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How would I go about solving the problem:

$$\sum_{n = -1}^{15}n 2^n$$

Edit: I've already looked at similar questions to these but none of the answers made sense to me. I'd really appreciate it if someone could explain it to me in a very easy to understand method. Thanks so much.

Raisa Llyn
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  • See https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence – lab bhattacharjee Aug 16 '15 at 03:02
  • do you mean $$\sum_{n=-1}^{15}n\cdot 2^n$$ – Dr. Sonnhard Graubner Aug 16 '15 at 03:03
  • Seems clear that there 17 terms in the sum: for n = -1, 0, 1, 2, etc. Also, seems clear the first few terms are -0.5, 0, 2, 8, 24, etc. Please make sure you have stated the problem correctly and/or say what you don't understand about the solution. – BruceET Aug 16 '15 at 03:42
  • The problem was stated correctly. I'm just going to ask a classmate about the problem because I don't understand derivatives (which is what the answers below have used). – Raisa Llyn Aug 16 '15 at 03:56
  • See the duplicate question where the accepted answer does not use derivatives. Indeed one should not use differentiation for this problem so that the result generalizes (to arbitrary integral domains for those who know what they are). – user21820 Aug 16 '15 at 05:37

3 Answers3

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Consider the finite series with general term $a^n$, for which there's a well-known formula.

Now differentiate both sides with respect to $a$, then multiply both sides by $a$, and finally substitute $a=2$.

Luis Mendo
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Hint: $$\sum_{n=a}^b n\cdot 2^n=(b-1) \cdot 2^{b+1}-(a-2) \cdot 2^a$$

Hint at proof: examine both $\sum_{n=a}^b 2^n$ and its sum and then take the derivative of both. You'll have to make a small extra step to finish.

Here are some intuitive proofs that do in fact work.

Community
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Zach466920
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  • How did you get that formula? Also, sorry for the newbie question, but what do you mean by a derivative? – Raisa Llyn Aug 16 '15 at 03:09
  • @RaisaLlyn it's gotten by taking the derivative of the function inside the series and the formula for its sum, it's a geometric series, and then comparing. If you're unfamiliar with the derivative, then feel free to just use the formula provided. – Zach466920 Aug 16 '15 at 03:11
  • I don't agree with saying "just use the formula" provided. It is what makes mathematics inaccessible to students. – user21820 Aug 16 '15 at 05:38
  • @user21820 I tried to provide a proof, please don't ignore that. Sometimes there aren't good non-hand-wavy proofs unless you have the skills. – Zach466920 Aug 16 '15 at 14:07
  • I don't mind hand-wavy proofs as long as they are explicitly stated to be non-rigorous and actually can be made rigorous. In fact I do like intuitive proofs that fall into this category, because going completely formal is not the way to do mathematics. I was only objecting to telling students to "just use it blindly" (it's an exaggerated paraphrase but I hope you get the point). That's all. =) – user21820 Aug 16 '15 at 14:13
  • @user21820 I added a very good link, also makes this post a duplicate :) – Zach466920 Aug 16 '15 at 14:21
  • Did you notice it's the same link as the one the question is marked as duplicate of? Anyway yes that's the way I would do it. In general you can check up summation by parts which is a 'more systematic' approach to such series than ad-hoc subtraction, and also applies immediately to sums of $k$-th powers. – user21820 Aug 16 '15 at 14:24
  • And I didn't realize you were the same one on Physics SE until just now haha.. – user21820 Aug 16 '15 at 14:26
  • @user21820 oh, on the bright side, we agree about math ;) – Zach466920 Aug 16 '15 at 14:27
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    Haha it is not like we can prove those quantum mechanics assumptions correct anyway, so whatever I said there has to be taken as conditional under assumption of currently accepted hypotheses. Then again, we cannot prove our axioms of PA correct either... They just seem right enough for much enough of the world so that's great! – user21820 Aug 16 '15 at 14:33
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Continuing from Luis Mendo's answer, consider more generally $$A=\sum_{i=a}^n x^i$$ which is a geometric progression and the well-known formula is $$A=\sum_{i=a}^n x^i=\frac{x^{n+1}-x^a}{x-1}$$ Now, consider $$B=\sum_{i=a}^n ix^i=x\times\sum_{i=a}^n ix^{i-1}=x \times\frac{dA}{dx}$$ So, take the derivative of $A$ with respect to $x$; this gives you the formula for $B$. Finally, replace, $a,n,x$ by their values.

Claude Leibovici
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