Difference between revisions of "1969 AHSME Problems/Problem 16"

Revision as of 12:14, 10 July 2015

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 $k^{n-2}b^{n}n$ 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 22: / Line 22: @@
 <cmath>-2k+n+1=0.</cmath>
 Solving for <math>n</math> gives
-<math>n=2k-1,</math>so the answer is <math>\fbox{C}</math>.
+<cmath>n=2k-1,</cmath> so the answer is <math>\fbox{C}</math>.
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Difference between revisions of "1969 AHSME Problems/Problem 16"

Revision as of 12:14, 10 July 2015

Problem

Solution

See also