How to find $\sum_{k=1}^{n}{3^kk}$

Question

I need hint to find this sum $$\sum_{k=1}^{n}{3^kk}$$ $\sum_{k=1}^{n}{3^k}$ can be easily calculated as it is sum of geometric progression.

You could also look at http://math.stackexchange.com/questions/120091/how-do-we-get-the-result-of-the-summation-sum-limits-k-1n-k-cdot-2k/120100#120100 — Ross Millikan, Dec 18 '13 at 21:52

score 6 · Accepted Answer · answered Dec 18 '13 at 21:50

6

Hint

What's the derivative of the geometric sum $$\sum_{k=1}^n x^{k}?$$

answered Dec 18 '13 at 21:50

score 2 · Answer 2 · answered Dec 18 '13 at 22:17

An alternative solution:$$\begin{align}\sum 3^kk=3+2\cdot3^2+3\cdot3^3+4\cdot3^4&\ldots\\=3+3^2+3^3+3^4&\ldots\\+3^2+3^3+3^4&\ldots\\ +3^3+3^4&\ldots\\\vdots\\=\sum3^k+3\sum3^k+3^2\sum3^k&\ldots\end{align}$$ So if $S_n=\sum_{k=1}^{n}3^kk$, then $$\begin{align}S_n&=\sum_{k=1}^n3^k\sum_{i=0}^{n-k}3^i\\&=\sum_{k=1}^n3^{k}\frac{3^{n-k+1}-1}{2}\\&=\frac12n3^{n+1}-\frac12\sum_{k=1}^n3^{k}\\&=\frac12n3^{n+1}-\frac32\frac{3^{n}-1}{2}\\ &=\frac{2n3^{n+1}-3^{n+1}+3}{4}\\ &=\frac34\left((2n-1)3^{n}+1\right)\end{align}$$

How to find $\sum_{k=1}^{n}{3^kk}$

2 Answers2