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If $T_{n} = (n^{2} + 1) n$ ! and $S_{n} = T_{1} + T_{2} + \dots + T_{n}$ , let $\frac{T_{50}}{S_{50}} = \frac{a}{b}$ , where a and b are relatively prime natural numbers, then value of $b - a$ is equal to

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answer is A.

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Detailed Solution

A We have

$T_{r} = [(r + 1) (r + 1)! - r . r!] - [(r + 1)! - r!]$

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