sum_{r=1}^{20} (r^{2}+1)(r!) is equal to: | vedclass

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(A) We need to evaluate the sum $S = \sum_{r=1}^{20} (r^{2}+1)r!$. Rewrite the term $(r^{2}+1)$ as $(r^{2}+r-r+1) = (r(r+1) - (r-1))$. Alternatively,o

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$22! - 2(21!)$

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(A) We need to evaluate the sum $S = \sum_{r=1}^{20} (r^{2}+1)r!$. Rewrite the term $(r^{2}+1)$ as $(r^{2}+r-r+1) = (r(r+1) - (r-1))$. Alternatively,observe that $(r^{2}+1)r! = (r^{2}+r-r+1)r! = (r(r+1) - (r-1))r! = r(r+1)! - (r-1)r!$. Let $T_r = r(r+1)! - (r-1)r!$. This is a telescoping sum: $S = \sum_{r=1}^{20} [r(r+1)! - (r-1)r!] = [1(2!) - 0(1!)] + [2(3!) - 1(2!)] + [3(4!) - 2(3!)] + \dots + [20(21!) - 19(20!)]$. All intermediate terms cancel out,leaving $S = 20(21!) - 0 = 20 	imes 21!$. We can rewrite $20 	imes 21!$ as $(22-2) 	imes 21! = 22 	imes 21! - 2 	imes 21! = 22! - 2(21!)$. Thus,the sum is $22! - 2(21!)$.

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

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