Jason99 6 Report post Posted May 2, 2017 (edited) \[ \begin{align} \sum_{n=1}^{k} \frac{n+2}{n! + (n+1)! + (n+2)!} &= \sum_{n=1}^{k} \frac{n+2}{n! ( 1 + n + 1 + (n+1)(n+2)) } \\ &= \sum_{n=1}^{k} \frac{n+2}{n! (n+2)(n+2) } \\ &= \sum_{n=1}^{k} \frac{1}{\frac{(n+2)!}{n+1}} \\ &= \sum_{n=1}^{k} \frac{n+1}{(n+2)!} \\ &= \sum_{n=1}^{k} \frac{n+2 - 1 }{(n+2)!} \\ &= \sum_{n=1}^{k} \frac{n+2}{(n+2)!} - \frac{1}{(n+2)!} \\ &= \frac{1}{(n+1)!} - \frac{1}{(k+2)!}\\ \end{align} \] \[ \begin{align} \sum_{n=1}^{k} d_n &= \sum_{n=1}^{k} \frac{n(n+1)}{2} \\ &= \sum_{n=1}^{k} \left(\frac{n^2}{2} + \frac{n}{2} \right) \\ &= \frac{\sum_{n=1}^{k} n^2}{2} + \frac{\sum_{n=1}^{k} n}{2} \\ &= \frac{\frac{k(k+1)(2k+1)}{6}}{2} + \frac{\frac{k(k+1)}{2}}{2} \\ &=\frac{k(k+1)(2k+1)}{12} + \frac{k(k+1)}{4} \\ &=\frac{k(k+1)(k+2)}{6} \\ \end{align} \] Edited May 2, 2017 by Jason99 Share this post Link to post Share on other sites
