Question

The coefficient of x² in the expansion of (1 + 2x)^m is 112. Find m

Answer 1

Let t_r+1 has x² in the expansion of (1 + 2x)^m.

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

t_r+1 = ⁿC_r a^n–r b^r  

Here, n = m, a = 1, b = 2x

∴ t_r+1 = ^mC_r (1)^m–r (2x)^r

= ^mC_r2^r · x^r

But t_r+1 has x²

∴ power of x = 2

∴ r = 2

∴ coefficient of x² = ^mC₂ · 2² 

But coefficient of x² is 112

∴ ^mC₂ · 2² = 112  

∴ `("m"!)/(2!("m" - 2)!) xx 4` = 112

∴ `("m"("m" - 1)("m" - 2)!)/(2 xx ("m" - 2)!) xx 4` = 112

∴ m(m – 1) = 56 = 8 × 7

∴ m(m – 1) = 8(8 – 1)

By comparing on both sides, m = 8.