frac{10001 times 100 !}{2 times 1 !+5 times 2 | vedclass

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(B) Let the general term of the denominator be $T_n = (n^2+1)n!$.We can write $T_n = (n^2+2n+1-2n)n! = (n+1)^2 n! - 2n \cdot n!$.This can be simplifie

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$\frac{10001}{10100}$

Vedclass · Answer

(B) Let the general term of the denominator be $T_n = (n^2+1)n!$.We can write $T_n = (n^2+2n+1-2n)n! = (n+1)^2 n! - 2n \cdot n!$.This can be simplified as $T_n = (n+1)(n+1)! - 2n \cdot n!$.Alternatively,note that $(n^2+1)n! = (n^2+n-n+1)n! = n(n+1)! - (n-1)n!$.Actually,the sum $S = \sum_{n=1}^{100} (n^2+1)n!$.Using the identity $(n^2+1)n! = (n+1)!n - (n)! (n-1)$ is not quite right.Let's use $(n^2+1)n! = (n+1)!n - n! + n! = (n+1)!n - (n-1)n!$.The sum is $\sum_{n=1}^{100} ((n+1)!n - n!(n-1)) = 100 \cdot 101! - 0 = 100 \cdot 101!$.The expression becomes $\frac{10001 	imes 100!}{100 	imes 101 	imes 100!} = \frac{10001}{10100}$.

$\frac{10001 \times 100 !}{2 \times 1 !+5 \times 2 !+10 \times 3 !+\ldots+10001 \times 100 !}=$

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