1- 2 3 2- - 3 3 3- - 4 3 4- - equals

The correct option is A 5e Here, T _ n = n+1 (n 2)! + 1 (n 1)! = 1 (n 3)! + 3 (n 2)! + 1 (n 1)! Therefore, ∑ T _ n =∑( 1 (n 3)! + 3 ...

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Sum of n Terms

1+ 2 3 2! + 3...

Question

$1 + \frac{2^{3}}{2!} + \frac{3^{3}}{3!} + \frac{4^{3}}{4!} + . . . . . . \infty$ equals

5 e

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4 e

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3 e

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2 e

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Solution

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