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

Kartik Watwani
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    $\sum_{n=0}^{n} \frac{1}{2^n}$ doesn't make any sense, you can't have the same variable as a counter and as a bound – Hippalectryon Jul 01 '15 at 12:19
  • you just have to find $u_1,u_2,...u_n$ from$ u_n= \sum_{n=0}^{n} \frac{1}{2^n}$ and then sum $u_1,u_2,,,,u_n$,and this is the question my mate ,it also has a answer which am posting. – Kartik Watwani Jul 01 '15 at 12:22
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    I am just saying that $\sum_{n=0}^n$ doesn't mean anything. You need to use another variable as the indices. Did you mean $\sum_{k=0}^n\frac{1}{2^k}$ maybe ? – Hippalectryon Jul 01 '15 at 12:24
  • @Hippalectryon ya you can use that ,it doesn't make any change ti question as well as answer. – Kartik Watwani Jul 01 '15 at 12:26
  • @Hippalectryon this is the question printed in the book i have not made any changes. – Kartik Watwani Jul 01 '15 at 12:27
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    You may have copied the question as it is from your book, but it still doesn't make sense that way. No books are exempt from errors/typos. In my answer I thus changed the indices to the most natural way they should probably be. (since my answer matches the expected answer, I guess it's fine) – Hippalectryon Jul 01 '15 at 12:29

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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 $$\displaystyle\sum_{k=1}^n u_k=2n-1-\frac{1}{2^n}$$

Hippalectryon
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