A question about combinatorics: $\sum_{\substack{0\le k\le n\\ k\text{ even}}}\frac1{k+1}\binom nk$

Question

Let $n\in\mathbb N$ be fixed. For $0\le k \le n$, et $C_k=\binom nk$. Evaluate: $$\sum_{\substack{0\le k\le n\\ k\text{ even}}} \frac{C_k}{k+1}.$$

My attempt : $(x+1)^n=\sum_{k=0}^{n}{\binom{n}{k}x^k}\implies \frac{(x+1)^{n+1}}{n+1}+C=\int (x+1)^ndx=\sum_{k=0}^{n}{\binom{n}{k}\frac{1}{k+1}x^{k+1}}.$

Putting $x=0$ we have $C=\frac{-1}{n+1}$.

so my answer is $$\frac{2^{n+1}-1}{n+1}$$

is its True ?

Any hints/solution will appreciated

thanks u

Answer 1

In your solution you have added all terms, whereas it should be done only for even ones:

$$\sum_{\substack{0\leq k \leq n\\k\text{ even}}}\frac 1{k+1}\binom nk=\sum_{k=0}^n\frac 1{k+1}\binom nk\left[\frac{x^{k+1}-(-x)^{k+1}}{2}\right]_0^1\\=\int_0^1\frac{(x+1)^n-(1-x)^n}{2}dx =\frac{2^n}{n+1}. $$

Answer 2

Hint : Observe that \begin{align*} 2\sum_{\substack{0\leq k \leq n\\k\text{ even}}} \frac{C_k}{k+1} &= \sum_{0\leq k\leq n} \frac{C_k}{k+1} +\sum_{0\leq k\leq n} \frac{C_k}{k+1} (-1)^k \end{align*}

You can apply the rest of your reasoning to the two sums above to get your desired result.