Difference between revisions of "1966 AHSME Problems/Problem 36"
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== Problem == | == Problem == | ||
− | Let <math>(1+x+x^2)^n=a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}</math> be an identity in <math>x</math>. If we let <math>s=a_0+a_2+a_4+\cdots +a_{2n}</math>, then <math>s</math> equals: | + | Let <math>(1+x+x^2)^n=a_0+a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}</math> be an identity in <math>x</math>. If we let <math>s=a_0+a_2+a_4+\cdots +a_{2n}</math>, then <math>s</math> equals: |
<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>\fbox{\textbf{(E)}}</math>. | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | ~ Minor edits by Qinglang, Nafer wrote a great solution, but put D instead of E | ||
== See also == | == 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]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:35, 29 July 2024
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 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 35 |
Followed by Problem 37 | |
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All AHSME Problems and Solutions |
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