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Binomial Theorem Question (1 Viewer)

Thread starter Hangm3
Start date Sep 19, 2016

Hangm3

ggnore

#1

Pls help with this question
In the expansion (1 + x + kx^2)9 in ascending powers of x, the coefficient of x^2 is zero. Find the value of k.

KingOfActing

lukewarm mess

#2

$\begin{align*}0 = [x^2](1+x+kx^2)^9 &= [x^2]\sum_{r = 0}^9 {9 \choose r}(x+kx^2)^r \\&= [x^2](1 + 9(x+kx^2) + 36(x+kx^2)^2 + \text{higher terms})\\&= 9k +36\\\therefore k = -4\end{align*}$

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