Prove the following relation:

Let $s_n=\displaystyle\sum_{k=0}^n\binom{n+k}k\frac1{2^k}$. Then

$$\begin{align*} s_{n+1}&=\sum_{k=0}^{n+1}\binom{n+k+1}k\frac1{2^k}\\ &=\sum_{k=0}^{n+1}\left(\binom{n+k}k+\binom{n+k}{k-1}\right)\frac1{2^k}\\ &=\binom{2n+1}{n+1}\frac1{2^{n+1}}+\sum_{k=0}^n\binom{n+k}k\frac1{2^k}+\sum_{k=0}^n\binom{n+1+k}k\frac1{2^{k+1}}\\ &=\binom{2n+1}{n+1}\frac1{2^{n+1}}+s_n+\frac12\sum_{k=0}^n\binom{n+1+k}k\frac1{2^k}\\ &=s_n+\frac12\left(s_{n+1}-\binom{2n+2}{n+1}\frac1{2^{n+1}}\right)+\binom{2n+1}{n+1}\frac1{2^{n+1}}\\ &=s_n+\frac12s_{n+1}+\binom{2n+1}{n+1}\frac1{2^{n+1}}-\binom{2n+2}{n+1}\frac1{2^{n+2}}\;, \end{align*}$$

and therefore

$$\begin{align*} s_{n+1}&=2s_n+\binom{2n+1}{n+1}\frac1{2^n}-\binom{2n+2}{n+1}\frac1{2^{n+1}}\\ &=2s_n+\frac1{2^{n+1}}\left(2\binom{2n+1}{n+1}-\binom{2n+2}{n+1}\right)\\ &=2s_n+\frac1{2^{n+1}}\left(2\binom{2n+1}{n+1}-\binom{2n+1}{n+1}-\binom{2n+1}n\right)\\ &=2s_n+\frac1{2^{n+1}}\left(\binom{2n+1}{n+1}-\binom{2n+1}n\right)\\ &=2s_n\;. \end{align*}$$

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.