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

Step 1: Claim: \[ \sum_{k=1}^{n}k\cdot k! = (n+1)!-1. \] Base case \( n=1 \): \( 1\cdot1!=1=2!-1. \)

Step 2: Assume the claim holds for \( n=m \). Then for \( n=m+1 \), \[ \sum_{k=1}^{m+1}k\cdot k!=(m+1)!-1+(m+1)(m+1)!=(m+1)!\big(1+(m+1)\big)-1=(m+2)!-1, \] so the pattern continues to hold.

Step 3: By induction the formula holds for every \( n \), so setting \( n=20 \) gives the required sum.
\[ \boxed{21!-1} \]