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The Sum $\sum_{r=1}^{20}(r^{2}+1)\times r!$ is equal to

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The "Method of Differences" is the most powerful tool for factorial sums. Always look for a way to write the coefficient as $(r+k) - (\dots)$ to shift the index of the factorial.

Updated On: Jun 6, 2026

$21\times21!$
$20\times21!$
21!
$20\times21\times20!$

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The Correct Option is B

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The Sum $\sum_{r=1}^{20}(r^{2}+1)\times r!$ is equal to

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The Correct Option is B

Solution and Explanation

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