Find a compact (close form) expression based upon the binomial theorem for the polynomials $\sum\limits_{0 \leq 2k \leq n} {n \choose 2k}{x}^{2k}$

Question

Find a compact (close form) expression based upon the binomial theorem for the polynomials

$\sum\limits_{0 \leq 2k \leq n} {n \choose 2k}{x}^{2k}$ and $\sum\limits_{0 \leq 2k + 1 \leq n} {n \choose 2k + 1}{x}^{2k + 1}$

Here is what I have so far: $$ = \sum\limits_{0 \leq 2k \leq n} {n \choose 2k}{x}^{2k}$$ $$ = \sum\limits_{0 \leq 2k \leq n} {n \choose 2k}{x}^{2k} {1}^{n - 2k}$$ $$ = {(x + 1)}^{n}$$ and $$ = \sum\limits_{0 \leq 2k + 1 \leq n} {n \choose 2k + 1}{x}^{2k + 1}$$ $$ = \sum\limits_{0 \leq 2k + 1 \leq n} {n \choose 2k + 1}{x}^{2k + 1} {1}^{n - 2k - 1}$$ $$ = {(x + 1)}^{n}$$

Should both answers be the same or have I made a mistake?

Answer 1

Note that $(1 +x)^{n} = \sum_{k=0}^n \binom{n}{k}x^k 1^{n-k}$. Note that this is not enough since we are interested only in the even powers of $x$.

Thus, we need to remove the odd powers of $x$. Consider, $(1 - x)^{n} = \sum_{k=0}^n \binom{n}{k}x^k (-1)^{k}$.

Thus, $(1 + x)^{n} + (1 - x)^{n} = \sum_{k=0}^n \binom{n}{k}x^k( 1 + (-1)^{k}) = 2\sum_{k=0}^{n/2} \binom{n}{2k}x^{2k}$.

Thus, we have $\sum_{0 \leq 2k \leq n} \binom{n}{2k}x^{2k} = \frac{1}{2}((1 + x)^{n} + (1 - x)^{n})$.

The formula for $\sum_{0 \leq 2k+1 \leq n} \binom{n}{2k+1}x^{2k+1}$ is obtaine in the same fassion, or using the notion that $\sum_{0 \leq 2k+1 \leq n} \binom{n}{2k+1}x^{2k+1} + \sum_{0 \leq 2k \leq n} \binom{n}{2k}x^{2k} = (1 + x)^n$.

Hope this helps.

Answer 2

The mistake OP made is that the first sum is the sum of the even powers while the second one is the sum of the odd powers as below: \begin{align} \sum_{0 \leqslant 2 k \leq n}\binom{n}{2 k} x^{2 k}&= \binom{n}{0}+\binom{n}{2}x^2+\binom{n}{4}x^4 +\dots+\binom{n}{2\lfloor\frac{n}{2}\rfloor} x^{2\lfloor\frac{n}{2}\rfloor} \\ &\neq (x+1)^n \\ \sum_{0 \leqslant 2 k +1\leq n}\binom{n}{2 k+1}x^{2 k+1} &= \binom{n}{1}x+\binom{n}{3}x^3+\binom{n}{5} x^5 +\dots+\binom{n}{ 2\lfloor\frac{n-1}{2}\rfloor+1}x^{2\lfloor\frac{n-1}{2}\rfloor+1} \\ &\neq (x+1)^n \end{align} Meanwhile $$\sum_{0 \leqslant 2 k \leq n}\binom{n}{2 k} x^{2 k}+ \sum_{0 \leqslant 2 k +1\leq n}\binom{n}{2 k+1}x^{2 k+1}=(x+1)^n$$