The sum ∑r=110(r2+1)×(r!) is equal to

∑r=110 r2+1r!=∑r=110 r2+r-(r-1)r! =∑r=110 r(r+1)r!-(r-1)r! =∑r=110 r(r+1)!-∑r=110 (r-1)r! =1(2!)+2(3!)+3(4!)+....10(11!) -1(2!)+2(3!)+....+9(10!) =10(11!)

The sum ∑r=110(r2+1)×(r!) is equal to

∑r=110 r2+1r!=∑r=110 r2+r-(r-1)r! =∑r=110 r(r+1)r!-(r-1)r! =∑r=110 r(r+1)!-∑r=110 (r-1)r! =1(2!)+2(3!)+3(4!)+....10(11!) -1(2!)+2(3!)+....+9(10!) =10(11!)

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The sum $\sum_{r = 1}^{10} (r^{2} + 1) \times (r!)$ is equal to

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$10 \times (11!)$

$(11!)$

$11 \times (11!)$

$101 \times (10!)$

answer is B.

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Detailed Solution

$\begin{array}{rcl} \sum_{r = 1}^{10} (r^{2} + 1) r! & = & \sum_{r = 1}^{10} (r^{2} + r - (r - 1)) r! \\ = & \sum_{r = 1}^{10} r (r + 1) r! - (r - 1) r! \\ = & \sum_{r = 1}^{10} r (r + 1)! - \sum_{r = 1}^{10} (r - 1) r! \\ = & (1 (2!) + 2 (3!) + 3 (4!) + . . . . 10 (11!)) \\ - (1 (2!) + 2 (3!) + . . . . + 9 (10!)) \\ = & 10 (11!) \end{array}$