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प्रश्न
Answer the following:
Find the term independent of x in the in expansion of `(1 - x^2) (x + 2/x)^6`
उत्तर १
`(1 - x^2) (x + 2/x)^6 = (x + 2/x)^6 - x^2(x + 2/x)^6` ...(1)
(r + 1)th term in the expansion of (a + b)n is given by
tr+1 = nCr, an–r br
For `(x + 2/x)^6`, a = x, b = `2/x`, n = 6
∴ tr+1 = `""^6"C"_"r" (x)^(6 - "r") (2/x)^"r"`
= 6Cr x6–r · 2r · x–r
= 6Cr · 2r · x6–2r ...(2)
For term independent of x in `(x + 2/x)^6`,
x6–2r = x0
∴ 6 – 2r = 0
∴ r = 3
For term independent of x in x2 `(x + 2/x)^6`,
x2 · x6–2r = x0
∴ x8–2r = x0
∴ 8 – 2r = 0
∴ r = 4
∴ from (1) and (2), the term independent of x in `(1 - x^2) (x + 2/x)^6` is
6C3 · 23 – 6C4 · 24
= `(6!)/(3!3!) xx 8 - (6!)/(4!2!) xx 16`
= `(6 xx 5 xx 4)/(1 xx 2 xx 3) xx 8 - (6 xx 5)/(1 xx 2) xx 16`
= 160 – 240
= – 80
उत्तर २
`(1 - x^2) (x + 2/x)^6`
= `(1 - x^2)[""^6"C"_0 x^6 + ""^6"C"_1 x^5(2/x) + ""^6"C"_2 x^4(2/x)^2 + ""^6"C"_3 x^3 (2/x)^3 + ""^6"C"_4 x^2(2/x)^4 + ""^6"C"_5 x(2/x)^5 + ""^6"C"_6 (2/x)^6]`
= `(1 - x^2)[""^6"C"_0 x^6 + ""^6"C"_1 (2x^4) + ""^6"C"_2 (4x^2) + ""^6"C"_3 (8) + ""^6"C"_4 (16/x^2) + ""^6"C"_5 (32/x^4) + ""^6"C"_6(64/x^6)]`
∴ the term independent of x
= 6C3 (8) – 6C4 (16)
= `(6!)/(3!3!) xx 8 - (6!)/(4!2!) xx 16`
= `(6 xx 5 xx 4)/(1 xx 2 xx 3) xx 8 - (6 xx 5)/(1 xx 2) xx 16`
= 160 – 240
= – 80
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