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I'm trying to find the closed-form expression for the following equation: $$E[D] = \sum_{i=1}^{11}(11-i)(1-p)^{i-1}p + \sum_{i=12}^{\infty}(i-11)(1-p)^{i-1}p$$

My initial thought was to distribute the $(1-p)^{i-1}p$ in each sum into the $(11-i)$ or $(i-11)$, so that we have the following equation: $$E[D]= 11p\sum_{i=1}^{11}(1-p)^{i-1} - p\sum_{i=1}^{11}i(1-p)^{i-1} + p\sum_{i=12}^{\infty}i(1-p)^{i-1} - 11p\sum_{i=12}^{\infty}(1-p)^{i-1}$$ but I'm not sure where to go from here. Can anyone help me figure out how to get rid of the summations?

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If you split up the polynomials of order 10, the leftover has the format $\sum_{i=1}^\infty i(1-p)^{i-1} = 1/p^2$, the standard binomial expansion:

$ E/p = \sum_{i=1}^{11} (11-i) (1-p)^{i-1} + \sum_{i\ge 12} (i-11)(1-p)^{i-1} $ $ = -\sum_{i=1}^{11} (i-11) (1-p)^{i-1} + \sum_{i\ge 12} (i-11)(1-p)^{i-1} $ $ = -2\sum_{i=1}^{11} (i-11) (1-p)^{i-1} + \sum_{i\ge 1} (i-11)(1-p)^{i-1} $ $ = -2\sum_{i=1}^{11} (i-11) (1-p)^{i-1} + \sum_{i\ge 1} i(1-p)^{i-1} -11 \sum_{i\ge 1} (1-p)^{i-1} $ $ = -2\sum_{i=1}^{11} (i-11) (1-p)^{i-1} + \frac{1}{p^2} -\frac{11}{p} $

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