JEE Main 2016 (Online) 10th AprilQuestion 1
The sum $\sum_{r=1}^{10}\left(r^{2}+1\right)\times(r!)$, is equal to :
Sequence and Series · Telescoping ($V_n$ Method)
Solution
Formulas used. (1) $(r+1)!=(r+1)\,r!$(2) Telescoping: $\sum_{r=1}^{N}(V_{r+1}-V_r)=V_{N+1}-V_1$Step 1 - Split the term. With $V_r=(r-1)\,r!$, $V_{r+1}=r\,(r+1)!$ and $$V_{r+1}-V_r=r!\left[r(r+1)-(r-1)\right]=r!\left(r^2+1\right).$$ Step 2 - Telescope. $$\sum_{r=1}^{10}(r^2+1)\,r!=V_{11}-V_1=10\times11!-0=10\times(11!).$$ Answer: (B)