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Does $(m + 1) + m2 + (m - 1)2^{2} \ldots + 2^{m}$ equal something simpler, where $m\in \mathbb{N}$? Excuse me if it is too simple, I am bit tired.

Thanks.

Lisa
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TKM
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  • instead of just downvoting, specify what is the problem, so I can learn. – TKM Aug 20 '14 at 20:07
  • I didn't downvote, but: A lot of people downvote if there is no evidence of effort on your part. – Thomas Andrews Aug 20 '14 at 20:14
  • Excuse me what is sth simpler? (not native english, sorry) – rlartiga Aug 20 '14 at 20:15
  • @rlartiga "something simpler" – RE60K Aug 20 '14 at 20:17
  • @Aditya thanks! – rlartiga Aug 20 '14 at 20:17
  • You've got bigger problems than texting slang. For one thing, while $m2$ is technically valid, it makes me wonder if it really means what you want it to mean. Do you mean twice $m$? That would normally be written as $2m$. Or do you mean $m$ squared? That would be written $m^2$. I doubt you mean $m_2$, but I can't rule that out. – Lisa Aug 20 '14 at 22:29
  • @Lisa, what else did you want to know at the problem where you placed a bounty? Note that, as far as I know, the person who originally answered is no longer active on MSE – Will Jagy Sep 23 '14 at 21:33
  • @Lisa, please take a look at http://math.blogoverflow.com/2014/08/23/binary-quadratic-forms-over-the-rational-integers-and-class-numbers-of-quadratic-%EF%AC%81elds/ perhaps you will get some ideas for a focused question. – Will Jagy Sep 23 '14 at 21:51

4 Answers4

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Here's a hint, and a good trick in general, called diagonalization.

Write your sum in the form:

1 + 1 + 1 + ... + 1
    2 + 2 + ... + 2 
        4 + ... + 4
            ...
                 2^m

And compute the sum by columns rather than rows.

forallepsilon
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2

You'll find essentially the same thing here.

But note that that it looks like your sum should end with $2^{m-1}$ and not $2^m$.

Community
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Frunobulax
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$$\sum_{i=0}^m (m+1-i)2^{i}=(m+1)\sum_{i=0}^m 2^i- \sum_{i=0}^m i2^{i}$$

At that point you should be able to calculate it.

rlartiga
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$$\begin{align}S=(m+1)+&2m\;\;\;\;\;\;\;\;+2^2(m-1)+2^2(m-2)+...2^{m-1}\\2S=\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;&2(m+1)+2^2m\;\;\;\;\;\;\;\;+2^3(m-1)+...(m-(m-2))2^{m-1}+2^m\end{align}$$ Subtract those: $$S=(m+1)-2-2^2...-2^{m-1}-2^m=(m+1)-2(2^m-1)$$

RE60K
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