Proving identity related to $\sum_{k=0}^{\large\lfloor{\frac{n}{2}}\rfloor}\binom{n}{2k}\frac{x^{2k}}{2k+1}$

Question

How it can be shown that:

$$\sum_{k=0}^{\large\lfloor{\frac{n}{2}}\rfloor}\binom{n}{2k}\frac{x^{2k}}{2k+1}=\frac{\left(x+1\right)^{\left(n+1\right)}-\left(1-x\right)^{\left(n+1\right)}}{2x\left(n+1\right)}$$

Where $\lfloor{x}\rfloor$ denotes the floor function.

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My try:

$$\begin{align}&\sum_{k=0}^{\large\lfloor{\frac{n}{2}}\rfloor}\binom{n}{2k}\frac{x^{2k}}{2k+1}\\ &=\frac{1}{n+1}\sum_{k=0}^{\large\lfloor{\frac{n}{2}}\rfloor}\binom{n+1}{2k+1}x^{2k}\\&=\frac{1}{\left(n+1\right)x}\sum_{k=0}^{\large\lfloor{\frac{n}{2}}\rfloor}\binom{n+1}{2k+1}\color{red}{x^{2k+1}} \end{align}$$

Which follows from the identity:

$$\sum_{k=0}^{\large\lfloor{\frac{n-1}{2}}\rfloor}\binom{n}{2k+1}=2^{n-1} \;\;\;\;\;\;\; (n \in \mathbb N_{\ge1})$$

But I don't know how to continue with that $\color{red}{x^{2k+1}}$.

Answer 1

$$\frac{(1+x)^n+(1-x)^n}{2}=\sum_{k=0}^{[n/2]} {n \choose 2k} x^{2k}$$ Integrate both sides Then $$C+\frac{(1+x)^{n+1}}{2(n+1)}-\frac{(1-x)^{n+1}}{2(n+1)}=\sum_{k-0}^{[n/2]} {n \choose 2k}\frac{x^{2k+1}}{2k+1}$$ Putting $x=0$ both sides we get constantant of integration $C=0$ Finally $$\sum_{k-0}^{[n/2]} {n \choose 2k}\frac{x^{2k+1}}{2k+1}=\frac{(1+x)^{n+1}-(1-x)^{n+1}}{2x(n+1)}$$