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RD Sharma Solutions for Class 11 Maths Exercise 18.2 Chapter 18 – Binomial Theorem
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1. Find the 11th term from the beginning and the 11th term from the end in the expansion of (2x – 1/x2)25.
Solution:
Given:
(2x – 1/x2)25
The given expression contains 26 terms.
So, the 11th term from the end is the (26 − 11 + 1) th term from the beginning.
In other words, the 11th term from the end is the 16th term from the beginning.
Then,
T16 = T15+1 = 25C15 (2x)25-15 (-1/x2)15
= 25C15 (210) (x)10 (-1/x30)
= – 25C15 (210 / x20)
Now, we shall find the 11th term from the beginning.
T11 = T10+1 = 25C10 (2x)25-10 (-1/x2)10
= 25C10 (215) (x)15 (1/x20)
= 25C10 (215 / x5)
2. Find the 7th term in the expansion of (3x2 – 1/x3)10.
Solution:
Given:
(3x2 – 1/x3)10
Let us consider the 7th term as T7
So,
T7 = T6+1
= 10C6 (3x2)10-6 (-1/x3)6
= 10C6 (3)4 (x)8 (1/x18)
= [10×9×8×7×81] / [4×3×2×x10]
= 17010 / x10
∴ The 7th term of the expression (3x2 – 1/x3)10 is 17010 / x10.
3. Find the 5th term in the expansion of (3x – 1/x2)10.
Solution:
Given:
(3x – 1/x2)10
The 5th term from the end is the (11 – 5 + 1)th, is., 7th term from the beginning.
So,
T7 = T6+1
= 10C6 (3x)10-6 (-1/x2)6
= 10C6 (3)4 (x)4 (1/x12)
= [10×9×8×7×81] / [4×3×2×x8]
= 17010 / x8
∴ The 5th term of the expression (3x – 1/x2)10 is 17010 / x8.
4. Find the 8th term in the expansion of (x3/2 y1/2 – x1/2 y3/2)10.
Solution:
Given:
(x3/2 y1/2 – x1/2 y3/2)10
Let us consider the 8th term as T8
So,
T8 = T7+1
= 10C7 (x3/2 y1/2)10-7 (-x1/2 y3/2)7
= -[10×9×8]/[3×2] x9/2 y3/2 (x7/2 y21/2)
= -120 x8y12
∴ The 8th term of the expression (x3/2 y1/2 – x1/2 y3/2)10 is -120 x8y12.
5. Find the 7th term in the expansion of (4x/5 + 5/2x) 8.
Solution:
Given:
(4x/5 + 5/2x) 8
Let us consider the 7th term as T7
So,
T7 = T6+1
∴ The 7th term of the expression (4x/5 + 5/2x) 8 is 4375/x4.
6. Find the 4th term from the beginning and 4th term from the end in the expansion of (x + 2/x) 9.
Solution:
Given:
(x + 2/x) 9
Let Tr+1 be the 4th term from the end.
Then, Tr+1 is (10 − 4 + 1)th, i.e., 7th, the term from the beginning.
7. Find the 4th term from the end in the expansion of (4x/5 – 5/2x) 9.
Solution:
Given:
(4x/5 – 5/2x) 9
Let Tr+1 be the 4th term from the end of the given expression.
Then, Tr+1 is (10 − 4 + 1)th term, i.e., the 7th term, from the beginning.
T7 = T6+1
∴ The 4th term from the end is 10500/x3.
8. Find the 7th term from the end in the expansion of (2x2 – 3/2x) 8.
Solution:
Given:
(2x2 – 3/2x) 8
Let Tr+1 be the 4th term from the end of the given expression.
Then, Tr+1 is (9 − 7 + 1)th term, i.e., 3rd term, from the beginning.
T3 = T2+1
∴ The 7th term from the end is 4032 x10.
9. Find the coefficient of:
(i) x10 in the expansion of (2x2 – 1/x)20
(ii) x7 in the expansion of (x – 1/x2)40
(iii) x-15 in the expansion of (3x2 – a/3x3)10
(iv) x9 in the expansion of (x2 – 1/3x)9
(v) xm in the expansion of (x + 1/x)n
(vi) x in the expansion of (1 – 2x3 + 3x5) (1 + 1/x)8
(vii) a5b7 in the expansion of (a – 2b)12
(viii) x in the expansion of (1 – 3x + 7x2) (1 – x)16
Solution:
(i) x10 in the expansion of (2x2 – 1/x)20
Given:
(2x2 – 1/x)20
If x10 occurs in the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(ii) x7 in the expansion of (x – 1/x2)40
Given:
(x – 1/x2)40
If x7 occurs at the (r + 1) th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
40 − 3r =7
3r = 40 – 7
3r = 33
r = 33/3
= 11
(iii) x-15 in the expansion of (3x2 – a/3x3)10
Given:
(3x2 – a/3x3)10
If x−15 occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(iv) x9 in the expansion of (x2 – 1/3x)9
Given:
(x2 – 1/3x)9
If x9 occurs at the (r + 1)th term in the above expression.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to contain x9, we must have
18 − 3r = 9
3r = 18 – 9
3r = 9
r = 9/3
= 3
(v) xm in the expansion of (x + 1/x)n
Given:
(x + 1/x)n
If xm occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(vi) x in the expansion of (1 – 2x3 + 3x5) (1 + 1/x)8
Given:
(1 – 2x3 + 3x5) (1 + 1/x)8
If x occurs at the (r + 1)th term in the given expression.
Then, we have:
(1 – 2x3 + 3x5) (1 + 1/x)8 = (1 – 2x3 + 3x5) (8C0 + 8C1 (1/x) + 8C2 (1/x)2 + 8C3 (1/x)3 + 8C4 (1/x)4 + 8C5 (1/x)5 + 8C6 (1/x)6 + 8C7 (1/x)7 + 8C8 (1/x)8)
So, ‘x’ occurs in the above expression at -2x3.8C2 (1/x2) + 3x5.8C4 (1/x4)
∴ Coefficient of x = -2 (8!/(2!6!)) + 3 (8!/(4! 4!))
= -56 + 210
= 154
(vii) a5b7 in the expansion of (a – 2b)12
Given:
(a – 2b)12
If a5b7 occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(viii) x in the expansion of (1 – 3x + 7x2) (1 – x)16
Given:
(1 – 3x + 7x2) (1 – x)16
If x occurs at the (r + 1)th term in the given expression.
Then, we have:
(1 – 3x + 7x2) (1 – x)16 = (1 – 3x + 7x2) (16C0 + 16C1 (-x) + 16C2 (-x)2 + 16C3 (-x)3 + 16C4 (-x)4 + 16C5 (-x)5 + 16C6 (-x)6 + 16C7 (-x)7 + 16C8 (-x)8 + 16C9 (-x)9 + 16C10 (-x)10 + 16C11 (-x)11 + 16C12 (-x)12 + 16C13 (-x)13 + 16C14 (-x)14 + 16C15 (-x)15 + 16C16 (-x)16)
So, ‘x’ occurs in the above expression at 16C1 (-x) – 3x16C0
∴ Coefficient of x = -(16!/(1! 15!)) – 3(16!/(0! 16!))
= -16 – 3
= -19
10. Which term in the expansion of 
contains x and y to one and the same power?
Solution:
Let us consider Tr+1 th term in the given expansion contains x and y to one and the same power.
Then we have,
Tr+1 = nCr xn-r ar
11. Does the expansion of (2x2 – 1/x) contain any term involving x9?
Solution:
Given:
(2x2 – 1/x)
If x9 occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to contain x9, we must have
40 – 3r = 9
3r = 40 – 9
3r = 31
r = 31/3
It is not possible since r is not an integer.
Hence, there is no term with x9 in the given expansion.
12. Show that the expansion of (x2 + 1/x)12 does not contain any term involving x-1.
Solution:
Given:
(x2 + 1/x)12
If x-1 occurs at the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to contain x-1, we must have
24 – 3r = -1
3r = 24 + 1
3r = 25
r = 25/3
It is not possible since r is not an integer.
Hence, there is no term with x-1 in the given expansion.
13. Find the middle term in the expansion of:
(i) (2/3x – 3/2x)20
(ii) (a/x + bx)12
(iii) (x2 – 2/x)10
(iv) (x/a – a/x)10
Solution:
(i) (2/3x – 3/2x)20
We have,
(2/3x – 3/2x)20 where n = 20 (even number)
So the middle term is (n/2 + 1) = (20/2 + 1) = (10 + 1) = 11. ie., 11th term
Now,
T11 = T10+1
= 20C10 (2/3x)20-10 (3/2x)10
= 20C10 210/310 × 310/210 x10-10
= 20C10
Hence, the middle term is 20C10.
(ii) (a/x + bx)12
We have,
(a/x + bx)12 where n = 12 (even number)
So the middle term is (n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. ie., 7th term
Now,
T7 = T6+1
= 924 a6b6
Hence, the middle term is 924 a6b6.
(iii) (x2 – 2/x)10
We have,
(x2 – 2/x)10 where n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle term is -8064x5.
(iv) (x/a – a/x)10
We have,
(x/a – a/x) 10 where n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle term is -252.
14. Find the middle terms in the expansion of:
(i) (3x – x3/6)9
(ii) (2x2 – 1/x)7
(iii) (3x – 2/x2)15
(iv) (x4 – 1/x3)11
Solution:
(i) (3x – x3/6)9
We have,
(3x – x3/6)9 where n = 9 (odd number)
So the middle terms are ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) = ((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and 6th.
Now,
T5 = T4+1
Hence, the middle terms are 189/8 x17 and -21/16 x19.
(ii) (2x2 – 1/x)7
We have,
(2x2 – 1/x)7 where n = 7 (odd number)
So the middle terms are ((n+1)/2) = ((7+1)/2) = 8/2 = 4 and
((n+1)/2 + 1) = ((7+1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5
The terms are 4th and 5th.
Now,
Hence, the middle terms are -560x5 and 280x2.
(iii) (3x – 2/x2)15
We have,
(3x – 2/x2)15 where n = 15 (odd number)
So the middle terms are ((n+1)/2) = ((15+1)/2) = 16/2 = 8 and
((n+1)/2 + 1) = ((15+1)/2 + 1) = (16/2 + 1) = (8 + 1) = 9
The terms are 8th and 9th.
Now,
Hence, the middle terms are (-6435×38×27)/x6 and (6435×37×28)/x9.
(iv) (x4 – 1/x3)11
We have,
(x4 – 1/x3)11
Where n = 11 (odd number)
So the middle terms are ((n+1)/2) = ((11+1)/2) = 12/2 = 6 and
((n+1)/2 + 1) = ((11+1)/2 + 1) = (12/2 + 1) = (6 + 1) = 7
The terms are 6th and 7th.
Now,
T7 = T6+1
Hence, the middle terms are -462x9 and 462x2.
15. Find the middle terms in the expansion of:
(i) (x – 1/x)10
(ii) (1 – 2x + x2)n
(iii) (1 + 3x + 3x2 + x3)2n
(iv) (2x – x2/4)9
(v) (x – 1/x)2n+1
(vi) (x/3 + 9y)10
(vii) (3 – x3/6)7
(viii) (2ax – b/x2)12
(ix) (p/x + x/p)9
(x) (x/a – a/x)10
Solution:
(i) (x – 1/x)10
We have,
(x – 1/x)10 where n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle term is -252.
(ii) (1 – 2x + x2)n
We have,
(1 – 2x + x2)n = (1 – x)2n where n is an even number.
So the middle term is (2n/2 + 1) = (n + 1)th term.
Now,
Tn = Tn+1
= 2nCn (-1)n (x)n
= (2n)!/(n!)2 (-1)n xn
Hence, the middle term is (2n)!/(n!)2 (-1)n xn.
(iii) (1 + 3x + 3x2 + x3)2n
We have,
(1 + 3x + 3x2 + x3)2n = (1 + x)6n where n is an even number.
So the middle term is (n/2 + 1) = (6n/2 + 1) = (3n + 1)th term.
Now,
T2n = T3n+1
= 6nC3n x3n
= (6n)!/(3n!)2 x3n
Hence, the middle term is (6n)!/(3n!)2 x3n.
(iv) (2x – x2/4)9
We have,
(2x – x2/4)9 where n = 9 (odd number)
So the middle terms are ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) = ((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and 6th.
Now,
T5 = T4+1
And,
T6 = T5+1
Hence, the middle term is 63/4 x13 and -63/32 x14.
(v) (x – 1/x)2n+1
We have,
(x – 1/x)2n+1 where n = (2n + 1) is an (odd number)
So the middle terms are ((n+1)/2) = ((2n+1+1)/2) = (2n+2)/2 = (n + 1) and
((n+1)/2 + 1) = ((2n+1+1)/2 + 1) = ((2n+2)/2 + 1) = (n + 1 + 1) = (n + 2)
The terms are (n + 1)th and (n + 2)th.
Now,
Tn = Tn+1
And,
Tn+2 = Tn+1+1
Hence, the middle term is (-1)n.2n+1Cn x and (-1)n+1.2n+1Cn (1/x).
(vi) (x/3 + 9y)10
We have,
(x/3 + 9y)10 where n = 10 is an even number.
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. i.e., 6th term.
Now,
T6 = T5+1
Hence, the middle term is 61236x5y5.
(vii) (3 – x3/6)7
We have,
(3 – x3/6)7 where n = 7 (odd number).
So the middle terms are ((n+1)/2) = ((7+1)/2) = 8/2 = 4 and
((n+1)/2 + 1) = ((7+1)/2 + 1) = (8/2 + 1) = (4 + 1) = 5
The terms are 4th and 5th.
Now,
T4 = T3+1
= 7C3 (3)7-3 (-x3/6)3
= -105/8 x9
And,
T5 = T4+1
= 9C4 (3)9-4 (-x3/6)4
Hence, the middle terms are -105/8 x9 and 35/48 x12.
(viii) (2ax – b/x2)12
We have,
(2ax – b/x2)12 where n = 12 is an even number.
So the middle term is (n/2 + 1) = (12/2 + 1) = (6 + 1) = 7. i.e., 7th term.
Now,
T7 = T6+1
Hence, the middle term is (59136a6b6)/x6.
(ix) (p/x + x/p)9
We have,
(p/x + x/p)9 where n = 9 (odd number).
So the middle terms are ((n+1)/2) = ((9+1)/2) = 10/2 = 5 and
((n+1)/2 + 1) = ((9+1)/2 + 1) = (10/2 + 1) = (5 + 1) = 6
The terms are 5th and 6th.
Now,
T5 = T4+1
And,
T6 = T5+1
= 9C5 (p/x)9-5 (x/p)5
Hence, the middle terms are 126p/x and 126x/p.
(x) (x/a – a/x)10
We have,
(x/a – a/x) 10 where n = 10 (even number)
So the middle term is (n/2 + 1) = (10/2 + 1) = (5 + 1) = 6. ie., 6th term
Now,
T6 = T5+1
Hence, the middle term is -252.
16. Find the term independent of x in the expansion of the following expressions:
(i) (3/2 x2 – 1/3x)9
(ii) (2x + 1/3x2)9
(iii) (2x2 – 3/x3)25
(iv) (3x – 2/x2)15
(v) ((√x/3) + √3/2x2)10
(vi) (x – 1/x2)3n
(vii) (1/2 x1/3 + x-1/5)8
(viii) (1 + x + 2x3) (3/2x2 – 3/3x)9
(ix) (∛x + 1/2∛x)18, x > 0
(x) (3/2x2 – 1/3x)6
Solution:
(i) (3/2 x2 – 1/3x)9
Given:
(3/2 x2 – 1/3x)9
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of x, we must have
18 – 3r = 0
3r = 18
r = 18/3
= 6
So, the required term is the 7th term.
We have,
T7 = T6+1
= 9C6 × (39-12)/(29-6)
= (9×8×7)/(3×2) × 3-3 × 2-3
= 7/18
Hence, the term independent of x is 7/18.
(ii) (2x + 1/3x2)9
Given:
(2x + 1/3x2)9
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of x, we must have
9 – 3r = 0
3r = 9
r = 9/3
= 3
So, the required term is the 4th term.
We have,
T4 = T3+1
= 9C3 × (26)/(33)
= 9C3 × 64/27
Hence, the term independent of x is 9C3 × 64/27.
(iii) (2x2 – 3/x3)25
Given:
(2x2 – 3/x3)25
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
= 25Cr (2x2)25-r (-3/x3)r
= (-1)r 25Cr × 225-r × 3r x50-2r-3r
For this term to be independent of x, we must have
50 – 5r = 0
5r = 50
r = 50/5
= 10
So, the required term is 11th term.
We have,
T11 = T10+1
= (-1)10 25C10 × 225-10 × 310
= 25C10 (215 × 310)
Hence, the term independent of x is 25C10 (215 × 310).
(iv) (3x – 2/x2)15
Given:
(3x – 2/x2)15
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
= 15Cr (3x)15-r (-2/x2)r
= (-1)r 15Cr × 315-r × 2r x15-r-2r
For this term to be independent of x, we must have
15 – 3r = 0
3r = 15
r = 15/3
= 5
So, the required term is the 6th term.
We have,
T6 = T5+1
= (-1)5 15C5 × 315-5 × 25
= -3003 × 310 × 25
Hence, the term independent of x is -3003 × 310 × 25.
(v) ((√x/3) + √3/2x2)10
Given:
((√x/3) + √3/2x2)10
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of x, we must have
(10-r)/2 – 2r = 0
10 – 5r = 0
5r = 10
r = 10/5
= 2
So, the required term is 3rd term.
We have,
T3 = T2+1
Hence, the term independent of x is 5/4.
(vi) (x – 1/x2)3n
Given:
(x – 1/x2)3n
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
= 3nCr x3n-r (-1/x2)r
= (-1)r 3nCr x3n-r-2r
For this term to be independent of x, we must have
3n – 3r = 0
r = n
So, the required term is (n+1)th term.
We have,
(-1)n 3nCn
Hence, the term independent of x is (-1)n 3nCn
(vii) (1/2 x1/3 + x-1/5)8
Given:
(1/2 x1/3 + x-1/5)8
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of x, we must have
(8-r)/3 – r/5 = 0
(40 – 5r – 3r)/15 = 0
40 – 5r – 3r = 0
40 – 8r = 0
8r = 40
r = 40/8
= 5
So, the required term is the 6th term.
We have,
T6 = T5+1
= 8C5 × 1/(28-5)
= (8×7×6)/(3×2×8)
= 7
Hence, the term independent of x is 7.
(viii) (1 + x + 2x3) (3/2x2 – 3/3x)9
Given:
(1 + x + 2x3) (3/2x2 – 3/3x)9
If (r + 1)th term in the given expression is independent of x.
Then, we have:
(1 + x + 2x3) (3/2x2 – 3/3x)9 =
= 7/18 – 2/27
= (189 – 36)/486
= 153/486 (divide by 9)
= 17/54
Hence, the term independent of x is 17/54.
(ix) (∛x + 1/2∛x)18, x > 0
Given:
(∛x + 1/2∛x)18, x > 0
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of r, we must have
(18-r)/3 – r/3 = 0
(18 – r – r)/3 = 0
18 – 2r = 0
2r = 18
r = 18/2
= 9
So, the required term is the 10th term.
We have,
T10 = T9+1
= 18C9 × 1/29
Hence, the term independent of x is 18C9 × 1/29.
(x) (3/2x2 – 1/3x)6
Given:
(3/2x2 – 1/3x)6
If (r + 1)th term in the given expression is independent of x.
Then, we have:
Tr+1 = nCr xn-r ar
For this term to be independent of r, we must have
12 – 3r = 0
3r = 12
r = 12/3
= 4
So, the required term is the 5th term.
We have,
T5 = T4+1
Hence, the term independent of x is 5/12.
17. If the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)18 are equal, find r.
Solution:
Given:
(1 + x)18
We know that the coefficient of the r term in the expansion of (1 + x)n is nCr-1
So, the coefficients of the (2r + 4) and (r – 2) terms in the given expansion are 18C2r+4-1 and 18Cr-2-1
For these coefficients to be equal, we must have
18C2r+4-1 = 18Cr-2-1
18C2r+3 = 18Cr-3
2r + 3 = r – 3 (or) 2r + 3 + r – 3 = 18 [Since, nCr = nCs => r = s (or) r + s = n]
2r – r = -3 – 3 (or) 3r = 18 – 3 + 3
r = -6 (or) 3r = 18
r = -6 (or) r = 18/3
r = -6 (or) r = 6
∴ r = 6 [since r should be a positive integer.]
18. If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion of (1 + x)43 are equal, find r.
Solution:
Given:
(1 + x)43
We know that the coefficient of the r term in the expansion of (1 + x)n is nCr-1
So, the coefficients of the (2r + 1) and (r + 2) terms in the given expansion are 43C2r+1-1 and 43Cr+2-1
For these coefficients to be equal, we must have
43C2r+1-1 = 43Cr+2-1
43C2r = 43Cr+1
2r = r + 1 (or) 2r + r + 1 = 43 [Since, nCr = nCs => r = s (or) r + s = n]
2r – r = 1 (or) 3r + 1 = 43
r = 1 (or) 3r = 43 – 1
r = 1 (or) 3r = 42
r = 1 (or) r = 42/3
r = 1 (or) r = 14
∴ r = 14 [since value ‘1’ gives the same term]
19. Prove that the coefficient of (r + 1)th term in the expansion of (1 + x)n + 1 is equal to the sum of the coefficients of rth and (r + 1)th terms in the expansion of (1 + x)n.
Solution:
We know, the coefficients of (r + 1)th term in (1 + x)n+1 is n+1Cr
So, the sum of the coefficients of the rth and (r + 1)th terms in (1 + x)n is
(1 + x)n = nCr-1 + nCr
= n+1Cr [since, nCr+1 + nCr = n+1Cr+1]
Hence proved.
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