Evaluating $\sum_{k=1}^{n} 2^{k}k$

Question

How can I evaulate the following series?

$$\sum_{k=1}^{n} 2^{k}k$$

I don't know where to begin. $2^k$ alone would be straight forward.

Answer 1

Notice that $\sum_{k=1}^nk2^k=\sum_{m=1}^n\sum_{k=m}^n2^k$. Can you conclude from here?

Answer 2

Hint: Prove by induction that $$\sum_{k=1}^{n}2^k\cdot k=2\left(1+2^n(n-1)\right)$$

Answer 3

$$ \sum_{k=1}^{n} 2^{k} = infinity$$

then your sum is greater so it to goes to infinity

The comparison test shows why.

https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-6/v/comparison-test-convergence

Answer 4

Call $a_n$ the expression you want to get. First, we have $$a_{n+1}=a_n+2^{n+1}(n+1)$$

Second, we have $2a_{n}=\sum_{k=1}^n2^{k+1}k$. So, $$2a_{n}+\sum_{k=1}^n2^{k+1}=\sum_{k=1}^n2^{k+1}(k+1)=a_{n+1}-2$$ So, $2a_n+2^{n+2}-4=a_{n+1}-2$. Hence $$2a_{n}+2^{n+2}-2=a_{n+1}$$

We have two linear equations with the variables $a_{n+1}$,$a_n$. Solve it.