Summing up $3+5+9+17+...$ 
Find the sum of sum of $3
+5+9+17+...$ till $n$ terms.

Using Method of differences, the sum of the series is
$$\sum\limits_{j=1}^n 2^{j-1}+n$$
I am facing difficulty in evaluating $$\sum\limits_{j=1}^n 2^{j-1}$$. How do I do that? 
Now I have $2^0 + 2^1 + 2^2 ... 2^{n-1}$
The sum of this series is : $2^n- 1$ as sum of GP is given by $a(1-r^n)/(1-r)$. 
Here $a = 1, r =2$

I had miscalculated earlier. After trying again, I found the error:
So I have $\sum\limits_{k=1}^n 2^{k} + n $ instead of $\sum\limits_{k=1}^n 2^{k-1} + n$
which gives $2^{n+1}+n-2$ using the formula for sum of $n$ terms of GP stated in the question. 
