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 == |
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Let <math>f(x)=(1+x+x^2)^n</math> then we have | Let <math>f(x)=(1+x+x^2)^n</math> then we have | ||
| − | <cmath>f(1)=a_0+a_1+a_2+...+ | + | <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-...+ | + | <cmath>f(-1)=a_0-a_1+a_2-...+a_{2n}=(1-1+1)^n=1</cmath> |
Adding yields | Adding yields | ||
<cmath>f(1)+f(-1)=2(a_0+a_2+a_4+...+a_{2n})=3^n+1</cmath> | <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>\ | + | Thus <math>s=\frac{3^n+1}{2}</math>, or <math>\fbox{\textbf{(E)}}</math>. |
~ Nafer | ~ 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
![\[f(1)=a_0+a_1+a_2+...+a_{2n}=(1+1+1)^n=3^n\]](http://latex.artofproblemsolving.com/5/6/7/5671c2d825943ceb3af8d69944ac957d27b9e5da.png)
![\[f(-1)=a_0-a_1+a_2-...+a_{2n}=(1-1+1)^n=1\]](http://latex.artofproblemsolving.com/7/c/1/7c14a7e9fcc8e6f137e822943a21873a823defba.png)
Adding yields
![\[f(1)+f(-1)=2(a_0+a_2+a_4+...+a_{2n})=3^n+1\]](http://latex.artofproblemsolving.com/8/0/3/8031c910f342615fdcf03b63c9d5e177acd7181e.png)
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 | |
| 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. 
