Question

Find the value of $1(1!)+2(2!)+3(3!)+\cdots+20(20!)$. 

• A. $20!-1$
• B. $21!-1$
• C. $22!-2$
• D. $21!$

Hint:

Memorize $n\cdot n!=(n+1)!-n!$; it turns many factorial sums into a one-line telescoping result.

Answer 1

The Correct Option is B

Use the identity $n\cdot n!=(n+1)!-n!$. Then the sum telescopes: \[ \sum_{n=1}^{20} n\cdot n! = \sum_{n=1}^{20} \big((n+1)!-n!\big) = \underbrace{(2!-1!)}_{n=1}+\underbrace{(3!-2!)}_{n=2}+\cdots+\underbrace{(21!-20!)}_{n=20} =21!-1!. \] Hence the value is $21!-1$.