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In the following expansion, find the indicated coefficient. x8 in (2x5-5x3)8 - Mathematics and Statistics

Sum

In the following expansion, find the indicated coefficient.

x8 in `(2x^5 - 5/x^3)^8`

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Solution

Let tr+1 contains x8 in the expansion of `(2x^5 - 5/x^3)^8`

We know that, in the expansion of (a + b)n,

tr+1 = nCr an–r br 

Here a = 2x5, b = `-5/x^3`, n = 8

∴ tr+1 = `""^8"C"_"r" (2x^5)^(8 - "r") (-5/x^3)^"r"`

= `""^8"C"_"r".2^(8-"r").x^(40-5"r").(-5)^"r"/x^(3"r")` 

8Cr .28–r. (–5)r. x40–8r 

But tr+1 contains x8

∴ power of x = 8

∴ 40 – 8r = 8

∴  r = 4

∴ coefficient of x8 = 8C4 . 28–4 . (–5)4 

= `(8 xx 7 xx 6 xx 5)/(1 xx 2 xx 3 xx 4) xx 2^4 xx 625`

= 70 × 16 × 625

= 700000

Concept: Binomial Coefficients
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.3 | Q 2. (ii) | Page 80
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