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.

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 $\text{Nmax}>1$.

share|improve this question
    
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

2 Answers 2

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 $S$

share|improve this answer
    
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 $\sum\limits_{n=1}^Nnx^{n}$.
share|improve this answer
    
Thank you very much. Did! –  Tran Tam Jul 1 '12 at 12:07

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.