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How do you find the finite sum of $ \sum_{k = 1}^{n} {k2^k}$ . Any hint?
No integration and differentiation.

math123abc
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  • i reckon you should write out first few terms, notice a pattern and then prove that using induction. – Aditya_math Nov 13 '20 at 21:25
  • like this ? https://www.youtube.com/watch?v=i7iKLZQ-vCk – math123abc Nov 13 '20 at 21:29
  • Let $S_n=1\cdot 2+2\cdot 2^2+3\cdot 2^3+\ldots+n\cdot 2^n$. Then: $S_n=(2+2^2+2^3+\ldots+2^n)+(1\cdot 2^2+2\cdot 2^3+3\cdot 2^4+\ldots+(n-1)2^n)$ i.e. $S_n=2^{n+1}-2+2S_{n-1}$. Is this of any help? –  Nov 13 '20 at 21:30
  • Now we have to solve it as a recurrence equation. – math123abc Nov 13 '20 at 21:35

4 Answers4

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Hint: Note that $$\sum_{k=1}^n k 2^k = \sum_{j=1}^n \sum_{k=j}^n 2^k = \sum_{j=1}^n2^{j}\sum_{k=0}^{n-j} 2^k $$ and you have a formula for $$\sum_{k=0}^{n-j} 2^k.$$

Zestylemonzi
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Hint

Let $$S_n=\sum_{k=1}^nk2^k$$

then

$$2S_n=\sum_{k=1}^n(k+1-1)2^{k+1}$$

$$=S_n+(n+1)2^{n+1}-\sum_{k=1}^{n+1}2^k$$

Think of geometric sum.

hamam_Abdallah
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Borrowing @hamam_Abdallah's notation, call the sum $S_n$. The first few terms suggest$$S_n=(n-1)2^{n+1}+2.$$There's an easy proof by induction. The base step $n=1$ works (or $n=0$, if you prefer); for the inductive step, note$$k2^{k+2}-(k-1)2^{k+1}=(k+1)2^{k+1}.$$

J.G.
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Hint: Generalise the problem to computing, for any $x\ne 1$, $$\sum_{k=1}^nk\,x^k=x\sum_{k=1}^nk\,x^{k-1}=x\Bigl(\sum_{k=1}^nx^k\Bigr)'.$$

Bernard
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