How can I solve this sum? $\sum_{k=1}^{n}k2^{k-1}$

Question

How can I solve this sum?

$\sum_{k=1}^{n}k2^{k-1}$

Is $\sum_{k=1}^{n}k2^{k-1}$ equals to $\sum_{k=1}^{n}k * \sum_{k=1}^{n}2^{k-1}$

The supposed equality $\sum_k a_kb_k=\sum_ka_k\sum_kb_k$ is not true, as you can check with simple examples — Marcel, Jun 05 '17 at 14:23
Elementary school! Well congrats on tackling a high school summation problem! — Franklin Pezzuti Dyer, Jun 05 '17 at 14:30
You have two questions here - the one in the title has the answer "No - it's false for $n=2$ (and indeed all $n>1$)", but the question of a closed form for the sum is quite different. Could you remove one of the questions from the question to make the title agree with the content? — Milo Brandt, Jun 05 '17 at 14:34
See also this one: https://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-limits-r-1d-r-cdot-2r — Hans Lundmark, Jun 05 '17 at 14:51

score 2 · Accepted Answer · answered Jun 05 '17 at 14:29

PARTIAL SOLUTION:

Here is how you would solve it. Set the sum equal to a variable $$S(n)=2^0+2*2^1+3*2^2+...+n*2^{n-1}$$ Then multiply both sides by $(1-2)$: $$S(n)(1-2)=(1-2)(2^0+2*2^1+3*2^2+...+n*2^{n-1})$$ Distribute: $$S(n)(1-2)=2^0+2*2^1-2^1+3*2^2-2*2^2+...+n*2^{n-1}-(n-1)*2^{n-1}-n*2^n$$ $$-S(n)=2^0+2^1+2^2+...+2^{n-1}-n2^n$$ Can you take it from here?

How can I solve this sum? $\sum_{k=1}^{n}k2^{k-1}$

1 Answers1