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How to sum the even-index coefficients of a polynomial

Evaluate at 1 and -1

For a polynomial whose powers have integer exponents, P(1) adds every coefficient. P(-1) adds coefficients of even powers and subtracts those of odd powers. Half their sum therefore keeps only the even-power coefficients.

Problem

Let $P (x) = (2 x^{2} - x - 1)^{8} = a_{0} x^{16} + a_{1} x^{15} + ... + a_{16}$ . Find (1) the sum of all coefficients and (2) $a_{0} + a_{2} + ... + a_{16}$ .

Answer

All coefficients:

0

. Even-index coefficients:

128

The leading exponent is 16, so even indices correspond to even powers.

Step-by-step solution

1. Add all coefficients

Set $x = 1$ : $P (1) = (2 - 1 - 1)^{8} = 0$ . Hence $a_{0} + a_{1} + ... + a_{16} = 0$ .

2. Alternate the signs

At $x = - 1$ , even powers contribute positively and odd powers negatively. Here $P (- 1) = (2 + 1 - 1)^{8} = 2^{8} = 256$ .

3. Isolate the required coefficients

Adding $P (1)$ and $P (- 1)$ cancels all odd-power terms. Thus the even-index sum is $(0 + 256) /2 = 128$ .

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