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Problem: compute the unique positive integer n such that $2*2^2+3*2^3+4*2^4+...+n*2^n=2^{n+10}$

My attempt: First what i did was label the sum of $2*2^2...n*2^n=S$ then i multiplied $S$ by $2$ and subtracted $2S-S$ to get $2^3-2^3-2^4...-2^{n}+n2^{n+1}=S=2^{n+10}$ after some factoring and isolating I get $2^4+2^{n+1}(n-1)=2^{n+10}$ now the answer is supposed to be 500ish but i don't really know what I'm doing wrong any tips and answers would be helpful. I thank you in advance for your help.

Davide Giraudo
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allan
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2 Answers2

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$S = 2 * 2^2 + 3 *2^3 ... + n*2^n$ $2S = 2*2^3 ... + (n-1)2^n + n*2^{n+1}$

$2S - S = -2^3 - 2^3 ... - 2^n + n*2^{n+1}$

so, $2S - S$ was wrong

then, $(n-1)2^{n+1} = 2^{n+10}$

$n = 513$

Davide Giraudo
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Gil
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  • Please, search before answering a duplicate question. – jjagmath Jul 06 '22 at 14:07
  • Also, avoid the use of $*$ to denote multiplication. That's a common practice in programming, not in Mathematics, where it has other meanings. Use \cdot ($\cdot$) or \times ($\times$). – jjagmath Jul 06 '22 at 14:09
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Let S denote $\sum_{k=2}^{n}k*2^k$

2S=$\sum_{k=2}^{n}k*2^{k+1}$

=$\sum_{k=3}^{n+1}(k-1)*2^k$

=$(\sum_{k=2}^{n}(k-1)*2^k)+(n+1-1)*2^{n+1}-(2-1)*2^2$

=$\sum_{k=2}^{n}k*2^k-\sum_{k=2}^{n}2^k+n*2^{n+1}-4$

$2S=S-2^2\sum_{k=0}^{n-2}2^k+n*2^{n+1}-4$

$S=n*2^{n+1}-4-4(\frac{1-2^{n-1}}{1-2})$

$S=n*2^{n+1}-4-4*2^{n-1}+4$

$2^{n+10}=n*2^{n+1}-4*2^{n-1}$

$2^{n+10}=n*2^{n+1}-2^{n+1}$

$2^{n+10}=(n-1)*2^{n+1}$

$2^9=n-1$

n=513

Aqeel
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