Find-the-value-of-n-1-n-2-1-n-2-x-n-n-

Question Number 61922 by Tawa1 last updated on 11/Jun/19

Find the value of : \underset{n = 1}{\sum^{\infty}} \frac{n^{2} + 1}{n + 2} . \frac{x^{n}}{n!}

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

l e t p (x) = \sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n + 2} \frac{x^{n + 2}}{n!} \Rightarrow \frac{p (x)}{x^{2}} = \sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n + 2} \frac{x^{n}}{n!} l e t f i n d p (x)

p^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n!} x^{n + 1} = x {\sum_{n = 1}^{\infty} \frac{n^{2}}{n!} x^{n} + \sum_{n = 1}^{\infty} \frac{x^{n}}{n!}}

\sum_{n = 1}^{\infty} \frac{x^{n}}{n!} = e^{x} - 1

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

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

= (x^{2} + x) e^{x} \Rightarrow p^{^{'}} (x) = x {(x^{2} + x) e^{x} + e^{x} - 1} = x {(x^{2} + x + 1) e^{x} - 1}

= (x^{3} + x^{2} + x) e^{x} - x \Rightarrow p (x) = \int (x^{3} + x^{2} + x) e^{x} d x - \frac{x^{2}}{2} + c

\int {x e}^{x} = {x e}^{x} - \int e^{x} d x = {x e}^{x} - e^{x} = (x - 1) e^{x}

\int x^{2} e^{x} d x = x^{2} e^{x} - \int 2 x e^{x} d x = x^{2} e^{x} - 2 (x - 1) e^{x} = (x^{2} - 2 x + 2) e^{x}

\int x^{3} e^{x} d x = x^{3} e^{x} - \int 3 x^{2} e^{x} d x = x^{3} e^{x} - 3 (x^{2} - 2 x + 2) e^{x}

= (x^{3} - 3 x^{2} + 6 x - 6) e^{x} \Rightarrow

p (x) = (x^{3} - 3 x^{2} + 6 x - 6 + x^{2} - 2 x + 2 + x - 1) e^{x} - \frac{x^{2}}{2} + c

= (x^{3} - 2 x^{2} + 5 x - 5) e^{x} - \frac{x^{2}}{2} + c

p (0) = 0 = - 5 + c \Rightarrow c = 5 \Rightarrow p (x) = (x^{3} - 2 x^{2} + 5 x - 5) e^{x} - \frac{x^{2}}{2} + 5 \Rightarrow

\sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n + 2} \frac{x^{n}}{n!} = \frac{p (x)}{x^{2}} = (x - 2 + \frac{5}{x} - \frac{5}{x^{2}}) e^{x} - \frac{1}{2} + \frac{5}{x^{2}} (x \neq 0) .

Commented by Tawa1 last updated on 12/Jun/19

God bless you sir

Commented by Mr X pcx last updated on 12/Jun/19

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

Answered by Smail last updated on 12/Jun/19

\frac{n^{2} + 1}{n + 2} = \frac{(n + 2) (n - 2) + 5}{n + 2} = n - 2 + \frac{5}{n + 2}

L e t f (x) = \underset{n = 1}{\sum^{\infty}} \frac{n^{2} + 1}{n + 2} \times \frac{x^{n}}{n!}

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

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

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

= {x e}^{x} - 2 (e^{x} - 1) + \frac{5}{x^{2}} p (x)

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

p^{'} (x) = \underset{n = 1}{\sum^{\infty}} \frac{x^{n + 1}}{n!} = x (e^{x} - 1)

p (x) = \int_{0}^{x} ({t e}^{t} - t) d t = {[e^{t} (t - 1) - \frac{t^{2}}{2}]}_{0}^{x}

p (x) = e^{x} (x - 1) - \frac{x^{2}}{2} + 1

f (x) = {x e}^{x} - 2 e^{x} + 2 + \frac{5}{x^{2}} (e^{x} (x - 1) - \frac{x^{2}}{2} + 1)

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

Commented by Tawa1 last updated on 12/Jun/19

God bless you sir

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