How can I prove this:
$\sum_{r=1}^{n / 2} \frac{{ }^n C_{2 r}}{2 r+1}+1=\frac{2^n}{n+1}$
Here, ${ }^n C_r=\left(\begin{array}{l} n \\ r \end{array}\right)$
I'm still in high school so a step-by-step proof without much advanced mathematics would be appreciated.
I was able to prove it using basic properties of binomial coefficients. One can also do it using integration which is quite unnecessary although clever.
Firstly, it is to be noted that: ${ }^n C_{2 r}={ }^{n+1} C_{2 r+1} \cdot \frac{2 r+1}{n+1}$
Using this property, the expression is now reduced to: $\frac{1}{n+1} \sum_{r=1}^{n / 2}\left({ }^{n+1} C_{2 r+1}\right)+1$
We also know that $\sum_{k=0}^n{ }^n C_{k+1}=2^n$
Comparing this with our original expression, we get: $\begin{aligned} & \frac{1}{n+1} \sum_{r=1}^{n / 2}{ }^{n+1} C_{2 r+1}=\frac{2^n-{ }^{n+1} C_1}{n+1}+1 \\\\ & \quad=\frac{2^n}{n+1} \end{aligned}$