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

Revision as of 10:32, 29 July 2024

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 1

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 $\boxed{E}$ .

~ Nafer

~ Minor edits by Qinglang, Nafer wrote a great solution, but put D instead of E

1966 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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

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 == See also ==
-{{AHSME box|year=1966|num-b=35|num-a=37}}
+{{AHSME 40p box|year=1966|num-b=35|num-a=37}}
 [[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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

Revision as of 10:32, 29 July 2024

Problem

Solution 1

See also