Question

$\sum_{r=1}^{20}$(r2+1)(r!) is equal to

• A. 22!–21!
• B. 22!–2(21!)
• C. 21!–2(20!)
• D. 21!–20!

Answer 1

Answer: (B)

22!–2(21!)

Answer 2

$\sum_{r=1}^{20}$(r2+1+2r−2r)r!=$\sum_{r=1}^{20}$((r+1)2−2r)r!
=$\sum_{r=1}^{20}$[(r+1)(r+1)!−rr!]−$\sum_{r=1}^{}$(r+1)r!=r!
=(2⋅2!–1!)+(3⋅3!–2⋅2!)+…+(21⋅21!–20⋅20!)−[(2!−1!)+(3!−2!)+…+(21!−20!)]
=(21⋅21!–1)−(21!−1)
=20⋅21!=(22−2)21!=22!–2(21!)
So, the correct option is (B): 22!–2(21!)