Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m (→‎Solution 1)
(2 intermediate revisions by one other user not shown)
Line 4: Line 4:
 
<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 ==
−
<math>\fbox{E}</math>
 
−
 
 
−
== Solution 2 ==
 
  
 
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  
Line 14: Line 11:
 
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>\boxed{D}</math>.
+
Thus <math>s=\frac{3^n+1}{2}</math>, or <math>\boxed{E}</math>.
  
 
~ Nafer
 
~ Nafer
 +
 +
~ Minor edits by Qinglang, Nafer wrote a great solution, but put D instead of E
  
 
== See also ==
 
== See also ==

Revision as of 01:26, 3 July 2021

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Problems/Problem_36&oldid=157273"