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CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 15 - MCQExams.com
CBSE
Class 11 Engineering Maths
Binomial Theorem
Quiz 15
Co-efficient of
x
n
−
1
in the expansion of,
(
x
+
3
)
n
+
(
x
+
3
)
n
−
1
(
x
+
2
)
+
(
x
+
3
)
n
−
2
(
x
+
2
)
2
+
.
.
.
+
(
x
+
2
)
n
is:
Report Question
0%
n
+
1
C
2
(
3
)
0%
n
−
1
C
2
(
5
)
0%
n
+
1
C
2
(
5
)
0%
n
C
2
(
5
)
In the expansion of
(
x
+
y
+
z
)
10
Report Question
0%
the coefficient of
x
2
y
2
z
3
is 0
0%
the coefficient of
x
8
y
z
is 90
0%
the number of terms is 66
0%
none of these
If in the expansion of
(
2
x
+
1
4
x
)
n
,
T
3
=
7
T
2
and sum of the binomial coefficients of second
and third terms is
36
,
then the value of
x
is -
Report Question
0%
−
1
/
3
0%
−
1
/
2
0%
1
/
3
0%
1
/
2
The first three terms in the expansion of
(
1
−
a
x
)
n
, where
n
is positive integer are,
1
−
4
x
+
7
x
2
, then the value of
a
Report Question
0%
4
0%
1
4
0%
8
0%
1
2
The value of x for which the sixth term in the expansion of
[
2
log
2
√
9
x
−
1
+
7
+
1
2
1
5
log
2
(
3
x
−
1
+
1
)
]
7
is equal to 84 is
Report Question
0%
4
0%
1 or 2
0%
0 or 1
0%
3.
Explanation
Here we have,
[
2
log
2
√
9
x
−
1
+
7
+
1
2
1
5
log
2
(
x
−
1
)
+
1
]
7
=
[
√
9
x
−
1
+
7
+
(
3
x
−
1
+
1
)
(
−
1
/
5
)
]
7
The sixth term in the expansion is given by,
T
6
=
7
C
5
(
√
9
x
−
1
+
7
)
2
⋅
{
(
3
x
−
1
+
1
)
(
−
1
/
5
)
}
5
⇒
84
=
21
⋅
(
9
x
−
1
+
7
)
⋅
(
3
x
−
1
+
1
)
−
1
⇒
84
21
=
9
x
−
1
+
7
3
x
−
1
+
1
⇒
4
⋅
(
3
x
−
1
+
1
)
=
(
3
x
−
1
)
2
+
7
⇒
(
3
x
−
1
)
2
−
4
×
3
x
−
1
+
3
=
0
⇒
(
3
x
−
1
−
1
)
(
3
x
−
1
−
3
)
=
0
⇒
3
x
−
1
=
1
or
3
⇒
3
x
−
1
=
3
0
or
3
1
⇒
x
−
1
=
0
or
1
⇒
x
=
1
,
2
Hence, option (B) is correct.
In the expansion of
(
1
+
)
50
,
the sum of the coefficient of odd powers of x is _______.
Report Question
0%
226
0%
249
0%
250
0%
251
Find the middle term(s) in the expansion of :
(
2
a
x
−
b
x
2
)
12
Report Question
0%
59136
a
6
b
6
x
6
0%
59163
a
5
b
5
x
5
0%
59631
a
7
b
7
x
7
0%
None of these
Explanation
Given that to find middle term in expansion of
(
20
x
−
b
x
2
)
12
n
=
12
⇒
even
for even number middle term
=
(
n
2
+
1
)
t
h
middle term
=
(
12
2
+
1
)
t
h
⇒
7
We know that
T
r
+
1
=
n
C
r
(
a
)
n
−
r
b
r
T
7
=
T
6
+
1
=
12
C
6
(
29
x
)
12
−
6
(
−
b
x
2
)
6
=
12
!
6
!
6
!
2
6
a
6
x
6
b
6
x
12
=
12
×
11
×
10
×
9
×
8
×
7
×
6
!
6
×
5
×
4
×
3
×
2
×
1
×
6
!
2
6
a
6
b
6
x
6
=
59136
a
6
b
6
x
6
The constant term in the expansion of
(
x
2
−
1
x
2
)
16
is
Report Question
0%
C
16
8
0%
C
16
7
0%
C
16
9
0%
C
16
10
Find the middle term(s) in the expansion of :
(
2
x
−
x
2
4
)
9
Report Question
0%
63
5
x
13
,
−
62
31
x
15
0%
63
4
x
13
,
−
63
32
x
14
0%
61
4
x
11
,
61
33
x
13
0%
None of these
Explanation
Given expansion term is
(
2
x
−
x
2
4
)
9
n
=
9
=
odd
middle term
=
(
9
+
1
2
)
t
h
+
(
9
+
3
2
)
t
h
=
5
t
h
or
6
t
h
term
T
5
=
T
4
+
1
=
9
C
4
(
2
x
)
5
(
−
x
2
4
)
4
=
9
!
4
!
5
!
2
5
×
x
5
(
−
1
)
4
×
x
8
4
4
=
9
×
8
×
7
×
6
×
5
!
4
×
3
×
2
×
5
!
×
2
5
.
x
13
.1
2
5
T
5
=
63
4
x
13
T
6
=
T
5
+
1
=
9
C
5
(
2
x
)
4
(
−
x
2
4
)
5
=
9
!
4
!
5
!
×
2
4
x
4
(
−
1
)
5
x
10
4
5
=
9
×
8
×
7
×
6
×
5
!
4
×
3
×
2
×
5
!
2
4
x
14
2
10
T
6
=
−
63
32
x
14
The coefficient of the middle term in the binomial expansion in powers of x of
(
1
+
α
x
)
4
and of
(
1
−
α
x
)
6
is the same if
α
equals :
Report Question
0%
3
5
0%
10
3
0%
−
3
10
0%
−
5
3
Explanation
Step 1: Calculating coefficient of individual term
The middle term of the binomial expansion
(
1
+
α
x
)
4
will be the
(
4
2
+
1
)
=
3rd term
The middle term of the binomial expansion
(
1
−
α
x
)
6
will be the
(
6
2
+
1
)
=
4th term
Coefficient of3rd term of
(
1
+
α
x
)
4
=
4
C
2
α
2
Coefficient of 4th term of
(
1
−
α
x
)
6
=
−
6
C
3
α
3
Step 2: Calculating
α
Equating both the coefficients
⟹
4
C
2
α
2
=
−
6
C
3
α
3
⟹
6
α
2
=
−
20
α
3
⟹
α
=
−
6
20
=
−
3
10
Hence, The value of
α
=
−
3
10
Find the coefficient of
x
7
in the expansion of
(
x
−
1
x
2
)
40
Report Question
0%
−
40
C
11
0%
−
40
C
10
0%
−
40
C
12
0%
None of these
Find the middle term in the expansion of
(
2
3
x
−
3
2
x
)
20
Report Question
0%
20
C
10
x
10
y
10
0%
20
C
11
x
11
y
11
0%
20
C
9
x
11
y
10
0%
None of these
Explanation
Given that,
(
2
x
3
−
3
y
2
)
20
n
=
20
⇒
even
For even values of
n
, middle term is
(
n
2
+
1
)
t
h
⇒
middle term
=
(
20
2
+
1
)
t
h
=
11
t
h
Now use the formula
T
r
+
1
=
n
C
r
(
a
)
n
−
r
(
b
)
r
Here,
r
=
10
,
a
=
2
x
3
,
b
=
3
y
2
T
11
=
T
10
+
1
=
20
C
10
(
2
x
3
)
20
−
10
(
3
y
2
)
10
T
11
=
20
C
10
(
2
x
3
)
10
(
3
y
x
)
10
T
11
=
20
C
10
x
10
y
10
Find the middle term(s) in the expansion of :
(
p
x
+
x
p
)
9
Report Question
0%
126
x
p
,
126
p
x
0%
126
p
,
126
x
0%
160
x
2
p
,
162
x
p
2
0%
None of these
Explanation
Given that to find middle term in expansion of
(
P
x
+
x
P
)
9
n
=
9
⇒
odd
For off values there are two middle terms
First middle term
=
(
n
+
1
2
)
t
h
term
⇒
(
9
+
1
2
)
t
h
term
=
5
t
h
term
second middle term
=
(
n
+
1
2
+
1
)
t
h
term
=
(
9
+
1
2
+
1
)
t
h
term
=
6
t
h
term
We know that
T
r
+
1
=
9
C
r
(
P
x
)
9
−
r
(
x
P
)
r
T
r
+
1
=
T
5
⇒
r
+
1
=
5
⇒
r
=
4
T
4
+
1
=
9
C
4
(
P
x
)
5
(
x
P
)
4
=
9
!
4
!
5
!
P
5
x
5
⇒
126
P
x
T
r
+
1
=
T
0
⇒
r
+
1
=
6
⇒
r
=
5
T
5
+
1
=
9
C
5
(
P
x
)
4
(
x
P
)
5
=
9
!
4
!
5
!
P
4
x
4
r
5
P
5
=
126
x
P
Find the middle term(s) in the expansion of :
(
3
x
−
x
3
6
)
9
Report Question
0%
189
8
x
15
,
−
21
16
x
17
0%
189
8
x
17
,
−
21
16
x
19
0%
189
7
x
15
,
−
23
13
x
19
0%
None of these
Explanation
Given that to find middle term in expansion of
(
3
x
−
x
3
6
)
9
n
=
9
⇒
odd
For odd values, there are two middle terms.
First middle term
=
n
+
1
2
t
h
term :
⇒
(
9
+
1
2
)
t
h
term
=
5
t
h
term
Second middle term
=
(
n
+
1
2
+
1
)
t
h
term;
⇒
(
9
+
1
2
+
1
)
t
h
term
=
6
t
h
term
We know that
T
r
+
1
=
9
C
r
(
3
x
)
r
(
−
x
3
6
)
r
=
9
C
r
(
3
r
)
x
r
(
−
1
)
x
3
r
(
1
6
)
r
T
r
+
1
=
T
5
⇒
r
+
1
=
5
⇒
r
=
4
T
4
+
1
=
9
C
4
(
3
x
)
9
−
4
(
−
x
3
6
)
4
T
5
+
1
=
9
C
5
(
3
x
)
9
−
5
(
−
x
3
6
)
5
=
9
!
5
!
4
!
3
5
x
5
x
12
6
4
=
9
!
4
!
5
!
3
4
x
4
x
15
6
5
=
126
3
5
x
12
×
x
5
64
=
−
21
16
x
19
=
189
7
x
17
Find the middle term in the expansion of :
(
x
2
−
2
x
)
10
Report Question
0%
−
8604
x
7
0%
−
8064
x
5
0%
−
804
x
4
0%
None of these
Explanation
Given to find middle term in the expansion of
(
x
2
−
2
x
)
10
n
=
10
⇒
even
for even numbers middle term is
(
n
2
+
1
)
t
h
middle term
=
(
10
2
+
1
)
t
h
=
6
t
h
We know that
T
r
+
1
=
n
C
r
(
a
)
n
−
r
(
b
)
r
T
6
=
T
5
+
1
=
10
C
5
(
x
2
)
5
(
−
2
x
)
5
=
−
10
!
5
!
5
!
x
10
2
5
x
5
T
6
=
−
8064
x
5
In the expansion of
(
1
+
a
x
+
b
x
2
)
(
1
−
3
x
)
15
, if coefficient of
x
2
is
0
then order pair
(
a
,
b
)
is equal to
Report Question
0%
(
28
,
325
)
0%
(
18
,
315
)
0%
(
28
,
315
)
0%
(
18
,
325
)
Explanation
Coefficient of
x
2
in
(
1
+
a
x
+
b
x
2
)
(
1
−
3
x
)
15
=
15
C
2
(
−
3
)
2
+
a
.
15
C
1
(
−
3
)
+
b
.
15
C
0
=
0
15.14
2
.
9
−
3.
a
.15
+
b
=
0
15
×
63
−
45
a
+
b
=
0
....(1)
Coefficient of
x
3
in
(
1
+
a
x
+
b
x
2
)
(
1
−
3
x
)
15
15
C
3
(
−
3
)
3
+
a
.
15
C
2
(
−
3
)
2
+
b
.
15
C
1
(
−
3
)
=
0
=
15.14.13
3
×
2
.
3
2
−
a
.3
.
15.14
2
+
15.
b
=
0
7.13.3
−
21
a
+
b
=
0
...(2)
by using (1) - (2)
672
−
24
a
=
0
⇒
a
=
28
Hence
b
=
315
The number of terms in the expansion of
{
(
2
x
+
3
y
)
9
+
(
2
x
−
3
y
)
9
}
is
Report Question
0%
10
0%
8
0%
4
0%
5
Explanation
.
Step 1: Find number of terms of expansion
Given equation is -
{
(
2
x
+
3
y
)
9
+
(
2
x
−
3
y
)
9
}
We know that for expansion of (x + y)
n
+ (x – y)
n
⇒
If n is odd then, number of terms =
n
+
1
2
Comparing with standard equation,n=9
∴
\dfrac{9+1}{2}
=\dfrac{10}{2}
=5
\textbf{Therefore,the expansion of}
\mathbf{ \left\{ ( 2x +3y)^9 + ( 2x - 3y)^9 \right\}}
\textbf{has 5 terms.}
The number of terms in the expansion of
\left\{( x+a)^{16} + (x -a)^{16} \right\}
is
Report Question
0%
7
0%
8
0%
9
0%
17
Explanation
\textbf{Step 1: Find number of terms of expansion}
\text{Given equation is -}
({x} + {16})^{16} +( {x} - {16})^{16}
\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n -$}
\text{Comparing with standard equation,n=16}
\Rightarrow \text{If n is even then, Number of terms = $\dfrac{n}{2} $+ 1}
\therefore \text{Numer of terms=}
\dfrac{16}{2} +1
=8+1
=9
\textbf{Therefore,the expansion of $\mathbf{ ({x} + {a})^{16} +( {x} - {a})^{16} }$ has 9 terms.}
The numbers of terms in the expansion of
( 3x +y)^{10}
is
Report Question
0%
9
0%
11
0%
10
0%
8
Explanation
\textbf{Step 1: Expand the Expression to find number of terms}
\text{Given-Equation is } (3x+y)^{10}
\text{Binomial Theorem is Given by-}
(b+a)^n=b^n+C_{1}^n(b)^{n-1}a+C_{2}^n(b)^{n-2}a^2+C_{3}^n(b)^{n-3}a^3+..............+C_{n}^n(a)^n
\text{Comparing with standard formula, n=10, a=3x and b=y}
(3x+y)^{10}=(3x)^{10}+C_{1}^{10}(3x)^{10-1}(y)+C_{2}^{10}(3x)^{10-2}(y)^2+C_{3}^{10}(2x)^{10-3}(3y)^3+.....+C_{10}^{10}(y)^{10}
\text{From expansion, total number of terms=n+1}
\therefore \text{Total number of terms= 10+1}
=11
\textbf{Therefore, total number of terms are 11}
\{ C_{ 0 }+3C_{ 1 }+5C_{ 2 }+...+(2n+1)C_{ n }\} =?
Report Question
0%
(n+1)2^{ n }
0%
(n+1)2^{ n-1 }
0%
(n-1)(n+2)
0%
none of these
Explanation
\textbf{Step 1: Find the value of given expression}
\text{Let X=} C_{ 0 }+3C_{ 1 }+5C_{ 2 }+...+(2n+1)C_{ n }
\Rightarrow X=(C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n })+2(C_{ 1 }+2C_{ 2 }+3C_{ 3 }+...+nC_{ n })
\text{In binomial expansion,sum of coefficient= } 2^n
\therefore (C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n }) = 2^n
(1)
\Rightarrow X=(C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n })+2(C_{ 1 }+2C_{ 2 }+3C_{ 3 }+...+nC_{ n })
=2^n +2\times\bigg(n+2\dfrac{n(n-1)}{2}+3\dfrac{n(n-1)(n-2)}{3\times 2}+........+n.1\bigg)
\textbf{[From equation 1]}
=2^n +n\bigg(1+(n-1)+\dfrac{(n-1)(n-2)}{ 2}+........+1\bigg)
=2^n+2n[ ^{n-1}C_{ 0 }+^{n-1}C_{ 1 }+3C_{ 2 }+......+^{n-1}C_{ n-1 }]
=2^{ n }+(2n).2^{ n-1 }
[\because \mathbf{ C_{ 0 }+C_{ 1 }+C_{ 2 }+...+C_{ n-1 } = 2^{n-1}}]
=2^{ n }+(n).2^{ n-1+1 }
=2^{ n }+(n).2^{ n }
=(n+1).2^{ n }
\textbf{Hence, value of given expression is }
\mathbf{(n+1).2^{ n } }
The sum of coefficients of the two middle terms in the expansion of
(1 + x)^{2n - 1}
is equal to
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0%
^{2n -1}C_n
0%
^{2n -1}C_{n+1}
0%
^{2n}C_{n-1}
0%
^{2n}C_n
The cofficient of
x^{32}
in the expansion of
( x^4 - \frac {1}{x^3})^{15}
is
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0%
273
0%
546
0%
1365
0%
1092
Explanation
\textbf{Step 1: Find value of general term for expansion}
\text{Given equation is}
\bigg ( x^4 -\dfrac {1}{x^3}\bigg)^{15}
\text{General term of expansion $ (p+q)^n$ is given by-}
T_n= T_{r +1 }=^nC_r p^{ (n-r)}.q^r
\text{Comparing given equation with standard form, p=x$^4$and q=-1/x$^3$ and n=15}
\Rightarrow T_n=T_{r+1}
= ^{15}C_r (x^4)^{(15-r)} .\bigg (\dfrac {-1}{x^3}\bigg)^r
= (-1)^r. ^{15}C_r. x^{(60-4r)} .\bigg (\dfrac {1}{x^3}\bigg)^r
=(-1)^r . ^{15}C_r x^{(60 -4r -3r)}
=(-1)^r . ^{15}C_r x^{(60 -7r)}
\textbf{Step 2: Find value of coefficient of term x}^{32}
\text{To obtain term of x}^{32}
\Rightarrow 60-7r=32
\Rightarrow
7r=28
\therefore
r=4
\text{For r=4,5th term is given by-}
T_5=T_{4+1} = (-1)^4.^{15}C_4 x^{(60-7 \times 4)} = ^{15}C_4 x^{32} .
\therefore
\text{Coefficient of } x^{32}
=^{15}C_4
= \dfrac { 15 \times 14 \times 13 \times 12 }{ 4 \times 3 \times 2\times 1}
= 1365
\textbf{Therefore,value of coefficient of term x$^{32}$ is 1365}
The number of terms in the expansion of
\displaystyle{(\sqrt{3} + ^4 \sqrt{5})^{124}}
which are integers, is equal to
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0%
nil
0%
30
0%
31
0%
32
The coefficient of
x^3
in the expansion of
(1+2x)^6(1-x)^7
is
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0%
43
0%
-43
0%
63
0%
-63
Explanation
\textbf{Step 1: Expand the Expression }
\text{Binomial Theorem is Given by-}
(1+p)^n=1+^nC_{1}(p)^1+^nC_{2}(p)^2+^nC_{3}(p)^3+..............+^nC_{n}(p)^n
\text{For expansion of} (1+2x)^6:
\text{By Comparing with Formula p=2x and n=6}
(1+2x)^6 =\{ 1+^{ 6 }C_{ 1 }.2x+^{ 6 }C_{ 2 }.(2x)^{ 2 }+^{ 6 }C_{ 3 }.(2x)^{ 3 }+...\}
\text{For expansion of} (1-x)^7 :
\text{By Comparing with Formula p=x and n=7}
(1-x)^7 = \{ 1-^{ 7 }C_{ 1 }.x+^{ 7 }C_{ 2 }.x^{ 2 }-^{ 7 }C_{ 3 }.x^{ 3 }+..........\}
\textbf{Step 2: Find the coefficient of } \mathbf{ x^3}
\text{Let P}= (1+2x)^6(1-x)^7
\Rightarrow P =\{ 1+^{ 6 }C_{ 1 }.2x+^{ 6 }C_{ 2 }.(2x)^{ 2 }+^{ 6 }C_{ 3 }.(2x)^{ 3 }+...\} \times \{ 1-^{ 7 }C_{ 1 }.x+^{ 7 }C_{ 2 }.x^{ 2 }-^{ 7 }C_{ 3 }.x^{ 3 }+..\}
\therefore \text{Coefficient of }x^3 = -^{7}C_3 + ^{7}C_2 \times 2 ^{6}C_1 + 4\times ^{6}C_2 \times (- ^{7}C_1 )+8\times ^{7}C_3
=1\times(-35)+(12\times21)+60 \times (-7) + (160 \times 1)
= (-35 +252-420+160)
=-43
\textbf{Therefore, coefficient of } \mathbf{ x^3 } \textbf{is equal to -43}
The number of terms in the expansion of
( 1 + \sqrt [3]{2}x)^9 + ( 1 - \sqrt [3]{2}x )^9
is
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0%
5
0%
7
0%
9
0%
10
Explanation
\textbf{Step 1: Find number of terms of expansion}
\text{Given equation is -}
( 1 + \sqrt [3]{2x})^9 + ( 1 + \sqrt [3]{2x} )^9
\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n$}
\text{Comparing with standard equation,n=9}
\Rightarrow \text{ If n is even then, Number of terms = }
\dfrac{n+1}{2}
\therefore \text{Numer of terms=}
\dfrac{9+1}{2}
=\dfrac{10}{2}
=5
\textbf{Therefore,the expansion of}
\mathbf{ ( 1 + \sqrt [3]{2x})^9 + ( 1 + \sqrt [3]{2x} )^9 }
\textbf{has 5 terms.}
If p and q are positive integers, then the coefficients of
x^p
and
x^q
in the expansion of
( 1+ x)^{p+q}
are
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0%
euqal
0%
equal with opposite signs
0%
reciprocal to each other
0%
none of these
Explanation
\textbf{Step 1: Find value of (p+1)th and (q+1)th term from expansion}
\text{General term of expansion $ (1+x)^n$ is given by-}
T_n= T_{r +1 }=^nC_r x^{ (n-r)}
\text{Comparing given equation with standard form, n=p+q}
\Rightarrow
T_{p+1} = ^{(p+q)}C_p . x^{p+1-1}
= ^{(p+q)}C_p . x^{p}
\therefore \text{Coefficient of x$^{p}$ is $ = ^{(p+q)}C_p $}
(1)
\Rightarrow T_{q+1} = ^{(p+q)}C_q. x^{q+1-1}
= ^{(p+q)}C_p . x^{q}
\therefore \text{Coefficient of x$^{q}$ is $ = ^{(p+q)}C_q $}
(2)
\Rightarrow ^{(p+q)}C_p = ^{(p+q)}C_{(p+q)-p}
\mathbf{[^{n}C_r = ^{n}C_{n-r}]}
=^{(p+q)}C_q
\therefore \text{Coefficient of x$^p$ and x$^q$ are same}
\textbf{[From equation (1) and (2)]}
\textbf{Therefore, coefficient of both terms are equal.}
The total number of terms in the expansion of
(a + y)^{100 } + ( a -y)^{100 }
is
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0%
50
0%
51
0%
101
0%
102
Explanation
\textbf{Step 1: Find number of terms of expansion}
\text{Given equation is -}
({a} + {y})^{100} +( {a} - {y})^{100}
\text{We know that for expansion of (x + y)$^n$ + (x – y)$^n -$ $\\$$\Rightarrow$ If n is even then, Number of terms = $\dfrac{n}{2} $+ 1}
\text{Comparing with standard equation,n=100}
\therefore \text{Numer of terms=}
\dfrac{100}{2} +1
=50+1
=51
\textbf{Therefore,the expansion of $\mathbf{ ({a} + {y})^{100} +( {a} - {y})^{100} }$ has 51 terms.}
If the coefficients of the second, third and fourth terms in the expansion of
(1+x)^{2n}
are in AP , then
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0%
n^2-3n+8=0
0%
2n^2 - 7n +5 = 0
0%
2n^2-9n+7 =0
0%
none of these
Explanation
\textbf{Step 1: Expand the Expression }
\text{Binomial Theorem is Given by-}
(1+x)^p=1+C_{1}^p(x)^1+C_{2}^p(x)^2+C_{3}^p(x)^3+..............+C_{p}^p(x)^p
\text{By Comparing with Formula p=2n}
\Rightarrow (1+x)^{-2n}
=C_{1}^{2n}(x)^1+C_{2}^{2n}(x)^2+C_{3}^{2n}(x)^3+..............+C_{2n}^{2n}(x)^{2n}
\textbf{Step 2: Find value of 2nd, 3rd and 4th term from expansion}
\text{General term of expansion$(1+x)^{2n}$ is given by-}
T_n= T_{r +1 }=^{2n}C_r x^{ r}
\Rightarrow T_2=T_{1+1}=^{2n}C_1x^1
\therefore \text{Coefficient of}
T_2 =^{2n}C_1x
=2n
\Rightarrow T_3=T_{2+1}=^{2n}C_2x^2
\therefore \text{Coefficient of}
T_3
=^{2n}C_2x^2
=\dfrac { 2n(2n-1) }{ 2 }
\Rightarrow T_4=T_{3+1} =^{2n}C_3x^3
\therefore \text{Coefficient of}
T_4
=^{2n}C_2x^3
=\dfrac { 2n(2n-1)(2n-2) }{ 6 }
\textbf{Step 3: Find condition for three terms to be in AP}
\text{Given-Coefficient of $T_2,T_3$ and $T_4$ are in AP }
\text{Condition for three terms a,b and c to be in AP- 2b=a+c }
\therefore 2 T_3=T_2 + T_4
2\bigg(\dfrac { 2n(2n-1) }{ 2 }\bigg) =2n+\dfrac { 2n(2n-1)(2n-2) }{ 6 }
\Rightarrow
3(2n-1)=3+2n^2-3n+1
\Rightarrow
6n-3=4+2n^2-3n
\therefore
2n^2 - 9n+7=0
.
\textbf{Therefore,option c is the correct answer}
The term independent of x in the expansion of
(x -\frac {1}{x})^{12}
is
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0%
924
0%
462
0%
231
0%
693
Explanation
\textbf{Step 1: Find value of general term}
\text{Given equation is}
\bigg ( x -\dfrac {1}{x}\bigg)^{12}
\text{General term of expansion $ (p+q)^n$ is given by-}
T_n= T_{r +1 }=^nC_r p^{ (n-r)}.q^r
\text{Comparing given equation with standard form, p=x and q=-1/x and n=12}
\Rightarrow T_n=T_{r+1} = ^{12}C_r. (x)^{(12-r)} .\bigg( \dfrac {-1}{x}\bigg)^r
= (-1)^r. ^{12}C_r. (x)^{(12-r)} .x^{-r}
= (-1)^r. ^{12}C_r. (x)^{(12-r-r)}
= (-1)^r. ^{12}C_r. (x)^{(12-2r)}
\textbf{Step 2: Find value of coefficient of term independent of x}
\text{To obtain term independent of x-:Coefficient of x will be 0}
\Rightarrow 12-2r=0
\Rightarrow
2r=12
\therefore
r=6
\text{For r=6,7th term is given by-}
T_7=T_{6+1} = (-1)^6.^{12}C_6 x^{(12-2x6)} = ^{12}C_6 x^0 .
\therefore
\text{Coefficient of } x^0
=^{12}C_6
= \dfrac { 12\times11\times10 \times 9 \times 8 \times 7}{6\times5 \times 4 \times 3 \times 2 \times 1}
= 924
\textbf{Therefore,value of coefficient of term independent of x will be 924}
The sum of possible values of
x
is
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0%
1
0%
3
0%
4
0%
None\ of \these
The middle term in the expansion of
(x/2+2)^{8}
is
1120
; then
x\in R
is equal to
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0%
-2
0%
3
0%
-3
0%
2
If the coefficients of
r^{th}
and
(r+1)^{th}
terms in the expansion of
(3+7x)^{29}
are equal, then
r
equals
Report Question
0%
15
0%
21
0%
14
0%
none of these
Explanation
Given expansion is
(3+7x)^{29}
General term
T_{r+1}=^{29}C_r.3^{29-r}.(7x)^{r}
For
r^{th}
term
Let put
r=r-1
\implies T_{r-1+1}=^{29}C_{r-1}3^{29-r+1}.(7x)^{r-1}~~~~-(1)
AS per question coefficient are equal
\implies ^{29}C_r.3^{29-r}.(7)^{r}=^{29}C_{r-1}.3^{30-r}.(7)^{r-1}
On solving the above expression we get
\dfrac{29-r+1}{r}\times7=3
\implies 210-7r=3r\\\implies 10r=210\\\implies r=21
The middle term in the expansion of
(a x)^n
is
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0%
56a^3 x^5
0%
-56a^3 x^5
0%
70a^4 x^4
0%
-70a^4 x^4
0:0:1
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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