# How can I evaluate$\sum\limits_{k=1}^n\,k\, 2^k$? [duplicate]

How can I evaluate the sum $$\displaystyle\sum_{k=1}^n\,k\, 2^k$$? I should probably use binomial coefficients, but how?

## marked as duplicate by Guy Fsone, user236182, Winther, Hans Lundmark, rtybaseJan 22 '18 at 8:45

• The relation $(k+1)2^{k+1} - k2^k = k2^k + 2^{k+1}$ can be used to evaluate the sum as a telescoping series + a geometrical series. – Winther Jan 22 '18 at 4:40
• @GuyFsone Thanks for your effort to improve posts on this sites by editing both post and the titles, that's a useful thing to do. However, it is better not to include stuff such as \displaystyle or \dfrac in the title. For more details see this post on meta: Guidelines for good use of $\LaTeX$ in question titles – Martin Sleziak Jan 24 '18 at 7:16
$$\sum_{k=1}^{n}k2^k= 2\sum_{k=1}^{n}k2^{k-1}= 2\left(\sum_{k=1}^{n}x^{k}\right)'\bigg|_{x=2}= 2\left(x\frac{x^{n}-1}{x-1}\right)'\bigg|_{x=2}\\= 2\left(\frac{((n+1)x^{n}-1)(x-1)-x^{n+1}+x}{(x-1)^2}\right)\bigg|_{x=2} =((n-1)2^{n+1}+2)$$