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smc

algebra senior

Problem

Let

(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}

be an identity in

x

. If we let

s = a_{0} + a_{2} + a_{4} + \dots + a_{2 n}

, then

s

equals:

(A)

2^{n}

(B)

2^{n} + 1

(C)

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

(D)

3^{n} /2

(E)

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

Solution

Let

f (x) = (1 + x + x^{2})^{n}

then we have

f (1) = a_{0} + a_{1} + a_{2} + ... + a_{2 n} = (1 + 1 + 1)^{n} = 3^{n}

f (- 1) = a_{0} - a_{1} + a_{2} - ... + a_{2 n} = (1 - 1 + 1)^{n} = 1

Adding yields

f (1) + f (- 1) = 2 (a_{0} + a_{2} + a_{4} + ... + a_{2 n}) = 3^{n} + 1

Thus

s = \frac{3 ^{n} + 1}{2}

, or

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

.

Final answer

E