1969 AHSME Problems/Problem 16

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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

$\fbox{E}$

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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