Show $\sum_{k = 1}^n (k+1)2^k = n2^{n+1}$ by mathematical induction.

Question

I made my base case $n = 1$. So $(1+1)*2^1 = 1(2^1+1)$ and $4 = 4$, which is true.

My inductive hypothesis is assuming $n = m$, the summation between $k = 1$ to $ m$ of $(k+1)*(2^k) = (m)*2^{m+1}$.

I need to show that $n = m + 1$ is true. So I did the summation $k = 1$ to $m + 1$ of $(k+1)*(2^k) = (m+1)*2^{(m+1)+1}$. But, I am unsure of how to proceed from here.

To prove a statement of the form $\sum_{k=1}^nf(k)=g(n)$ by induction, it suffices to prove (i) $f(1)=g(1)$ and (ii) $f(n+1)=g(n+1)-g(n)$. — Angina Seng, May 07 '20 at 03:09
Does this answer your question? How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$? — Jyrki Lahtonen, May 07 '20 at 11:00
@JyrkiLahtonen: it's a lovely set of answers to a related question, but none of them answers this question. — Brian M. Scott, May 07 '20 at 12:26

score 4 · Accepted Answer · answered May 07 '20 at 03:13

4

You have to show that $\sum_{k=1}^{m+1}(k+1)2^k=(m+1)2^{m+2}$, presumably using the induction hypothesis that $\sum_{k=1}^m(k+1)2^k=m2^{m+1}$. I’ll get you started:

$$\begin{align*} \sum_{k=1}^{m+1}(k+1)2^k&=\sum_{k=1}^m(k+1)2^k+\left((m+2)2^{m+1}\right)\\ &\overset{(*)}=m2^{m+1}+(m+2)2^{m+1}\\ &=\ldots \end{align*}$$

The starred step is where the induction hypothesis is used.

answered May 07 '20 at 03:13

Brian M. Scott

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In the interest of saving your time and effort I recommend using Approach0 before posting. Many time duplicates of a FAQ often end up deleted, and that tool often helps when the obvious search string contains LaTeX-code. – Jyrki Lahtonen May 07 '20 at 11:09
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That is why you donvoted this answer? – James May 07 '20 at 20:31

Show $\sum_{k = 1}^n (k+1)2^k = n2^{n+1}$ by mathematical induction.

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