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

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(B) Given ${T_n} = ({n^2} + 1)n!$.We can rewrite ${T_n}$ as:${T_n} = (n^2 + n - n + 1)n! = (n(n+1) - (n-1))n! = n(n+1)! - (n-1)n!$.This is a telescopi

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$9$

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(B) Given ${T_n} = ({n^2} + 1)n!$.We can rewrite ${T_n}$ as:${T_n} = (n^2 + n - n + 1)n! = (n(n+1) - (n-1))n! = n(n+1)! - (n-1)n!$.This is a telescoping series form.${S_n} = \sum_{k=1}^{n} {T_k} = \sum_{k=1}^{n} [k(k+1)! - (k-1)k!]$.${S_n} = [1(2!) - 0(1!)] + [2(3!) - 1(2!)] + ... + [n(n+1)! - (n-1)n!]$.${S_n} = n(n+1)!$.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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