1.40!+2.51!+3.62!+……=

1.40+2.51+3.62+…… =∑1∞ n2+3nn-1 = ∑1∞ (n2-1)+1+3(n-1)+3n-1 =∑1∞ n+1n-2+4n-1+3n-2 =∑1∞ 1n-3+3n-2+7e =e+3e+7e=11e

1.40!+2.51!+3.62!+……=

1.40+2.51+3.62+…… =∑1∞ n2+3nn-1 = ∑1∞ (n2-1)+1+3(n-1)+3n-1 =∑1∞ n+1n-2+4n-1+3n-2 =∑1∞ 1n-3+3n-2+7e =e+3e+7e=11e

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$\frac{1.4}{0!} + \frac{2.5}{1!} + \frac{3.6}{2!} + \dots \dots =$

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17e

15e

11e

answer is B.

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Detailed Solution

$\begin{array}{rcl} \frac{1.4}{0} + \frac{2.5}{1} + \frac{3.6}{2} + \dots \dots \\ = & \sum_{1}^{\infty} \frac{n^{2} + 3 n}{n - 1} \\ = & \sum_{1}^{\infty} \frac{(n^{2} - 1) + 1 + 3 (n - 1) + 3}{n - 1} \\ = & \sum_{1}^{\infty} (\frac{n + 1}{n - 2} + \frac{4}{n - 1} + \frac{3}{n - 2}) \\ = & \sum_{1}^{\infty} (\frac{1}{n - 3} + \frac{3}{n - 2}) + 7 e \\ = & e + 3 e + 7 e = 11 e \end{array}$