Question: Find $\sum_{k=1}^n u_k $ if $ u_n= \sum_{k=0}^{n} \frac{1}{2^k}$
My try : $u_1=\frac{1}{2^0}+\frac{1}{2^1} , u_2=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2} , u_n=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2} +...\frac{1}{2^n} $
Now, $\sum_{n=1}^n u_n=u_1+u_2+u_3+....u_n$
$\sum_{n=1}^n u_n=[\frac{1}{2^0}+\frac{1}{2^1}]+[\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}]+....+[\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2} +...\frac{1}{2^n}]$
$\sum_{n=1}^n u_n=\frac{n}{2^0}+\frac{n}{2^1}+\frac{n-1}{2^2}+...+\frac{1}{2^n}$
$\sum_{n=1}^n u_n=n[\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2} +...\frac{1}{2^n}]-[\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+....+\frac{n-1}{2^n}]$
$\sum_{n=1}^n u_n=[n-\frac{n}{2^n}]-[\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+....+\frac{n-1}{2^n}]$
I solved the first first part using sum of GP and got $[n-\frac{n}{2^n}]$ but i am unable to solve the second part,So how could i do it or is there any other way to solve this question.
Answer to the question:$2^{-n}+2n-1$
Assuming your $\sum_{n=a}^n$ (which don't make sense) are $\sum_{k=a}^n$ :
$\displaystyle u_n=\sum_{k=0}^n\frac{1}{2^k}=\dfrac{1-\frac{1}{2^{n+1}}}{1-\frac{1}2}=2-\frac{1}{2^n}$ hence $\displaystyle\sum_{k=1}^n u_k=2n-\sum_{k=1}^n\frac{1}{2^k}=2n-\frac{1}2\dfrac{1-\frac{1}{2^n}}{1-\frac{1}2}$ thus $$\displaystyle\sum_{k=1}^n u_k=2n-1-\frac{1}{2^n}$$