# 1969 AHSME Problems/Problem 16

## Problem

When $(a-b)^n,n\ge2,ab\ne0$, is expanded by the binomial theorem, it is found that when $a=kb$, where $k$ is a positive integer, the sum of the second and third terms is zero. Then $n$ equals:

$\text{(A) } \tfrac{1}{2}k(k-1)\quad \text{(B) } \tfrac{1}{2}k(k+1)\quad \text{(C) } 2k-1\quad \text{(D) } 2k\quad \text{(E) } 2k+1$

## Solution

Since $a=kb$, we can write $(a-b)^n$ as $(kb-b)^n$. Expanding, the second term is $-k^{n-1}b^{n}{{n}\choose{1}}$, and the third term is $k^{n-2}b^{n}{{n}\choose{2}}$, so we can write the equation $$-k^{n-1}b^{n}{{n}\choose{1}}+k^{n-2}b^{n}{{n}\choose{2}}=0$$ Simplifying and multiplying by two to remove the denominator, we get $$-2k^{n-1}b^{n}n+k^{n-2}b^{n}n(n-1)=0$$ Factoring, we get $$k^{n-2}b^{n}n(-2k+n-1)=0$$ Dividing by $k^{n-2}b^{n}$ gives $$n(-2k+n-1)=0$$ Since it is given that $n\ge2$, $n$ cannot equal 0, so we can divide by n, which gives $$-2k+n-1=0$$ Solving for $n$ gives $$n=2k+1$$ so the answer is $\fbox{E}$.

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