MCQ
$\sum_{r=1}^{20}\left(r^{2}+1\right)(r !)$ is equal to:
  • A
    $22\,!-21 !$
  • $22\, !-2(21 \,!)$
  • C
    $21\, !-2 (20\,!)$
  • D
    $21 \,!-20\, !$

Answer

Correct option: B.
$22\, !-2(21 \,!)$
b
$\sum_{x=1}^{20}\left(r^{2}+1\right) r !$

$\sum_{x=1}^{20}\left((r+1)^{2}-2 r\right) r !$

$\sum_{x=1}^{20}((r+1)(r+1) !-r \cdot r !)-\sum_{r=1}^{20} r \cdot r !$

$\sum_{x=1}^{20}((r+1)(r+1) !-r \cdot r !)-\sum_{r=1}^{20}((r+1) !-r !)$

$=(21.21-1)-(\lfloor 21-1)$

$=20.21 !=22 !-2 \cdot 21 !$

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