Prove-that-i-n-0-n-2-n-2e-ii-n-0-n-3-n-5e-iii-n-0-n-4-n-15e-

Question Number 24540 by Tinkutara last updated on 20/Nov/17

P r o v e t h a t

(i) \underset{n = 0}{\sum^{\infty}} \frac{n^{2}}{n!} = 2 e .

(i i) \underset{n = 0}{\sum^{\infty}} \frac{n^{3}}{n!} = 5 e .

(i i i) \underset{n = 0}{\sum^{\infty}} \frac{n^{4}}{n!} = 15 e .

Commented by prakash jain last updated on 20/Nov/17

Write a polynomial in n P (n) o f

d e g r e e k a s

P (n) = a_{0} + a_{1} n + a_{2} n (n - 1) + . . + a_{k} n (n - 1) \dots (n - k + 1)

s o l v e f o r a_{0}, . . a_{k}

a n d c o n t i n u e a s i n c a s e o f n^{2} i n a n s w e r

Answered by prakash jain last updated on 20/Nov/17

n^{2} = n + n (n - 1)

\underset{n = 0}{\sum^{\infty}} \frac{n + n (n - 1)}{n!}

= \underset{n = 1}{\sum^{\infty}} \frac{1}{(n - 1)!} + \underset{n = 2}{\sum^{\infty}} \frac{1}{(n - 2)!}

= e + e = 2 e

Commented by Tinkutara last updated on 21/Nov/17

Thank you very much Sir!

Answered by ajfour last updated on 20/Nov/17

(i) {x e}^{x} = x + \frac{x^{2}}{1!} + \frac{x^{3}}{2!} + \frac{x^{4}}{3!} + \dots .

\frac{d}{d x} ({x e}^{x}) = 1 + \frac{2 x}{1!} + \frac{3 x^{2}}{2!} + \frac{4 x^{3}}{3!} + \dots

\Rightarrow e^{x} + {x e}^{x} = \frac{1^{2}}{1!} + \frac{2^{2} x}{2!} + \frac{3^{2} x^{2}}{3!} + \frac{4^{2} x^{3}}{4!} + . .

W i t h x = 1 w e g e t

2 e = \underset{n = 0}{\sum^{\infty}} \frac{n^{2}}{n!} .

Commented by Tinkutara last updated on 21/Nov/17

Thank you very much Sir!