Question:

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

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

Updated On: Jul 16, 2026

$20!-1$
$21!-1$
$22!-2$
$21!$

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The Correct Option is B

Approach Solution - 1

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 $21!-1$.

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Approach Solution -2

Instead of expanding the general telescoping sum symbolically, check the pattern using small cases first and then match it against the options.

Option A ($ 20!-1 $): With $ 2 $ terms, the sum is $ 1\cdot1!+2\cdot2!=1+4=5 $. The matching factorial expression is $ 3!-1=5 $, i.e. one factor higher than the last term, not equal to it. So the pattern is $ (n+1)!-1 $, meaning the full sum should end in $ 21!-1 $, not $ 20!-1 $; this option is rejected.
Option B ($ 21!-1 $): Testing further, with $ 3 $ terms the sum is $ 1+4+3\cdot6=23 $, and $ 4!-1=23 $, matching again. Extending this verified pattern $ (n+1)!-1 $ to $ 20 $ terms gives $ 21!-1 $, consistent with the required sum.
Option C ($ 22!-2 $): This does not fit the verified pattern $ (n+1)!-1 $ for any of the tested small cases; rejected.
Option D ($ 21! $): The pattern consistently gives one less than the factorial, i.e. $ (n+1)!-1 $, not the bare factorial $ 21! $; this fails the small-case checks and is rejected.

The pattern $ (n+1)!-1 $, verified for $ n=1,2,3 $, extends to give $ 21!-1 $ for the full sum up to $ 20 $ terms.

Hence, the correct answer is $ 21!-1 $.