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Could you please help me to calculate this finite sum ? \begin{align} S=\sum _{n=1}^{\text{Nmax}} n*P^n \end{align} where $P\leq 1$ and $\text{Nmax}>1$.

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Could you please help me calculate your accept rate? Zero percent?? May I suggest you learn about accepting answers, and why it's considered a good thing to do. –  Gerry Myerson Jul 1 '12 at 11:17
    
Hi Gerry Myerson, I am sorry to say that I don't know what accept rate is. Could you please show me ? Thanks –  Tran Tam Jul 1 '12 at 12:06
    
Have a look at meta.math.stackexchange.com/questions/3286/… –  Gerry Myerson Jul 2 '12 at 0:15
    
Thanks a lot for your hint. –  Tran Tam Jul 2 '12 at 10:41
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2 Answers

up vote 2 down vote accepted

$$S=P.\sum_{n=1}^N nP^{n-1}=P.\frac{d}{dP}\sum_{n=1}^NP^n=P.\frac{d}{dP}\frac{P(1-P^N)}{1-P}$$ So $$S=\frac{(1-P)(1-(N+1)P^N)+P(1-P^N)}{(1-P)^2}$$

Simplify to get the answer.

Another way to do this ( and this does not need derivatives ) is

$$S=1.P+2.P^2+3.P^3\cdots +(N-1)P^{N-1}+N.P^N$$ Now $$S.P=1.P^2+2.P^3\cdots+(N-1)P^N+(N+1)P^{N+1}$$ So $$S-SP=P+P^2+P^3+\cdots+P^N+(N+1)P^{N+1}$$ Which implies $$S(1-P)=\frac{P(1-P^N)}{1-P}+(N+1)P^{N+1}$$ Now simplify to get the value of $S$

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Thank you very much –  Tran Tam Jul 1 '12 at 12:07
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  1. Start from the geometric series $\sum\limits_{n=1}^Nx^n=x\dfrac{1-x^N}{1-x}$.
  2. Differentiate both sides to get an expression of $\sum\limits_{n=1}^Nnx^{n-1}$.
  3. Multiply the result by $x$ to deduce $\sum\limits_{n=1}^Nnx^{n}$.
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Thank you very much. Did! –  Tran Tam Jul 1 '12 at 12:07
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