Assertion(A) : The value of 21!+62!+123!+204!+……∞=3eReason (R) : The nth term of the series 2, 6, 12, 20, ...... is n(n + 1)

21+62+123+204+……∞ =∑1∞ n(n+1)n =∑1∞ (n+1)n-1 =∑1∞ (n-1)+2n-1 =∑1∞ 1n-2+2n-1 =e+2e=3e

Assertion(A) : The value of 21!+62!+123!+204!+……∞=3eReason (R) : The nth term of the series 2, 6, 12, 20, ...... is n(n + 1)

21+62+123+204+……∞ =∑1∞ n(n+1)n =∑1∞ (n+1)n-1 =∑1∞ (n-1)+2n-1 =∑1∞ 1n-2+2n-1 =e+2e=3e

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Assertion(A) : The value of $\frac{2}{1!} + \frac{6}{2!} + \frac{12}{3!} + \frac{20}{4!} + \dots \dots \infty = 3 e$
Reason (R) : The $n^{th}$ term of the series 2, 6, 12, 20, ...... is n(n + 1)

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Detailed Solution

$\begin{array}{rcl} \frac{2}{1} + \frac{6}{2} + \frac{12}{3} + \frac{20}{4} + \dots \dots \infty \\ = & \sum_{1}^{\infty} \frac{n (n + 1)}{n} \\ = & \sum_{1}^{\infty} \frac{(n + 1)}{n - 1} \\ = & \sum_{1}^{\infty} \frac{(n - 1) + 2}{n - 1} \\ = & \sum_{1}^{\infty} (\frac{1}{n - 2} + \frac{2}{n - 1}) \\ = & e + 2 e = 3 e \end{array}$