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This sum: $$\sum_{n=0}^\infty n\cdot\left(\frac{1}{2}\right)^{n-1}$$ can be calculated using this $\sum_{n=0}^\infty a^n=\frac{1}{1-a}$ ; $a<1$ cause we can calculate the sum $\sum_{n=0}^\infty \left(\frac{1}{2}\right)^{n}$ and then calculate it's derivative.

Can we calculate this sum: $$\sum_{n=0}^\infty (n-4)\cdot \left(\frac{1}{2}\right)^n$$ using a similar method?

If not, how can I calculate it?


marked as duplicate by Jyrki Lahtonen Mar 13 '18 at 21:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Hint: \begin{eqnarray*} \sum_{n=0}^{\infty} n x^n =\frac{x}{(1-x)^2}. \end{eqnarray*} $\endgroup$ – Donald Splutterwit Mar 13 '18 at 19:39
  • $\begingroup$ thank you so much! i got it. $\endgroup$ – M.Noussa Mar 13 '18 at 19:40
  • $\begingroup$ @M.Noussa Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Mar 17 '18 at 22:58

Note that

$$\sum_{n=0}^\infty (n-4)\cdot \left(\frac{1}{2}\right)^n=\sum_{n=0}^\infty n\cdot \left(\frac{1}{2}\right)^n-4\sum_{n=0}^\infty \left(\frac{1}{2}\right)^n$$

then use

$$\sum_{n=0}^{\infty} n x^n =\frac{x}{(1-x)^2}$$


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