nice-calculus-n-1-n-2-n-4-n-

Question Number 138026 by mnjuly1970 last updated on 09/Apr/21

\dots \dots \dots n i c e \dots \dots \dots c a l c u l u s \dots \dots \dots . .

Θ = \underset{\overset{n = 1}{}}{\sum^{\infty}} (\frac{n^{2}}{n! . 4^{n}}) = ? ?

Answered by Dwaipayan Shikari last updated on 09/Apr/21

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

\underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n}}{n!} = {x e}^{x} \Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n^{2} x^{n}}{n!} = x^{2} e^{x} + {x e}^{x}

x = \frac{1}{4} \Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n^{2}}{n! 4^{n}} = \frac{5}{16} e^{1 / 4}

Commented by mnjuly1970 last updated on 09/Apr/21

t h a n k s a l o t \dots .

Answered by mathmax by abdo last updated on 09/Apr/21

let f (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n!} = e^{x} \Rightarrow f^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{{nx}^{n - 1}}{n!}

\Rightarrow {xf}^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{{nx}^{n}}{n!} \Rightarrow {({xf}^{^{'}} (x))}^{^{'}} = Σ \frac{n^{2} x^{n - 1}}{n!} \Rightarrow

x (f^{^{'}} (x) + {xf}^{(2)} (x)) = Σ \frac{n^{2} x^{n}}{n!} \Rightarrow x (e^{x} + {xe}^{x}) = Σ (\dots) \Rightarrow

Σ \frac{n^{2} x^{n}}{n!} = (x^{2} + x) e^{x} x = \frac{1}{4} \Rightarrow \sum_{n = 1}^{\infty} \frac{n^{2}}{n! 4^{n}} = (\frac{1}{4^{2}} + \frac{1}{4}) e^{\frac{1}{4}}

= \frac{5}{16} e^{\frac{1}{4}}

Commented by mnjuly1970 last updated on 09/Apr/21

g r a t e f u l s i r

Answered by Ñï= last updated on 09/Apr/21

\underset{n = 1}{\sum^{\infty}} \frac{n^{2}}{n! 4^{n}} = \frac{1}{4} \underset{n = 0}{\sum^{\infty}} \frac{n + 1}{n! 4^{n}} = \frac{1}{4} (x D + 1) e^{x} ∣_{x = \frac{1}{4}}

= \frac{1}{4} (x + 1) e^{x} ∣_{x = \frac{1}{4}} = \frac{5}{16} e^{1 / 4}

Commented by mnjuly1970 last updated on 09/Apr/21

t v h a n k s a l o t . .

Answered by mnjuly1970 last updated on 09/Apr/21