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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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Have a look at… – Gerry Myerson Jul 2 '12 at 0:15
Thanks a lot for your hint. – Tran Tam Jul 2 '12 at 10:41
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
  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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