Binomial coefficients in Geometric summation Guys please help me find the sum given below.
$$\sum_{k=j}^i\binom{i}{k}\binom{k}{j}\cdot 2^{k-j}$$
(NOTE):The two coefficients are multiplied by 2 power (k-j)
I am using the formula: $\binom{r}{m}\binom{m}{q}=\binom{r}{q}\binom{r-q}{m-q}$
But not able to reach something fruitful.
Thanks in advance.

Well you were totally in right direction
$\binom{i}{j}\cdot \sum_{k=j}^{k=i}$$\binom{i-j}{k-j}\cdot 2^{k-j}$ can be written as 
$\binom{i}{j} \cdot \sum_{e=0}^{e=i-j}$$\binom{i-j}{e} \cdot 2^e$ .Apply binomial theorem to get the answer $\binom ij \cdot (1+2)^{i-j}$
