Difference between revisions of "1966 AHSME Problems/Problem 36"
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<math>\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}</math> | <math>\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}</math> | ||
− | == Solution == | + | == Solution 1 == |
− | |||
− | == | + | Let <math>f(x)=(1+x+x^2)^n</math> then we have |
+ | <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{E}</math>. | ||
− | + | ~ Nafer | |
− | + | ||
+ | ~ Minor edits by Qinglang, Nafer wrote a great solution, but put D instead of E | ||
== See also == | == See also == |
Revision as of 02:26, 3 July 2021
Problem
Let be an identity in . If we let , then equals:
Solution 1
Let then we have Adding yields Thus , or .
~ Nafer
~ Minor edits by Qinglang, Nafer wrote a great solution, but put D instead of E
See also
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.