Mathematics question

Question

\(\sum\limits_{r = 1}^{20} (r^2 + 1)r!\) is equal to:

 

Answer 1

\(\sum\limits_{r = 1}^{20} (r^2 + 1)r!\)

\(=\sum\limits_{r = 1}^{20} (r^2 + 3r - 3r + 3 - 2)r!\)

\(=\sum\limits_{r = 1}^{20} (r^2 + 3r + 2 - 2r - 2 - r - 1 + 2)r!\)

\(=\sum\limits_{r = 1}^{20} ((r +2) (r +1) - 2(r + 1) - (r + 1) + 2)r!\)

\(=\sum\limits_{r = 1}^{20} ((r + 2)! - (r + 1)!) - 2\sum\limits_{r = 1}^{20} ((r + 1)! - r!)\)

\(= (22! - 2!) - 2(21! - 1!)\)

\(= 22! -2 (21!)\)