1.31!+2.42!+3.53!+4.64!+……∞ =……

1.31+2.42+3.53+4.64+……∞ =∑1∞ n(n+2)n =∑1∞ n+2n-1 =∑1∞(n-1)+3n-1 =∑1∞ 1n-2+3n-1 =e+3e=4e

1.31!+2.42!+3.53!+4.64!+……∞ =……

1.31+2.42+3.53+4.64+……∞ =∑1∞ n(n+2)n =∑1∞ n+2n-1 =∑1∞(n-1)+3n-1 =∑1∞ 1n-2+3n-1 =e+3e=4e

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$\frac{1.3}{1!} + \frac{2.4}{2!} + \frac{3.5}{3!} + \frac{4.6}{4!} + \dots \dots \infty = \dots \dots$

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Detailed Solution

$\begin{array}{rcl} \frac{1.3}{1} + \frac{2.4}{2} + \frac{3.5}{3} + \frac{4.6}{4} + \dots \dots \infty \\ = & \sum_{1}^{\infty} \frac{n (n + 2)}{n} \\ = & \sum_{1}^{\infty} \frac{n + 2}{n - 1} \\ = & \sum_{1}^{\infty} \frac{(n - 1) + 3}{n - 1} \\ = & \sum_{1}^{\infty} (\frac{1}{n - 2} + \frac{3}{n - 1}) \\ = & e + 3 e = 4 e \end{array}$