Prove the following relation: I must prove the relation $$\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}=2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}.$$
I got this far before I got stuck:
$\begin{eqnarray*}
\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k} & = & \sum_{k=0}^{n+1}\left\{\binom{n+k}{k}+\binom{n+k}{k-1}\right\}\frac1{2^k}\\
& = & \sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^n\binom{n+k}{k-1}\frac1{2^k}+\binom{2n+1}{n}\frac1{2^k}.
\end{eqnarray*}$
If I can combine the second and third terms and get something same as first term, I am done but I could not do that.

You correctly used Pascal's identity, but then you goofed going to the next line. (Should have an $n$ in that last exponent of $2$, not a $k$.) I recommend going a different way, though.
$\begin{eqnarray*}
\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k} & = & \sum_{k=0}^{n+1}\left\{\binom{n+k}{k}+\binom{n+k}{k-1}\right\}\frac1{2^k}\\
& = & \sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^{n+1}\binom{n+k}{k-1}\frac1{2^k}\\
& = & \sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\sum_{k=1}^{n+1}\binom{n+k}{k-1}\frac1{2^k}\\
& = & \sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^n\binom{n+k+1}{k}\frac1{2^{k+1}}\\
& = & -\binom{2n+2}{n+1}\frac1{2^{n+2}}+\sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^{k+1}}\\
& = & -\binom{2n+2}{n+1}\frac1{2^{n+2}}+\sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}+\frac12\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}.
\end{eqnarray*}$
You see how we have half the original sum on the right-hand side now? If we subtract that and then multiply by $2$, we have 
$\begin{eqnarray*}
\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k} & = & -\binom{2n+2}{n+1}\frac1{2^{n+1}}+2\sum_{k=0}^{n+1}\binom{n+k}{k}\frac1{2^k}\\
& = & -\binom{2n+2}{n+1}\frac1{2^{n+1}}+2\binom{2n+1}{n+1}\frac1{2^{n+1}}+2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}.
\end{eqnarray*}$
Finally, applying Pascal's identity to $\binom{2n+2}{n+1}$, and using the fact that $\binom{2n+1}{n}=\binom{2n+1}{n+1},$ the extraneous binomial coefficients cancel out, and we're left with $$\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}=2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k},$$ as desired.
