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Answer the following: Find the term independent of x in the in expansion of (1-x2)(x+2x)6 - Mathematics and Statistics

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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`

योग

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 · 236C4 · 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

shaalaa.com

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

shaalaa.com
General Term in Expansion of (a + b)n
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Methods of Induction and Binomial Theorem - Miscellaneous Exercise 4.2 [पृष्ठ ८६]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4.2 | Q II. (23) | पृष्ठ ८६

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