The n^{th} term of the series frac{2}{1!} + f | vedclass

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(B) The given series is $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$The numerators are $2, 7, 15, 26, \dots$Let the numerator

Vedclass · Accepted Answer

$\frac{n(3n + 1)}{2(n!)}$

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(B) The given series is $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$The numerators are $2, 7, 15, 26, \dots$Let the numerator be $a_n$. The differences between consecutive terms are $5, 8, 11, \dots$,which form an Arithmetic Progression $(AP)$.Thus,the $n^{th}$ term of the numerator is the sum of an $AP$ with $a = 2$,$d = 3$ for $n$ terms:$a_n = \frac{n}{2}[2(2) + (n - 1)3] = \frac{n}{2}[4 + 3n - 3] = \frac{n(3n + 1)}{2}$.Therefore,the $n^{th}$ term of the series is $T_n = \frac{a_n}{n!} = \frac{n(3n + 1)}{2(n!)}$.Hence,the correct option is $B$.

The $n^{th}$ term of the series $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$ is

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