JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Sequences and Series

Look for $f(r)=r!g(r)$ with $g$ quadratic such that $f(r+1)-f(r)=r!\left(r^3+6r^2+2r+5\right)$. Since $f(r+1)-f(r)=r!\left[(r+1)g(r+1)-g(r)\right]$, the requirement is

Writing $g(r)=\alpha r^2+\beta r+\gamma$ and comparing coefficients gives $\alpha =1$, $\beta =4$, $\gamma =-5$, and the constant terms agree as well, so $g(r)=r^2+4r-5$ works. The sum therefore telescopes:

Hence $\alpha =160$.