Question

Answer the following:

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

Answer 1

Here a = x², b = ax, n = 10

We have, t_r+1 = ⁿC_r, a^n–r .b^r

= ¹⁰C_r. (x²)^10–r . (ax)^r

= ¹⁰C_r. a^r . x^20-2r . x^r

= ¹⁰C_r a^r . x^20–r

To get the coefficient of x¹⁶, we must have

x^20–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