If T_n = (n^2 + 1)n! and S_n = T_1 + T_2 + T_ | vedclass

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(B) Given $T_n = (n^2 + 1)n!$.We can rewrite $T_n$ as:$T_n = (n^2 + n - n + 1)n! = (n(n+1) - (n-1))n! = n(n+1)! - (n-1)n!$.Let $f(n) = n(n+1)!$. Then

Vedclass · Accepted Answer

$9$

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(B) Given $T_n = (n^2 + 1)n!$.We can rewrite $T_n$ as:$T_n = (n^2 + n - n + 1)n! = (n(n+1) - (n-1))n! = n(n+1)! - (n-1)n!$.Let $f(n) = n(n+1)!$. Then $T_n = f(n) - f(n-1)$.Thus,$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (f(k) - f(k-1)) = f(n) - f(0)$.Since $f(0) = 0(1!) = 0$,we have $S_n = n(n+1)!$.Now,$\frac{T_{10}}{S_{10}} = \frac{(10^2 + 1)10!}{10(11)!} = \frac{101 	imes 10!}{10 	imes 11 	imes 10!} = \frac{101}{110}$.Here,$a = 101$ and $b = 110$,which are relatively prime.Therefore,$b - a = 110 - 101 = 9$.

If $T_n = (n^2 + 1)n!$ and $S_n = T_1 + T_2 + T_3 + ...... + T_n$. Let $\frac{T_{10}}{S_{10}} = \frac{a}{b}$,where $a$ and $b$ are relatively prime natural numbers,then the value of $(b - a)$ is:

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