Prove that $\displaystyle \sum_{r=0}^n {n+r\choose r} \frac{1}{2^{r}}= 2^{n}$
what i try
$$\binom{n}{n}+\binom{n+1}{1}\frac{1}{2}+\binom{n+2}{2}\frac{1}{2^2}+\binom{n+3}{3}\frac{1}{2^3}+\cdots +\binom{n+n}{n}\frac{1}{2^n}$$
$$\binom{n}{n}+\binom{n+1}{n}\frac{1}{2}+\binom{n+2}{n}\frac{1}{2^2}+\binom{n+3}{n}\frac{1}{2^3}+\cdots +\binom{n+n}{n}\frac{1}{2^n}.$$
coefficient of $x^n$ in
$$(1+x)^n+(1+x)^{n+1}\frac{1}{2}+(1+x)^{n+2}\frac{1}{2^2}+\cdots\cdots +(1+x)^{2n}\frac{1}{2^n}.$$
How do i solve ithelp me plesse