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Show with induction that

$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$

My solution:

Base case 1: n = 1

LHS = $(1+1)\cdot2^{1-1} = 2$

RHS = $1\cdot2^{1}= 2$

Case 2: n = p

When $LHS_{P}$ = $RHS_{P}$

$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(p+1)\cdot2^{p-1}=p\cdot2^{p}$

Case 3: n = p + 1

$LHS_{P+1}$ = $LHS_{P}$ + (p+2)$\cdot2^{p}$

$RHS_{P+1}$ = (p+1)$\cdot2^{p+1}$

So i need to to prove that:

$RHS_{P+1}$ = $RHS_{P}$ + (p+2)$\cdot2^{p}$

$RHS_{P+1}$ = $p\cdot2^{p}$ + (p+2)$\cdot2^{p}$

Am I thinking right here?

$RHS_{P+1}$ =

$(p+1)2^{p+1}=$

$(p+1)\cdot2^{p}\cdot2$ = ?

(Can I get this equal to $p\cdot2^{p}$ + (p+2)$\cdot2^{p}$ ? )

Daniel Andersson
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  • I would just try to simplify $(n+1)2^{n+1}-n2^n$. – Angina Seng Apr 12 '19 at 19:37
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Apr 12 '19 at 19:41

4 Answers4

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Hint: You have to prove that $$n\cdot2^n +(n+2)\cdot2^n=(n+1)\cdot2^{n+1}$$

Dr. Sonnhard Graubner
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Hint:

$$\underbrace{2 \cdot 2^0 + 3 \cdot 2^1 + \cdots (p-2) \cdot 2^{p-1} }_{ \text{use the induction hypothesis in this part} }+ (p-1) \cdot 2^{p}. $$

Community
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Adler Marques
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The solution without induction: $$2\cdot2^0+3\cdot2^1+...+(n+1)2^{n-1}=\frac{1}{2}\left(2\cdot2^1+3\cdot2^2+...+(n+1)2^n\right)=$$ $$=\frac{1}{2}\left(x^2+x^3+...+x^{n+1}\right)'_{x=2}=\frac{1}{2}\left(\frac{x^2(x^n-1)}{x-1}\right)'_{x=2}=$$ $$=\frac{1}{2}\left(\frac{\left((n+2)x^{n+1}-2x\right)(x-1)-\left(x^{n+2}-x^2\right)}{(x-1)^2}\right)_{x=2}=$$ $$=\frac{1}{2}\left((n+2)2^{n+1}-4-2^{n+2}+4\right)=\frac{1}{2}\cdot n2^{n+1}=n2^n.$$

Michael Rozenberg
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$LHS_{p+1} = RHS_{p+1}$

$p\cdot2^{p} + (p+2)\cdot2^{p}=(p+1)\cdot2^{p+1}$

$2^{p}(p+(p+2) = (p+1)2^{p}\cdot2$

$2p+2=2p+2$

Daniel Andersson
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