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 $-k^{n-1}b^{n}{n}\choose{1}+k^{n-2}b^{n}{n}\choose{2}=0$ (Error compiling LaTeX. Unknown error_msg) 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 $k^{n-2}b^{n}n$ 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{C}$.

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1969 AHSME Problems/Problem 16

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