Q-find-the-sum-S-2-3-2-3-3-3-4-3-4-then-find-n-1-n-4-n-

Question Number 78941 by M±th+et£s last updated on 21/Jan/20

Q . f i n d t h e s u m

S = \frac{2^{3}}{2!} + \frac{3^{3}}{3!} + \frac{4^{3}}{4!} + \dots .

t h e n f i n d \underset{n = 1}{\sum^{\infty}} \frac{n^{4}}{n!}

Answered by mind is power last updated on 21/Jan/20

= \sum_{n ⩾ 1} \frac{n^{3}}{(n - 1)!} = \sum_{k ⩾ 0} \frac{{(k + 1)}^{3}}{k!} = \sum_{k ⩾ 0} \frac{(k^{3} + 3 k^{2} + 3 k + 1)}{k!}

= \sum_{k ⩾ 1} \frac{k^{2}}{(k - 1)!} + 3 \sum_{k ⩾ 1} \frac{k}{(k - 1)!} + 3 \sum_{k ⩾ 1} \frac{1}{(k - 1)!} + \sum_{k ⩾ 1} \frac{1}{k!}

= \sum_{k ⩾ 0} \frac{{(k + 1)}^{2}}{k!} + 3 \sum_{k ⩾ 0} \frac{k + 1}{k!} + 3 \sum_{k ⩾ 0} \frac{1}{k!} + \sum_{k ⩾ 0} \frac{1}{k!} - 1

= \sum_{k ⩾ 0} \frac{k^{2} + 2 k + 1}{k!} + 3 \sum_{k ⩾ 0} \frac{k}{k!} + \sum_{k ⩾ 0} \frac{7}{k!} - 1

= \sum_{k ⩾ 1} \frac{k}{(k - 1)!} + 2 \sum_{k ⩾ 1} \frac{1}{(k - 1)!} + \sum_{k ⩾ 0} \frac{1}{k!} + 3 \sum_{k ⩾ 1} \frac{1}{(k - 1)!} + 7 \sum_{k ⩾ 0} \frac{1}{k!} - 1

= \sum_{k ⩾ 0} \frac{k + 1}{k!} + 2 \sum_{k ⩾ 0} \frac{1}{k!} + \sum_{k ⩾ 0} \frac{1}{k!} + 10 \sum_{k ⩾ 0} \frac{1}{k!} - 1

= \sum_{k ⩾ 1} \frac{1}{(k - 1)!} + \sum_{k ⩾ 0} \frac{1}{k!} + 2 \sum_{k ⩾ 0} \frac{1}{k!} + \sum_{k ⩾ 0} \frac{1}{k!} + 10 \sum_{k ⩾ 0} \frac{1}{k!} - 1

= 15 \sum_{k ⩾ 0} \frac{1}{k!} - 1 = 15 e - 1

Answered by Smail last updated on 22/Jan/20

\underset{n = 1}{\sum^{\infty}} \frac{n^{4}}{n!} = \underset{n = 1}{\sum^{\infty}} \frac{n^{3}}{(n - 1) (n - 2) (n - 3) (n - 4)!}

\frac{n^{3}}{(n - 1)!} = \frac{1}{(n - 1)!} + \frac{7}{(n - 2)!} + \frac{6}{(n - 3)!} + \frac{1}{(n - 4)!}

S = \underset{n = 4}{\sum^{\infty}} (\frac{1}{(n - 1)!} + \frac{7}{(n - 2)!} + \frac{6}{(n - 3)!} + \frac{1}{(n - 4)!}) + (1 + \frac{16}{2} + \frac{27}{2})

S = \frac{45}{2} + \underset{n = 4}{\sum^{\infty}} \frac{1}{(n - 1)!} + \underset{n = 4}{\sum^{\infty}} \frac{7}{(n - 2)!} + \underset{n = 4}{\sum^{\infty}} \frac{6}{(n - 3)!} + \underset{n = 4}{\sum^{\infty}} \frac{1}{(n - 4)!}

S = \frac{45}{2} + \underset{n = 3}{\sum^{\infty}} \frac{1}{n!} + \underset{n = 2}{\sum^{\infty}} \frac{7}{n!} + \underset{n = 1}{\sum^{\infty}} \frac{6}{n!} + \underset{n = 0}{\sum^{\infty}} \frac{1}{n!}

= \frac{45}{2} + (e - 1 - 1 - \frac{1}{2}) + 7 (e - 1 - 1) + 6 (e - 1) + e

= \frac{45}{2} + 15 e - \frac{5}{2} - 14 - 6

\underset{n = 1}{\sum^{\infty}} \frac{n^{4}}{n!} = 15 e

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