This time I will provide both the problem and solution in the continuing series from Dr. Andreescu’s lecture on Telescoping Sums and Products.
Example 2. Evaluate
Example 2. Solution
We can write
![\displaystyle\sum^{n}_{k=1}k!(k^2+k+1)=\displaystyle\sum^{n}_{k=1}[(k+1)^2-k)]k!](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Csum%5E%7Bn%7D_%7Bk%3D1%7Dk%21%28k%5E2%2Bk%2B1%29%3D%5Cdisplaystyle%5Csum%5E%7Bn%7D_%7Bk%3D1%7D%5B%28k%2B1%29%5E2-k%29%5Dk%21&bg=ffffff&fg=333333&s=0 "\displaystyle\sum^{n}_{k=1}k!(k^2+k+1)=\displaystyle\sum^{n}_{k=1}[(k+1)^2-k)]k!")
![=\displaystyle\sum^{n}_{k=1}[(k+1)!(k+1)-k!k]](https://s0.wp.com/latex.php?latex=%3D%5Cdisplaystyle%5Csum%5E%7Bn%7D_%7Bk%3D1%7D%5B%28k%2B1%29%21%28k%2B1%29-k%21k%5D&bg=ffffff&fg=333333&s=0 "=\displaystyle\sum^{n}_{k=1}[(k+1)!(k+1)-k!k]")
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