Maharashtra State BoardHSC Science (General) 11th
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Answer the following: If the coefficient of x16 in the expansion of (x2 + ax)10 is 3360, find a - Mathematics and Statistics

Sum

Answer the following:

If the coefficient of x16 in the expansion of (x2 + ax)10 is 3360, find a

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Solution

Here a = x2, b = ax, n = 10

We have, tr+1 = nCr, an–r .br

= 10Cr. (x2)10–r . (ax)r

= 10Cr. ar . x20-2r . xr

= 10Cr ar . x20–r

To get the coefficient of x16, we must have

x20–r = x16

∴ 20 – r = 16

∴ r = 4

∴ coefficient of x16 = 10C4 . a4 

Given coefficient of x16 = 3360

10C4a4 = 3360

∴ `(10!)/(4!6!) "a"^4` = 3360

∴ `(10 xx 9 xx 8 xx 7 xx 6!)/(4 xx 3 xx 2 xx 1 xx 6!) "a"^4` = 3360

∴ 210. a4 = 3360

∴ a4 = 16

∴ a = ± 2

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
Miscellaneous Exercise 4 | Q II. (12) | Page 86
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