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Question Number 138003 by benjo_mathlover last updated on 09/Apr/21

	Answered by EDWIN88 last updated on 09/Apr/21

	$b e g i n f r o m \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n!} = e^{x} - 1 . T a k i n g d e r i v a t i v e$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n - 1}}{n!} = e^{x}; \underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n + 1}}{n!} = x^{2} e^{x}$ $\frac{d}{d x} \underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n + 1}}{n!} = (2 x + x^{2}) e^{x}; w e g e t$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n (n + 1) x^{n}}{n!} = (2 x + x^{2}) e^{x}$ $m u l t p l y i n g w i t h x$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n (n + 1) x^{n + 1}}{n!} = (2 x^{2} + x^{3}) e^{x}$ $\frac{d}{d x} \underset{n = 1}{\sum^{\infty}} \frac{n (n + 1) x^{n + 1}}{n!} = (4 x + 5 x^{2} + x^{3}) e^{x}$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n {(n + 1)}^{2} x^{n}}{n!} = (4 x + 5 x^{2} + x^{3}) e^{x}$ $r e p l a c i n g x b y x^{2} g i v e s$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n {(n + 1)}^{2} x^{2 n}}{n!} = (4 x^{2} + 5 x^{4} + x^{6}) e^{x^{2}}$ $m u l t i p l y i n g b y x a g a i n g i v e s$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n {(n + 1)}^{2} x^{2 n + 1}}{n!} = (4 x^{3} + 5 x^{5} + x^{7}) e^{x^{2}}$ $t a k i n g d e r i v a t i v e a g a i n g i v e s$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n {(n + 1)}^{2} (2 n + 1) x^{2 n}}{n!} = (12 x^{2} + 33 x^{4} + 17 x^{6} + 2 x^{8}) e^{x^{2}}$ $s u b s t i t u t i n g x = \frac{1}{2}; w e g e t r e s u l t$ $\underset{n = 1}{\sum^{\infty}} \frac{n {(n + 1)}^{2} (2 n + 1)}{4^{n} . n!} = \frac{683}{128} e^{1 / 4}$
	Answered by Ñï= last updated on 09/Apr/21

	$\underset{n = 1}{\sum^{\infty}} \frac{n {(n + 1)}^{2} (2 n + 1)}{4^{n} n!} = \frac{1}{4} \underset{n = 0}{\sum^{\infty}} \frac{{(n + 2)}^{2} (2 n + 3)}{4^{n} n!}$ $= \frac{1}{4} {(x D + 2)}^{2} (2 x D + 3) e^{x} ∣_{x = \frac{1}{4}}$ $= \frac{1}{4} {(x D + 2)}^{2} (2 x + 3) e^{x} ∣_{x = \frac{1}{4}}$ $= \frac{1}{4} (x D + 2) (2 x^{2} + 9 x + 6) e^{x} ∣_{x = \frac{1}{4}}$ $= \frac{1}{4} (2 x^{3} + 17 x^{2} + 33 x + 12) e^{x} ∣_{x = \frac{1}{4}}$ $= \frac{1}{4} (\frac{1}{32} + \frac{17}{16} + \frac{33}{4} + 12) e^{\frac{1}{4}}$ $= \frac{683}{128} e^{\frac{1}{4}}$
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