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Help the King out...

$$2+8+24+64+...+(n)(2^n)=2(1+(n-1)(2^n))$$

I am at the step where I am proving $P(k+1)$ to be true:

$$2(1+(k-1)(2^k))+(k+1)((2)^{k+1}))=2(1+((k+1)-1)(2^{k+1}))$$

Martin Sleziak
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King Squirrel
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2 Answers2

3

$\sum^{n}_{k=1}k2^k$ is by the induction hypothesis, $2(1+(n-1)2^n)$. In $n+1$, then, we need to prove that $\sum^{n+1}_{k=1}k2^k=(n+1)2^{n+1}+\sum^{n}_{k=1}k2^k=2(1+n2^{n+1})$, i.e., $$n2^{n+1}+2^{n+1}+\sum^{n}_{k=1}k2^k=2(1+n2^{n+1})=2+2n2^{n+1}\\ \implies 2+(n-1)2^{n+1}=\sum^{n}_{k=1}k2^k\\ \implies 2(1+(n-1)2^{n})=\sum^{n}_{k=1}k2^k$$ This is the induction hypothesis. Q.E.D.

2

See part I of my answer here for the background to the following systematic approach.

We have here $f(k) = k\cdot2^k$ and $g(n) = 2 + (n-1)\cdot 2^{n+1}$

Inductive step:

1: Assume true for $n$, that is $\sum_{k=1}^{n}f(k) = g(n)\tag{1}$

2: Let $m = n + 1$

$\begin{align}f(m) &= m\cdot2^m\\\\ g(m) - g(m-1) &= \left(2 + (m-1)\cdot2^{m+1}\right) - \left(2 + (m-2)\cdot2^{m}\right)\\&=(2m-2 - m + 2)\cdot2^m\\&=m\cdot2^m\\\\\therefore f(m) &= g(m) - g(m-1)\\\text{i.e } \color{blue}{f(n+1)}&=\color{blue}{g(n+1) - g(n)}\tag{2}\end{align}$

We thus have $$\begin{align}\sum_{k=1}^{n+1}f(k)=\sum_{k=1}^{n}f(k) + f(n+1) = \underbrace{g(n)}_{\text{from }(1)} + \underbrace{g(n+1) - g(n)}_{\text{from }(2)} = g(n+1)\end{align}$$

implies that the statement is true for $n+1$.

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