Difference between revisions of "1966 AHSME Problems/Problem 36"

Revision as of 21:06, 23 December 2019

Problem

Let $(1+x+x^2)^n=a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}$ be an identity in $x$ . If we let $s=a_0+a_2+a_4+\cdots +a_{2n}$ , then $s$ equals:

$\text{(A) } 2^n \quad \text{(B) } 2^n+1 \quad \text{(C) } \frac{3^n-1}{2} \quad \text{(D) } \frac{3^n}{2} \quad \text{(E) } \frac{3^n+1}{2}$

Solution

$\fbox{E}$

Solution 2

Let $f(x)=(1+x+x^2)^n$ then we have $f(1)=a_0+a_1+a_2+...+a_2n=(1+1+1)^n=3^n$ $f(-1)=a_0-a_1+a_2-...+a_2n=(1-1+1)^n=1$ Adding yields $f(1)+f(-1)=2(a_0+a_2+a_4+...+a_2n)=3^n+1$ Thus $s=\frac{3^n+1}{2}$ , or $\boxed{D}$ .

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 35	Followed by Problem 37
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
All AHSME Problems and Solutions

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

@@ Line 9: / Line 9: @@
 == Solution 2 ==
-Let f(x)=(1+x+x^2)^n then we have
+Let <math>f(x)=(1+x+x^2)^n</math> then we have
-<math>s=a_0+a_1+a_2+...a_{2n}=f(1)=(1+1+1)^n=3^n</math>
+<cmath>f(1)=a_0+a_1+a_2+...+a_2n=(1+1+1)^n=3^n</cmath>
+<cmath>f(-1)=a_0-a_1+a_2-...+a_2n=(1-1+1)^n=1</cmath>
+Adding yields
+<cmath>f(1)+f(-1)=2(a_0+a_2+a_4+...+a_2n)=3^n+1</cmath>
+Thus <math>s=\frac{3^n+1}{2}</math>, or <math>\boxed{D}</math>.
 == See also ==

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Difference between revisions of "1966 AHSME Problems/Problem 36"

Revision as of 21:06, 23 December 2019

Contents

Problem

Solution

Solution 2

See also