find-the-value-of-n-0-n-3-5-n-

Question Number 61937 by Tawa1 last updated on 12/Jun/19

find the value of \underset{n = 0}{\sum^{\infty}} \frac{n^{3} + 5}{n!}

Commented by mr W last updated on 12/Jun/19

s e e m s t o b e 10 e .

o t h e r s s h o u l d p r o v e .

Commented by Tawa1 last updated on 12/Jun/19

Yes, 10 e is correct, but workings sir

Commented by maxmathsup by imad last updated on 12/Jun/19

l e t S = \sum_{n = 0}^{\infty} \frac{n^{3} + 5}{n!} \Rightarrow S = \sum_{n = 0}^{\infty} \frac{n^{3}}{n!} + 5 \sum_{n = 0}^{\infty} \frac{1}{n!}

w e h a v e \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} = e^{x} w i t h r a d i u s R = \infty \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{n!} = e

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

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

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

= \sum_{n = 0}^{\infty} \frac{1}{n!} + 4 e = 5 e \Rightarrow S = 5 e + 5 e = 10 e .

Commented by Tawa1 last updated on 12/Jun/19

God bless you sir

Commented by maxmathsup by imad last updated on 12/Jun/19

y o u a r e w e l c o m e .

Answered by mr W last updated on 12/Jun/19

f (x) = \underset{n = 0}{\sum^{\infty}} \frac{(n^{3} + 5) x^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{n^{3} x^{n}}{n!} + 5 \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 1}{\sum^{\infty}} \frac{n^{2} x^{n}}{(n - 1)!} + 5 \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{{(n + 1)}^{2} x^{n + 1}}{n!} + 5 \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{(n^{2} + 2 n + 1) x^{n + 1}}{n!} + 5 \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{n^{2} x^{n + 1}}{n!} + 2 \underset{n = 0}{\sum^{\infty}} \frac{{n x}^{n + 1}}{n!} + \underset{n = 0}{\sum^{\infty}} \frac{x^{n + 1}}{n!} + 5 \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n + 1}}{(n - 1)!} + 2 \underset{n = 1}{\sum^{\infty}} \frac{x^{n + 1}}{(n - 1)!} + x \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} + 5 \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{(n + 1) x^{n + 2}}{n!} + 2 \underset{n = 0}{\sum^{\infty}} \frac{x^{n + 2}}{n!} + (x + 5) \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{(n + 1) x^{n + 2}}{n!} + 2 x^{2} \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} + (x + 5) \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{(n + 1) x^{n + 2}}{n!} + (2 x^{2} + x + 5) \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{{n x}^{n + 2}}{n!} + \underset{n = 0}{\sum^{\infty}} \frac{x^{n + 2}}{n!} + (2 x^{2} + x + 5) \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 1}{\sum^{\infty}} \frac{x^{n + 2}}{(n - 1)!} + x^{2} \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} + (2 x^{2} + x + 5) \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= \underset{n = 0}{\sum^{\infty}} \frac{x^{n + 3}}{n!} + (3 x^{2} + x + 5) \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= x^{3} \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} + (3 x^{2} + x + 5) \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= (x^{3} + 3 x^{2} + x + 5) \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

= (x^{3} + 3 x^{2} + x + 5) e^{x}

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{(n^{3} + 5) x^{n}}{n!} = (x^{3} + 3 x^{2} + x + 5) e^{x}

w i t h x = 1 :

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{n^{3} + 5}{n!} = (1^{3} + 3 \times 1^{2} + 1 + 5) e^{1} = 10 e

w i t h x = 2 :

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{2^{n} (n^{3} + 5)}{n!} = (2^{3} + 3 \times 2^{2} + 2 + 5) e^{2} = 27 e^{2}

w i t h x = \frac{1}{2} :

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{(n^{3} + 5)}{n! 2^{n}} = (\frac{1}{2^{3}} + 3 \times \frac{1}{2^{2}} + \frac{1}{2} + 5) e^{\frac{1}{2}} = \frac{51 \sqrt{e}}{8}

Commented by mr W last updated on 12/Jun/19

t h a n k y o u s i r s!

Commented by MJS last updated on 12/Jun/19

great job!

Commented by Tawa1 last updated on 12/Jun/19

God bless you sir .

Commented by Prithwish sen last updated on 12/Jun/19

excellent sir

Commented by malwaan last updated on 13/Jun/19

F A N T A S T I C s i r!

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