Rajasthan Board RBSE Class 11 Maths Chapter 7 Binomial Theorem Ex 7.2
Question 1.
In the following expansions, find the term as stated :
(i) 5th term of (a + 2x3)17
(ii) 9th term of \({ \left( \frac { x }{ y } -\frac { 3y }{ { x }^{ 2 } } \right) }^{ 17 }\)
(iii) 6th term of \({ \left( \frac { 2 }{ \sqrt { x } } -\frac { { x }^{ 2 } }{ 2 } \right) }^{ 9 }\)
Solution:
(i) 5th term of (a + 2x3)17
We know that (r + 1 )th term in the expansion of (a + b)n
Tr+1 = nCr an-r br
Here a = ‘a’, b = 2x3, n = 17
r + 1 = 5 ⇒ r = 4
Hence, 5th term = T5
⇒ T5 = T4+1
⇒ T5 = 17C4 a17-4(23x3)4
⇒ T5 = 2380 × a13 × 24 × x12
⇒ T5 = 2380 × 16 × a13x12
⇒ T5 = 38080 a13 x12
Hence,
5th term = 38080 a13 x12 or 17C4 a13 × 16x12
Question 2.
Find the coefficient of:
(i) x-7 in the expansion of \({ { \left( ax-\frac { 1 }{ { bx }^{ 2 } } \right) }^{ 8 } }\)
(ii) x4 in the expansion of \({ { \left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } \right) }^{ 15 } }\)
(iii) x6 in the expansion (a – bx2)10
Solution:
(i) Coefficient of x-7 in the expansion of
In this term for coefficient of x-7 power of x should be
-7 = 8 – 3r = – 7
Hence, 8 – 3r = – 7
⇒ -3r = – 7 – 8 = – 15
r = 5
Hence, coefficient of x-7 in the expansion
(ii) Coefficient of x4 in the expansion of \({ { \left( { x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } \right) }^{ 15 } }\)
In this term, for coefficient of x4, power of x should be 4
So, 60 – 7r = 4
⇒ -7r = 4 – 60 = – 56
⇒ r = 8
Hence, coefficient of x4 in the expansion = 15C8 = 6435
(iii) Coefficient of x6 in the expansion of (a – b2)10
In this term, for coefficient of x6, power of x should be 6.
So, 2r = 6 ⇒ r = 3
Hence, coefficient of x6 in the expansion
= (-1)3 10C3 x a10-3 b3
= – 1 × 120 × a7 b3
= – 120 a7 b3
Question 3.
In the following expansions find the term independent of x :
Solution:
(i) Term independent of x in the expansion of
for term independent of x
x12-3r = x0
⇒ 12- 3r = 0
⇒ -3r = -12
⇒ r = 4
Hence, term independent of x = (-1)4 12C4
= 1 × 495 = 495
(ii) Term independent of x in the expansion of
For the independent of x
Hence, term independent of x
= (-1)2 10C2 32
= 1 × 45 × 9 = 405
(iii) Term independent of x in the expansion of
\({ { \left( \sqrt { \frac { x }{ 3 } } +\frac { 3 }{ 2{ x }^{ 2 } } \right) }^{ 10 } }\)
Let (r + 1)th term of expansion is constant i.e., independent of x.
(r + 1)th term is constant i.e., power of x should be zero.
∴ 5- r/2 -2r = 0
(iv) Term independent of x in the expansion of
In this expansion for term independent of x, power of.v should be 0
20 – 4r = 0
⇒ – 4r = -20
⇒ r = 5
∴ Term independent of x in the expansion
= (-1)5 10C5 =-252
Question 4.
In the following expansion, find the middle term :
Solution:
(i) Middle term in the expansion of \({ { \left( { \frac { x }{ 2 } }+2y \right) }^{ 6 } }\) In the expansion of \({ { \left( { \frac { x }{ 2 } }+2y \right) }^{ 6 } }\) middle term will be \({ T }_{ { { \left( { \frac { n }{ 2 } }+1 \right) } } }\) th term where x = 6 which is even number.
(ii) Middle term in the expansion of \({ { \left( { 3a }-\frac { { a }^{ 3 } }{ 6 } \right) }^{ 9 } }\)
Given expression is\({ { \left( { 3a }-\frac { { a }^{ 3 } }{ 6 } \right) }^{ 9 } }\)
Where n = 9 which is an odd number
∴ Middle term \(\frac { n+1 }{ 2 }+1 \)th term and \(\frac { n+1 }{ 2 } \)th term
i.e., Middle term is 5th and 6th term (T6 and T6)
Hence, two middle term are \(\frac { 189 }{ 8 } \) a17 and – \(\frac { 21 }{ 16 } \) a19
(iii) Middle term in the expansion of \({ { \left( x+\frac { 1 }{ x } \right) }^{ 2n } }\)
Given expansion is \({ { \left( x+\frac { 1 }{ x } \right) }^{ 2n } }\)
Where ‘n ’ = 2n, which is an even number
(iv) Middle term in the expansion of \({ { \left( 3x-\frac { 2 }{ { x }^{ 2 } } \right) }^{ 15 } }\)
Given expansion is \({ { \left( 3x-\frac { 2 }{ { x }^{ 2 } } \right) }^{ 15 } }\)
where n = 15, which is an odd number
∴ Middle term \(\frac { n+1 }{ 2 } \)th term and \(\frac { n+1 }{ 2 }+1 \)th term
i,e„ 8th and 9th term (T8 and T9)
Question 5.
Prove that if n is even then in expansion of (1 + x)n, coefficient of middle term will be \(\frac { 1.3.5…..(n-1) }{ 2.4.6…..n } { 2 }^{ n }\)
If n is odd, then coefficient of both the middle term will be \(\frac { 1.3.5…..n }{ 2.4.6…..(n+1) } { 2 }^{ n }\)
Solution:
If n is even in (1 + x) term we take (1 + x)2n in place of (1 + x)n then ‘n’ = 2n is also even
Now, middle term = \(\left( \frac { 2n }{ 2 } +1 \right) \)th term
= (n + 1)th
Similarly, if n is odd, then we take (1 + x)2n+1 in place of (1 + x)n, then ‘n’ = 2n + 1 is also odd. Now middle term are
Question 6.
If in the expansion of \({ \left( ax+\frac { 1 }{ bx } \right) }^{ 11 }\)
coefficient of x7 and x7 are equal then prove that ab – 1 – 0.
Solution:
(r + 1)th term in the expansion of \({ \left( ax+\frac { 1 }{ bx } \right) }^{ 11 }\)
In this term, for coefficient of x7, power of x should be 7.
Hence, 11 – 2r = 7
⇒ 2r = 11-7
⇒ 2r = 4,
⇒ r = 2
Coefficient of x7 in the expansion
Question 7.
In the expansion of (1 + y)n if coefficients of 5th, 6th and 7th terms are in A.P., then find the value of n.
Solution:
Question 8.
In Binomial expansion (x – a)n second, third and fourth terms are 250, 720 and 1080 respectively. Find x, a and n.
Solution:
Second term in expansion of (x + a)n
T2 = nC1 xn-1.a
According to question,
Similarly, on dividing (3) by (2), we get
Question 9.
If in the expansion of (1+ a)n coefficient of three consecutive terms are in ratio 1 : 7 ; 42, then find the value of n.
Solution:
Let in the expansion of (1 + x)n three consecutive terms are (r – 1 )th term, rth term and (r + 1)th term.
Here, (r – 1)th term = nCr – 2ar-2 and its coefficient ,
=nCr – 2
Similarly, coefficient of rth term and (r + 1)th term are nCr -1 and nCr respectively.
Now, given ratio of coefficient is 1 : 7 : 42
or 6r = n – r + 1
or n – 7r + 1 = 0 …….(2)
On solving (1) and (2), we get
n = 55
Question 10.
Find the positive value of m for which in the expansion of (1 + x)m coefficient x2 is 6.
Solution:
From Binomial theorem
(1 + x)m = mC0 + mC1x + mC2 x2 + ……
Then, coefficient of x2 in the expansion of (1 + x )m = mC2
