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I have problem with finding the formula for the sum of $n$ initial terms of $$a(n)=n \cdot 2^n.$$

I only wrote the first few terms $\displaystyle a(1)=2$, $a(2)=8$, $a(3)=24$, $a(4)=64$

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I don't understand, what are you asking for? You already wrote out the first couple of terms, and you have the generic form of an $n$-th term. Do you want to sum the first $n$ terms? – gt6989b Feb 18 '14 at 17:14
yes, that's what I am asking for – Gregor Feb 18 '14 at 17:16
up vote 2 down vote accepted


Note that if we take the usual geometric series $$ \sum_{k=0}^n x^k = \frac{1 - x^{k+1}}{1-x}, \quad \forall x \neq 0, $$ we can differentiate it with respect to $x$ to get $$ \frac{d}{dx} \sum_{k=0}^n x^k = \sum_{k=1}^n k x^{k-1} = \frac{1}{x} \sum_{k=1}^n k x^k $$ on the left-hand side. Now differentiate the right-hand side using the quotient rule and you should be able to take it from there.

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seems a bit complicated – Gregor Feb 18 '14 at 17:24
No one said life was supposed to always be easy :-). On the other hand, this is a very useful and often needed result. It helps to derive it once and keep in your toolbox forever. – gt6989b Feb 18 '14 at 17:31
OK, thanks for help – Gregor Feb 18 '14 at 17:42
Complicated? gt gave a beautiful and, in retrospect, obvious solution. When solutions look like this, you're living the good life. – Addem Feb 18 '14 at 17:47

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