1969 AHSME Problems/Problem 16

Revision as of 13:04, 10 July 2015 by FlamezofDeath (talk | contribs) (Solution)

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}$.

See also

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