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

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

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

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(B) The given series is $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$Observe the numerators: $2, 7, 15, 26, \dots$Let $a_n$ be the numerator of the $n^{th}$ term. The differences between consecutive terms are $5, 8, 11, \dots$,which form an Arithmetic Progression $(AP)$ with first term $a = 5$ and common difference $d = 3$.The $n^{th}$ term of the numerator sequence can be expressed as the sum of the first $n$ terms of this $AP$ plus the initial value:$a_n = 2 + \sum_{k=0}^{n-1} (5 + 3k) = 2 + \frac{n}{2} [2(5) + (n-1)3] = 2 + \frac{n}{2} [10 + 3n - 3] = 2 + \frac{n(3n + 7)}{2} = \frac{4 + 3n^2 + 7n}{2}$.Alternatively,looking at the pattern provided in the options:$T_n = \frac{2 + 5 + 8 + \dots + (3n - 1)}{n!} = \frac{\frac{n}{2}[2(2) + (n-1)3]}{n!} = \frac{\frac{n}{2}[4 + 3n - 3]}{n!} = \frac{n(3n + 1)}{2(n!)}$.

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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