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Suppose we are asked to solve the following summation

$$\sum_{i=1}^{n} (i2^{i-1})$$

Below is my attempt at solving this..

I know that the quotient rule for exponents can help me simplify the summation.

$$\sum_{i=1}^{n} (i2^{i}/2)$$

now i can extract the constant from the summation..

$$ 1/2\sum_{i=1}^{n} (i2^{i})$$

this is where i am stuck.. can anyone give me some advice on what to do next?

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2 Answers 2

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If you don't want to do calculus, you need to solve for it instead of directly calculating it. Define $S$ as,

$$S=\sum_{i=0}^n(i2^{i-1}),$$

$$2S=\sum_{i=0}^n(i2^i)=\sum_{i=1}^{n+1}[(i-1)2^{i-1}],$$

$$2S-S=n2^n+\sum_{i=1}^n\left\{[(i-1)-i]2^{i-1}\right\}=n2^n-\sum_{i=1}^n 2^{i-1}=n2^n-2^n+1=(n-1)2^n+1.$$

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$$\sum_{i=0}^ni2^{i-1}=\left(\sum_{i=0}^nx^i\right)'_{x=2}=\left(\frac{x^{n+1}-1}{x-1}\right)'_{x=2}=$$ $$=\left(\frac{(n+1)x^{n}(x-1)-x^{n+1}+1}{(x-1)^2}\right)_{x=2}=(n+1)2^{n}-2^{n+1}+1.$$

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  • $\begingroup$ Im sorry, there was a typo in the summation, $i$ is supposed to start at 1, not 0 $\endgroup$ Jan 27, 2018 at 7:33
  • $\begingroup$ @Soon_to_be_code_master What difference does it make? $\endgroup$ Jan 27, 2018 at 7:34
  • $\begingroup$ They are the same. $\endgroup$ Jan 27, 2018 at 7:35
  • $\begingroup$ I was wondering if you could explain more in detail on the steps you took to get to $(n+1)2^n−2^{n+1}+1.$. more specifically, how did you get to$((n+1)x^n(x−1)−x^{n+1}+1)/ ((x - 1)^2)$ $\endgroup$ Jan 27, 2018 at 8:39
  • $\begingroup$ I just substituted $x=2$ in $\frac{(n+1)x^{n}(x-1)-x^{n+1}+1}{(x-1)^2}.$ About your second question. I used $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}.$ $\endgroup$ Jan 27, 2018 at 8:42

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