Only Even

Let $f(x) = ( x^2 + 2x-1)^8 = a_{16}x^{16} + a_{15}x^{15} + ... + a_2x^2 + a_1x + a_0$

We wish to find $a_{16} + ... + a_2 + a_0$ .

Note that $f(-1) = a_{16} - a_{15} + ... + a_2 - a_1 + a_0$ and that $f(1) = a_{16} + a_{15} + ... + a_2 + a_1 + a_0$

Consequently, $f(-1) + f(1) = 2 (a_{16} + ... + a_2 + a_0) = 2^8 + (-2)^8 = 2*256$

We then have: $a_{16} + ... + a_2 + a_0 = 256$

Suppose that $f(x)=(x^2+2x-1)^8 = \sum_{i=0}^{16} a_n x^n.$

Now note that

$f(1)=2^8 = \sum_{i=0}^{16} a_n$

and

$f(-1)=2^8 = \sum_{i=0}^{16} a_n (-1)^n.$

Thus, because the odd coefficients cancel out,

$\frac 12 (f(1)+f(-1)) = \sum_{i=0}^8 a_{2n}$

and the desird answer is $\frac 12 (2^8+2^8)=256$

We first note that for a given polynomial, $f(x) = a_m x^m +a_{m-1}x^{m-1}+...+a_0$ , we have the following;

• $f(1) = a_m + a_{m-1} + ... + a_0$
• $f(-1) = a_m(-1)^m + a_{m-1}(-1)^{m-1} + ... + a_0$ .

Thus, combining these two properties, we see that $\frac{f(1)+f(-1)}{2}$ gives exactly the sum of the coefficients of the even powers of $x$ (as the even power terms will double up while the odd power terms will cancel due to the powers of $-1$ ).

Hence, noting that $g(x) = (x^2 +2x - 1)^8$ is a polynomial, we have that the sum of the coefficients of the even powers of $x$ is $\frac{g(1)+g(-1)}{2} = \frac{2^8 + (-2)^8}{2} = 2^8 = 256$

since odd polynomials change there sign on reversing the value of the variable hence even ones preserve their signs to calculate the sum of coefficients of even ones we just need to add the values of polynomial at unity and negative unity and average it out which gives (256+256)/2=256

We'll expand the binomial and we'll get:

(x^2+2x-1)^8 = a0 + a1 x + a2 x^2 + ... + a16*x^16

a0,a1,a2,...,a16 are the coefficients of polynomial.

The sum of even coefficients is:

a0 + a2 + ... + a14 + a16

if we want to determine the summof all coefficients of a polynomial, we'll have to make the variable x = 1 and x=-1

We'll put x = 1 and we'll calculate:

(1+2-1)^8 = a0 + a1 + a2 + ... + a16

a0 + a1 + ... + a16 = 2^8 (1)

Now, we'll put x = -1

(1-2-1)^8 = a0 - a1 + ... - a15 + a16

a0 - a1 + ... - a15 + a16 = 2^8 (2)

We'll add (1) + (2):

(a0 +a1 + ... + a16) + (a0 -a1 + ... - a15 + a16) = 2^9 (the odd coefficients will be eliminated)

We'll eliminate and combine like terms:

2(a0 + a2 + ... + a14 + a16) = 2^9

a0 + a2 + ... + a14 + a16 = 2^9/2

a0 + a2 + ... + a14 + a16 = 2^8

The sum of even coefficientas of the given polynomial is: a0 + a2 + ... + a14 + a16 = 2^8

First split the expression $x^2 + 2x - 1$ into $x^2 - 1$ and $2x$ . Then use binomial expansion to expand the expression, to get ${8 \choose 0}(x^2 - 1)^8 + {8 \choose 1}(x^2 - 1)^7 (2x)^1 + {8 \choose 2}(x^2 - 1)^6 (2x)^2 + {8 \choose 3}(x^2 - 1)^5 (2x)^3 + {8 \choose 4}(x^2 - 1)^4 (2x)^4 + {8 \choose 5}(x^2 - 1)^3 (2x)^5 + {8 \choose 6}(x^2 - 1)^2 (2x)^6 + {8 \choose 7}(x^2 - 1)^1 (2x)^7 + {8 \choose 8}(2x)^8$ . Since the question asks for the coefficients of the terms with even powers of $x$ , the terms with only odd powers of $x$ need not be considered. It can be seen that any power of $x^2 - 1$ will only have even powers of $x$ . Thus, we only need to look at the power of $2x$ . From this, we only need to consider the terms of the expansion which have an even power of $2x$ . However, the coefficients of the terms of the expansion of $x^2 - 1$ will add up to zero. For example, $(x^2 - 1)^2$ is equals to $x^4 - 2x^2 + 1$ , and $1 - 2 + 1$ is equals to zero. Therefore, the only term to consider is the last term, which is ${8 \choose 8}(2x)^8$ , as the power of $x^2 - 1$ in this case will be zero. Working out the expression gives $256x^8$ , so the sum of the coefficients of the even powers of $x$ is $256$ .

=256

We show a more general result. For a polynomial $P(x)$ , the sum of the coefficients of the terms with even powers is $\frac{1}{2}(P(1)+P(-1))$ . We prove this as follows:

Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ . Note that $x^{2b}=1$ for $x=1, -1$ , and $x^{2b-1}=-(-x)^{2b-1}$ . In particular, this means that

$P(1)+P(-1)=2(a_0+a_2+a_4+\cdots)$ , and so the sum of the coefficients of the terms with even powers of x is $\frac{1}{2}(P(1)+P(-1))$ , as desired.

We can extend this concept to the roots of unity filter . Let $1, \omega, \omega^2, \omega^3, \cdots, \omega^{k-1}$ be the $k$ th roots of unity. Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ . Note for any term with an index divisible by $k$ will have $x^k$ evaluate to 1 for each of the roots of unity. For an index not divisible by $k$ , say $i$ , the term evaluates to

$a_i(\omega^i+\omega^{2i}+\cdots+\omega^{i(n-1)})=a_i(\frac{\omega^{in}-1}{\omega^i-1})$ . Recall that $i$ is not divisible by $k$ , thus this evaluates to 0.

Therefore, the sum of the coefficients in a polynomial $P(x)$ whose indices are divisible by k is

$\frac{1}{k}(P(1)+P(\omega)+P(\omega^2)+\cdots+P(\omega^{k-1}))$ .

Note that this agrees with our above answer. The 2nd roots of unity are $1, -1$ , so we seek $\frac{1}{2}(P(1)+P(-1))$ as above.

This is a powerful technique with several non-obvious applications. For example, the problem

"Compute $\binom{2013}{0}+\binom{2013}{3}+\cdots+\binom{2013}{2013}$ "

has a nice solution using the roots of unity filter. We leave this as an exercise for the reader to practice their skills, and suggest as a hint to replace 2013 with the more general $3n$ .

Let $E$ be the sum of the even coefficients of the polynomial. Let $D$ be the sum of the odd coefficients of the polynomial. Setting $x = 1$ gives the sum of all the coefficients. So $D + E = (1 + 2 - 1)^{8} = 2^{8}$ . Setting $x = -1$ will give the sum of all the even coefficients minus the sum of all the odd coefficients. So $E - D = (1 - 2 - 1)^{8} = (-2)^{8} = 2^{8}$ . Solving these two equations gives $E = 2^{8}, D= 0$ . Thus, the sum of the even coefficients is $2^8 = 256$ .