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:
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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 ==
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== Solution 1 ==
<math>\fbox{E}</math>
 
  
== Solution 2 ==
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Let <math>f(x)=(1+x+x^2)^n</math> then we have
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<cmath>f(1)=a_0+a_1+a_2+...+a_{2n}=(1+1+1)^n=3^n</cmath>
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<cmath>f(-1)=a_0-a_1+a_2-...+a_{2n}=(1-1+1)^n=1</cmath>
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Adding yields
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<cmath>f(1)+f(-1)=2(a_0+a_2+a_4+...+a_{2n})=3^n+1</cmath>
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Thus <math>s=\frac{3^n+1}{2}</math>, or <math>\fbox{\textbf{(E)}}</math>.
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~ Nafer
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~ 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}}   
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{{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 $(1+x+x^2)^n=a_0+a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}$ 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 $\fbox{\textbf{(E)}}$.

~ Nafer

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

See also

1966 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 35
Followed by
Problem 37
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