Summing up $3+5+9+17+...$

Question

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 $a = 1, r =2$

Answer 1

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 $$S_n=2^{n+1}+n-2.$$

Answer 2

Without words:

$$n+2^{n+1}-2$$

Answer 3

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.