Question -
Answer -
(i) x10 inthe expansion of (2x2 – 1/x)20
Given:
(2x2 –1/x)20
If x10 occursin the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(ii) x7 inthe expansion of (x – 1/x2)40
Given:
(x – 1/x2)40
If x7 occursat 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 inthe expansion of (3x2 – a/3x3)10
Given:
(3x2 –a/3x3)10
If x−15 occursat the (r + 1)th term in the given expression.
Then, we have:
Tr+1 = nCr xn-r ar
(iv) x9 inthe expansion of (x2 – 1/3x)9
Given:
(x2 –1/3x)9
If x9 occursat 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 inthe expansion of (x + 1/x)n
Given:
(x + 1/x)n
If xm occursat 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 occursat 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 theabove 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 inthe expansion of (a – 2b)12
Given:
(a – 2b)12
If a5b7 occursat 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 occursat 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 theabove expression at 16C1 (-x) – 3x16C0
∴ Coefficient of x =-(16!/(1! 15!)) – 3(16!/(0! 16!))
= -16 – 3
= -19