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 $a = 1, r =2$

Maybe do you mean the following?
$$3+5+9+17+...=\sum_{k=1}^n2^k+n=\frac{2(2^n-1)}{2-1}+n=2^{n+1}-2+n.$$
If so let $S_n=3+5+...+(2^n+1).$
Thus, $$S_n-S_1=\sum_{k=2}^n(S_k-S_{k-1})=\sum_{k=2}^n(2^k+1)=\frac{4(2^{n-1}-1)}{2-1}+n-1,$$
which gives $$S_n=2^{n+1}+n-2.$$
