Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\displaystyle a(1)=2$, $a(2)=8$, $a(3)=24$, $a(4)=64$

share|improve this question
    
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 at 17:14
    
yes, that's what I am asking for –  Gregor Feb 18 at 17:16

1 Answer 1

up vote 2 down vote accepted

Hint

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.

share|improve this answer
    
seems a bit complicated –  Gregor Feb 18 at 17:24
1  
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 at 17:31
    
OK, thanks for help –  Gregor Feb 18 at 17:42
1  
Complicated? gt gave a beautiful and, in retrospect, obvious solution. When solutions look like this, you're living the good life. –  Addem Feb 18 at 17:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.