1966 AMC 12 Problem 36

Question

Let

\begin{aligned} (1+x+x^2)^n &=a_0+a_1x\\ &\quad+a_2x^2+\cdots\\ &\quad+a_{2n}x^{2n} \end{aligned}

be an identity in

x.

If we let

s=a_0+a_2+a_4+\cdots+a_{2n},

then

s

equals:

2^n

2^n+1

\dfrac{3^n-1}{2}

\dfrac{3^n}{2}

\dfrac{3^n+1}{2}

Evaluate the polynomial at

x=1

and

x=-1

Adding those two values cancels all odd-degree coefficients

Let

P(x)=(1+x+x^2)^n.

Then

P(1)=3^n

is the sum of all coefficients, while

P(-1)=1

is the even-coefficient sum minus the odd-coefficient sum. Therefore

s=\frac{P(1)+P(-1)}2=\frac{3^n+1}{2}.

Therefore, the correct answer is E.

Problem 36 in other years