dfrac{x}{3}-\dfrac{2}{x^{2}}\right)^{10}
contains
x^{4}
then
r
is equal to
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0%
2
0%
1
0%
3
0%
5
If
C_{0}
,
C_{1}
,
C_{2}
,
...,
C_{n}
are the binomial coefficients and
n
is odd, then
2C_{1}
+
2^{3}.C_{5}+
...
+2^{n}\ ^{n}C_{n}
equals
Report Question
0%
\dfrac {3^{n}+(-1)^{n}}{2}
0%
\dfrac {3^{n}-(-1)^{n}}{2}
0%
\dfrac {3^{n}+1}{2}
0%
\dfrac {3^{n}-1}{2}
\begin{array} { l } { \text { Assertion } ( \mathrm { A } ) : \text { The expansion of } ( 1 + x ) ^ { n } = } \\ { C _ { 0 } + C _ { 1 } x + C _ { 2 } x ^ { 2 } + \ldots + C _ { n } x ^ { n } } \\ { \text { Reason (R): If } x = - 1 , \text { then the above expansion is } } \\ { \text { zero } } \end{array}
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0%
\begin{array} { l } { \text { Both A and R are true and } \mathrm { R } \text { is the correct } } \\ { \text { explanation of } \mathrm { A } } \end{array}
0%
\begin{array} { l } { \text { Both A and R are true and R is not the } } \\ { \text { correct explanation of } \mathrm { A } } \end{array}
0%
A
is true, but
R
is false
0%
A
is false, but
R
is true
Explanation
We Know that Assertion is nothing but the standard binomial expansion of
(1+x)^n
=C_{0}+C_{1}X+C_{2}X^2+ \dots +C_{n}x^n
So if
x=-1
then
(1+(-1))^n=(1-1)^n=0
Ans:
Opt:[B]
If the ratio
T_2 : T_3
in the expansion of
(a+b)^n
and
T_3 : T_4
in the expansion of
(a+b)^{n+3}
are equal , then n
=
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0%
3
0%
4
0%
5
0%
6
The coefficient of
{x^5}
in the expansion of
{\left( {1 + {x^2}} \right)^5}{\left( {1 - x} \right)^4}
is
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0%
{4.^6}{C_4}
0%
{2.^6}{C_4}
0%
{2.^6}{C_2}
0%
{4.^6}{C_2}
In the binomial expansion of
( a - b ) ^ { n } n > 0
and the sum
5 ^ { t h }
and
6 ^ { \text { th } }
terms is zero, then
\frac { a } { b }
equal to
Report Question
0%
\frac { 5 } { n - 4 }
0%
\frac { 6 } { n - 5 }
0%
\frac { n - 5 } { 6 }
0%
\frac { n - 4 } { 5 }
The co-efficient of
x
in the expansion of
\left( 1 - 2 x ^ { 3 } + 3 x ^ { 5 } \right) \left( 1 + \dfrac { 1 } { x } \right) ^ { 8 }
is
Report Question
0%
56
0%
65
0%
154
0%
62
Explanation
(1-2x^3+3x^5) \left (1+\dfrac {1}{x}\right)^8
find coefficient of
x
\Rightarrow \left( 1-2{ x }^{ 3 }+3{ x }^{ 5 } \right) \left( 1+8\left( \dfrac { 1 }{ x } \right) +_{ }^{ 8 }{ { C }_{ 2 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 2 }+_{ }^{ 8 }{ { C }_{ 3 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 3 }+_{ }^{ 8 }{ { C }_{ 4 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 4 }+_{ }^{ 8 }{ { C }_{ 5 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 5 }+_{ }^{ 8 }{ { C }_{ 6 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 6 }+_{ }^{ 8 }{ { C }_{ 7 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 7 }+{ \left( \dfrac { 1 }{ x } \right) }^{ 8 } \right)
(By Binomial theorem)
=(1+(2.^BC_2 +3^8C_4)x+-----)
=
co-efficient of
x
is
-2\times ^8C_2+3\times ^8C_4=\dfrac {-2\times 8!}{6!\ 2!}+3\times \dfrac {8!}{4!\ 4!}
=\dfrac {-2\times 8\times 7}{2\times 1}dfrac {3\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 1}=-56+210
=154
If the sum of odd terms and the sum of even terms in
(x+a)^{n}
are
p
and
q
respectively then
p^{2}+q^{2}=
Report Question
0%
\dfrac{(x+a)^{2n}-(x-a)^{2n}}{2}
0%
(x+a)^{2n}-(x-a)^{2n}
0%
\dfrac{(x+a)^{2n}+(x-a)^{2n}}{2}
0%
(x+a)^{2n}+(x-a)^{2n}
The sum of the series
\dfrac {2\left(\dfrac {n}{2}\right)!\left(\dfrac {n}{2}\right)!}{(n!)}[C^{2}_{0}-2C^{2}_{1}+3C^{2}_{2}...+(-1)^{n}(n+1)C^{2}_{n}]
where
n
is an even positive integer, is equal to
Report Question
0%
0
0%
(-1)^{\dfrac {n}{2}}(n+1)
0%
(-1)^{\dfrac {n}{2}}(n+2)
0%
(-1)^{n}n
Integral part of
(8+3\sqrt{7})^{n}
is
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0%
an even number
0%
an odd number
0%
an even or odd number depending upon the value of n
0%
nothing can be said
If
A
and
b
are coefficients of
( 1 + x ) ^ { 2 }
and
( 1 + x ) ^ { 2 n - 1 }
respectively, then
Report Question
0%
d = B
0%
A = 2 B
0%
2 A = B
0%
A + B = 0
The total number of terms in the expansion of
( x + a ) ^ { 200 } + ( x - a ) ^ { 200 }
after simplification is
Report Question
0%
101
0%
102
0%
201
0%
202
The sum of the co-efficients of all odd degree terms in the expansion of
{ \left( x+\sqrt { { x }^{ 3 }-1 } \right) }^{ 5 }+{ \left( x-\sqrt { { x }^{ 3 }-1 } \right) }^{ 5 }
(x> 1)
is:
Report Question
0%
2
0%
-1
0%
0
0%
1
Explanation
Sum of the coefficient of odd term is given by
= 2\left[_{ }^{ 5 }{ { C }_{ 0 }^{ } } + {_{ }^{ 5}}{ { C }_{ 2 }^{ } } + {_{ }^{ 5 }}{ { C }_{ 4 }^{ } }\right]
=2\left[1+10-10+5-10+5\right]
=2(1+5+5-10)
=2
.
Sum of the coefficient of integral powers of x in
{\left(1-2\sqrt{x}\right)}^{50}
is
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0%
\dfrac{{3}^{50}+1}{2}
0%
\dfrac{{3}^{50}}{2}
0%
{2}^{49}-1
0%
{2}^{49}+1
Find the middle terms(s) in the expansion of
\left ( 3x-\dfrac{2}{x^{2}} \right )^{15}
.
Report Question
0%
\dfrac {-6435\times 3^7\times 2^7}{x^6}, \dfrac {6437\times 3^7 \times 2^8}{x^9}
0%
\dfrac {-6435\times 3^8\times 2^7}{x^6}, \dfrac {6437\times 3^7 \times 2^8}{x^9}
0%
\dfrac {-6435\times 3^8\times 2^7}{x^6}, \dfrac {6437\times 3^7 \times 2^7}{x^9}
0%
\dfrac {-6435\times 3^8\times 2^7}{x^6}, \dfrac {6437\times 3^8 \times 2^8}{x^9}
Sum of coefficients of
{ x }^{ 2r }
,
r= 1,2,3,
....... in
(1+x{ ) }^{ n }
is
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0%
({ 2 }^{ n-1 }-1)
0%
({ 2 }^{ n-1 }+1)
0%
({ 2 }^{ n-2 }+1)
0%
({ 2 }^{ n-2 }-1)
The sum of coefficients of integral powers of
x
in the binomial expansion of
(1-2\sqrt x)^{50}
is
Report Question
0%
\frac{1}{2}(3^{50}-1)
0%
\frac{1}{2}(2^{50}+1)
0%
\frac{1}{2}(3^{50}+1)
0%
\frac{1}{2}(3^{50})
Coefficient of
x^{n}
in expansion of
\dfrac{(1+2x)^{2}}{(1-x)^{3}}
is
Report Question
0%
2n
0%
\dfrac{3}{2}(3n^{2}+n)
0%
n^{2}+n-1
0%
None\ of\ these
Coefficient of
x^{r}
in the expansion of
(1-2x)^{-1/2}
is
Report Question
0%
\dfrac{(2r)!}{(r!)^{2}}
0%
\dfrac{(2r)!}{2^{r}(r!)^{2}}
0%
\dfrac{(2r)!}{(r!){2^{2r}}}
0%
\dfrac{(2r)!}{2^{r}(r+1)!(r+1)}
Explanation
Coefficient of
{x^r}
in the expansion of
{\left( {1 - 2x} \right)^{ - \frac{1}{2}}}
by using binomial expansion
\begin{array}{l} \Rightarrow \frac { { \left[ { \frac { 1 }{ 2 } \left( { \frac { 1 }{ 2 } +1 } \right) \left( { \frac { 1 }{ 2 } +2 } \right) ...........\left( { \frac { 1 }{ 2 } +r-1 } \right) \left( { 2x } \right) } \right] } }{ { \left( { r! } \right) } } \\ \Rightarrow \frac { { \frac { 1 }{ 2 } .\frac { 3 }{ 2 } .\frac { 5 }{ 2 } .........\left( { r-\frac { 1 }{ 2 } } \right) { 2^{ r } }{ x^{ r } } } }{ { \left( { r! } \right) } } =\frac { { 2r! } }{ { { { \left( { r! } \right) }^{ 2 } } } } \end{array}
The coefficient of
x^{99}
in
(x+1)(x+3)(x+5).....(x+199)
is
Report Question
0%
1+2+3+...+99
0%
1+3+5+...+199
0%
1.3.5............199
0%
None\ of\ these
Explanation
\text x^{99} \text { in }(x+1)(x+3)(x+5) \ldots(x+199) \\
\text { Multiplying }(x+1)(x+3) \\
\text { will give } x^{2}+(1+3) x+(1)(3) \\
\text { so coefficient } x \text { is }(1+3) \\
\Rightarrow \operatorname{2n}(x+1)(x+3)(x+5)
So coefficient of
x^{2}=(1+3+5)
So coefficient of
x^{n}
in
(n+1)
terms will be sum of coefficients
So we can say it will be sum of roots of equation
So coefficient of
x^{99}
will be roots in 100 terms
1+3+5+\ldots+199
The sum of coefficient of integral powers of
x
in the binomial expansions
{\left( {1 - 2\sqrt x } \right)^{50}}
is:
Report Question
0%
\frac{1}{2}\left( {{3^{50}} - 1} \right)
0%
\frac{1}{2}\left( {{2^{50}} + 1} \right)
0%
\frac{1}{2}\left( {{3^{50}} + 1} \right)
0%
\frac{1}{2}\left( {{3^{50}}} \right)
The middle term in the expansion of
{\left(3x-\dfrac{{x}^{3}}{6}\right)}^{9}
is
Report Question
0%
\dfrac{21}{16}{x}^{19}
0%
\dfrac{-21}{16}{x}^{19}
0%
\dfrac{21}{16}{x}^{-19}
0%
\dfrac{-21}{16}{x}^{-19}
The sum of the co-efficient of all odd degree terms in the expansion of
\left( x + \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } + \left( x - \sqrt { x ^ { 3 } - 1 } \right)
Report Question
0%
0
0%
1
0%
2
0%
- 1
Sum of the series
)^{100}C_1)^2 + 2(^{100}C_2)^2 + 3(^{100}C_3)^2+.......+100(^{100}C_{100})^2
equals
Report Question
0%
\frac{2^{99}[1.3.5 ... ... (199)]}{99!}
0%
100. ^{200}C_{100}
0%
50. ^{200}C_{100}
0%
100. ^{199}C_{99}
The sum of the series
^{2020}C_0- ^{2020}C_1+^{2020}C_2-^{2020}C_3+.....+^{2020}C_{1010}
is
Report Question
0%
\dfrac{1}{2}^{2020}C_{1010}
0%
^{2020}C_{1010}
0%
Zero
0%
\dfrac{-1}{2}^{2020}C_{1010}
Explanation
(1-x)^{m}={ }^{n} C_{0}-{ }^{n} C_{1} x+\ldots+{ }^{n} C_{n} x^{n}\\
\text { Put } x=1 \text { and } n=2020\\
0={ }^{2020} C_{0}-{ }^{2020} C_{1}+{ }^{2020} C_{2}+\ldots { }^{2020} C_{2020}\\
\text { also }^{n} C_{r}={ }^{n} C_{n-r}\\
\text { }^{2020} C_{2020}={ }^{2020} \mathrm{Co}\\
\left[2\left[{ }^{2020} \mathrm{C}_{0}-{ }^{2020} \mathrm{C}_{1}+\ldots\right]+{ }^{2020} \mathrm{C}_{1010}\right]=0\\
\text { So } { }^{2020} \mathrm{C}_{0}-{ }^{2020} \mathrm{C}_{1}+\ldots-{ }^{2020} \mathrm{C}_{1009}=\\
\frac{-1}{2}^{2020} C_{1010}
The value of
\dfrac{1}{12!}+\dfrac{1}{10!2!}+\dfrac{1}{8!4!}+...+\dfrac{1}{12!}
Report Question
0%
\dfrac{{2}^{12}}{12!}
0%
\dfrac{{2}^{11}}{12!}
0%
\dfrac{{2}^{11}}{11!}
0%
None of these
If
{ \left( 1+x+2{ x }^{ 2 } \right) }^{ 20 }=0+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+......+{ a }_{ 40 }{ x }^{ 40 }
then
{ a }_{ 1 }+{ a }_{ 3 }+{ a }_{ 5\quad }+......+{ a }_{ 37\quad }
equals -
Report Question
0%
{ 2 }^{ 19}({ 2 }^{ 20 }-21)
0%
{ 2 }^{ 20 }({ 2 }^{ 19 }-19)
0%
{ 2 }^{ 19 }({ 2 }^{ 20 }+21)
0%
None of these
The greatest terms of the expansion
(2x+5y)^{13}
when
x=10
,
y=2
is?
Report Question
0%
^{13}C_5\cdot 20^8\cdot 10^5
0%
^{13}C_6\cdot 20^7\cdot 10^4
0%
^{13}C_4\cdot 20^9\cdot 10^4
0%
None of these
The coefficient of x
^9
in (x - 1) (x - 4) (x - 9)........(x - 100) is
Report Question
0%
-235
0%
235
0%
385
0%
None of these
Find the value of
\dfrac{1}{\left(n-1\right)!}+\dfrac{1}{\left(n-3\right)!3!}+\dfrac{1}{\left(n-5\right)!5!}+...
Report Question
0%
\dfrac{{2}^{n-1}}{(n-1)!}
0%
\dfrac{{2}^{n}}{n!}
0%
\dfrac{{2}^{n-1}}{n!}
0%
None of these
The constant term in the expansion of
{ (1+x) }^{ n }{ (1+\frac { 1 }{ x } ) }^{ n }
is
Report Question
0%
{ C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }+......+(n+1){ C }_{ n }^{ 2 }
0%
{ ({ C }_{ 0 }+{ C }_{ 1 }+.....+{ C }_{ n }) }^{ 2 }
0%
{ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+.....+{ C }_{ n }^{ 2 }
0%
None of these
The largest coefficient in the expansion of
\left(4+3x\right)^{25}
is
Report Question
0%
^{25}C_{11}3^{25}\left(\dfrac{4}{3}\right)^{14}
0%
^{25}C_{11}4^{25}\left(\dfrac{3}{4}\right)^{11}
0%
^{25}C_{14}4^{14}3^{11}
0%
^{25}C_{14}4^{11}.3^{14}
Explanation
=(4+3 x)^{25}\\
\left({}^{25} c_{0}(4)+{ }^{25} c_{1}(4)^{2}(3 x)+\cdots^{25} c_{25}\right.\\
(4+3 x)^{25}\\
\text { Let Tr the term is largest }\\
\therefore \quad T_{r-1}<T_{r}>T_{r+1}\\
=\frac{T_{2}}{T_{r_{-1}}}>1 \quad \frac{T_{r+1}}{T_{r}}<1
T_{r+1^{}}= {}^{25} C_{r}(4)^{25-r}(3 x)^{k}
T_{r-1}=25_{C_{r-2}}(4)^{25-r+2}(3 x)^{r-2}
T_{r}={ }^{25} C_{r-1}\left(4\right)^{26-r}(3 x)^{r-1}
\therefore \frac{25 C_{r-1}(4)^{26-r}(3 x)^{r-1}}{25 C_{r-2}(4)^{27-r}(3 x)^{r-2}}>1
\frac{(25-r)}{(x-1)} \times \frac{3}{4}>1
{}75-3 r > 4 r -4\\
79>7 r\\
11.1>r
\Rightarrow r=12
largest term =
\left({ }^{25}c_{11}(3)^{25}\left(\frac{4}{3}\right)^{14}\right.)
The sum of the co-efficient of all odd degree terms in the expansion of
\left( x + \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } + \left( x - \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } , ( x > 1 )
is :
Report Question
0%
- 1
0%
1
0%
0
0%
2
If the last term in the binomial expansion of
\left(2^{1/3}-\dfrac {1}{\sqrt {2}}\right)^{n}
is
\left(\dfrac {1}{3^{5/3}}\right)^{\log_{3}8}
, then the
5^{th}
terms form the beginning is:
Report Question
0%
210
0%
420
0%
103
0%
None\ of\ these
Coefficient of
x^{25}
in
(1+x+x^{2}+x^{3}+....+x^{10})^{7}
is
Report Question
0%
^{31}C_{15}-7.^{20}C_{14}
0%
^{31}C_{14}-7.^{20}C_{14}
0%
31
0%
None\ of\ these
The coefficient of
x^{8}
in
(1+2x^{2}-x^{3})^{9}
is
Report Question
0%
1680
0%
2140
0%
2520
0%
2730
The coefficients of
x^{10}
in the expansion of
(1+x)^{15}+(1+x)^{16}+(1+x)^{17}+....+(1+x)^{30}
is
Report Question
0%
^{31}C_{10}-^{15}C_{10}
0%
^{31}C_{11}-^{15}C_{11}
0%
^{30}C_{10}-^{15}C_{10}
0%
^{31}C_{10}-^{14}C_{11}
The sum of the coefficient in the expansion of
(a+2b+c)^{11}
is-
Report Question
0%
4^{11}
0%
32
0%
31
0%
None\ of\ these
State true or false.
The general term for
3,7,13,21,31,43
........ is
n^2−(n−1),n=1,2,3,...
Report Question
0%
True
0%
False
The coefficient of
x^{3}
in the expansion of
(1+2x+3x^{2})^{10}
is
Report Question
0%
Less than
200
0%
Less than
400
but greater than
200
0%
1400
0%
1500
The coefficient of
{x}^{n}
in the expansion of
\dfrac { 1 }{ \left( 1-x \right) \left( 1-2x \right) \left( 1-3x \right) }
is
Report Question
0%
\dfrac { 1 }{ 2 } \left( { 2 }^{ n+2 }-{ 3 }^{ n+3 }+1 \right)
0%
\dfrac { 1 }{ 2 } \left( { 2 }^{ n+2 }-{ 2 }^{ n+3 }+1 \right)
0%
\dfrac { 1 }{ 2 } \left( { 2 }^{ n+2 }-{ 3 }^{ n+2 }+1 \right)
0%
None\ of\ these
Explanation
\text { We know that } \\
\frac{1}{1-a} \text { is sum of } \infty GP=1+a+a^{2}+a^{3} \ldots\infty \\
\text { Similary: }
\frac{1}{(1-x)(1-2 x)(1-3 x)}={\left(1+x+x^{2}+\ldots\infty\right)\left(1+(2 x)+(2 x)^{2}\ldots\infty\right)}\\
{\left(1+(3 x)+(3 x)^{2}+(3 x)^{3}\ldots\infty\right.)}
From this equation, we observe that
coefficient of
x^{n}
is
: \rightarrow
\frac{1}{2}\left[2^{n+2}-3^{n+3}+1\right]
Option (a)
The number of rational terms in the expansion of
{ \left( 1+\sqrt { 2 } +\sqrt [ 3 ]{ 3 } \right) }^{ 6 }
is
Report Question
0%
6
0%
7
0%
5
0%
8
The coefficient of
x^4
in the expansion of
(1+x+x^2+x^3)^{11}
is
Report Question
0%
990
0%
495
0%
330
0%
none of these
Explanation
=\left(1+x+x^{2}+x^{3}\right)^{11}
=\left((1+x)\left(1+x^{2}\right)\right)^{11}
=(1+x)^{ 11}\left[1+^{11} c_{1} x^{2}+{ }^{11} c_{2} x^{4}+\cdots^{11} c_{11}\left(x^{2}\right)^{11}\right.]
=\left[1+{ }^{11} c_{1} x+{ }^{11} c_{2} x^{2}+\cdots\right]\left[1+{ }^{11} c_{1} x^{2}+{ }^{11} c_{2} x^{4}+\cdots\right]
={ }^{11} c_{0} \cdot{ }^{11} c_{2} \cdot x^{4}+{ }^{11} c_{2} \cdot{ }^{11} c_{1} \cdot x^{4}+{ }^{11} c_{4}{ }^{11} c_{0} x^{4}
=\quad 990 x^{4}
Option
(A)
The coefficient of x
^{24}
in the expansion of
(1 +3x + 6x
^2
+ 10x
^3
+ -----------+
\infty
)
^{2/3}
=
Report Question
0%
300
0%
250
0%
25
0%
205
Explanation
\left(1+3 x+6 x^{2}+10 x^{3}+\ldots\infty\right)^{\frac{2}{3}} \\
\text { We know that, } \\
{ (1+\alpha) }^{n}=1+n \alpha+\frac{n(n-1) \alpha^{2}}{2 !}+\ldots \\
\text { comparing with given eqn, }
\frac{n(n-1)}{2} \cdot \alpha^{2}=6 x^{2} \\
=n(n-1) \frac{9 x^{2}}{n^{2}}=6 x^{2} \\
\Rightarrow n=-3
n \alpha=3 x
\Rightarrow\alpha=\frac{3 x}{n}
\alpha=-x
\therefore \text {Expansion is} (1-x)^{-\not 3 \times \frac{2}{\not3}}
\quad =(1-x)^{-2}
=1+2 x+\frac{2 \times 3}{2 !} x^{2}+\frac{2 \times 3 \times 4}{3 !} x^{3}+\ldots
\therefore
coeff. of
x^{24}
is 2
= \frac{2 \times 3 \times 4 \times\ldots25}{24 !}
=25
The co-efficient of
x^{k}
in expansion of
1+\left(1+x\right)+\left(1+x\right)^{2}++\left(1+x\right)^{n}
is :
\left(n>k\right)
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0%
^{n}C_{k}
0%
^{n+1}C_{k}
0%
^{n+1}C_{k+1}
0%
None\ of\ these
The number of terms in the expansion of
{ \left[ { a }^{ 3 }+\dfrac { 1 }{ { a }^{ 3 } } +1 \right] }^{ 100 }
is
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0%
201
0%
300
0%
200
0%
^{ 100 }{ C }_{ 3 }
Explanation
\left(a^{3}+\frac{1}{a^{3}}+1\right)^{100} \\
\frac{1}{a^{300}}\left(1+a^{3}+a^{6}\right)^{100}
Here, terms will be in the form,
a^{0} ; a^{3}, a^{6}\cdots
and highest power term will be
\left(a^{6}\right)^{100}
This forms a A.P of powers,
\therefore \quad 600=0+(n-1) 3
\therefore \quad n=201 \\
\text { Option (a) }
For
x\in R, x\neq -1
if
(1+x)^{2016}+x(1+x)^{2015}+x(1+x)^{2014}+.+x^{2016}=\sum _{ i=0 }^{ 2016 }{ { a }_{ i }{ x }^{ i } }
, then
a_{17}
is equal to
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0%
\dfrac{2017!}{ 17!2000!}
0%
\dfrac{2016!}{ 17!1999!}
0%
\dfrac{2017!}{2000!}
0%
\dfrac{2016!}{ 16!}
The middle term in the expansion of
\left(1-\dfrac{1}{x}\right)^{n}\left(1-x\right)^{n}
is
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0%
^{2n}C_{n}
0%
-^{2n}C_{n}
0%
-^{2n}C_{n-1}
0%
none\ of\ these
If the middle term in the expansion of
( 1 + x ) ^ { 2 n }
is the greatest term, then
x
lies in the interval ___________________.
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0%
\left( \frac { n } { n + 1 } , \frac { n + 1 } { n } \right)
0%
\left( \frac { n + 1 } { n } , \frac { n } { n + 1 } \right)
0%
( n - 2 , n )
0%
( n - 1 , n )
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