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Question Number 104517 by Quvonchbek last updated on 22/Jul/20

	Answered by 1549442205PVT last updated on 22/Jul/20

	$It is easy to see that the sequence of numbers :$ $11, 19, 29, 41, . . . has general term is$ $n^{2} + 3 n + 1 (n \in N^{}) . Hence,$ $S = \underset{n = 1}{\overset{\infty}{Σ}} \frac{n^{2} + 3 n + 1}{n!} = \underset{k = 1}{\overset{\infty}{Σ}} \frac{n^{2}}{n!} + \underset{n = 1}{\overset{\infty}{Σ}} \frac{3 n}{n!} + \underset{n = 1}{\overset{\infty}{Σ}} \frac{1}{n!} ()$ $On the other hands, we have$ $e = \underset{n = 0}{\overset{\infty}{Σ}} \frac{1}{n!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + . . . Hence,$ $\underset{n = 1}{\sum^{\infty}} \frac{3 n}{n!} = \frac{3}{1!} + \frac{3.2}{2!} + \frac{3.3}{3!} + \frac{3.4}{4!} + . . .$ $= 3 + \frac{3}{1!} + \frac{3}{2!} + \frac{3}{3!} + . . . = 3 (1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . . .)$ $= 3 e (1) . Also we have also$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{n!} = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . . . = e - 1 (1)$ $\underset{n = 1}{\sum^{\infty}} \frac{n^{2}}{n!} = \frac{1}{1!} + \frac{4}{2!} + \frac{9}{3!} + \frac{16}{4!} + . . .$ $= \frac{1}{1!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + . . .$ $We have also that$ $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + . . .$ $= 1 + \frac{2 - 1}{1!} + \frac{3 - 2}{2!} + \frac{4 - 3}{3!} + \frac{5 - 4}{4!} + . . .$ $= \frac{1}{1!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \frac{5}{4!} + . . . - (1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . . .)$ $= 1 + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \frac{5}{4!} + . . . - e$ $\Rightarrow 2 e = \frac{1}{1!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \frac{5}{4!} + . . . (2)$ $From (*), (1) and (2) we get$ $S = Σ \frac{n^{2} + 3 n + 1}{n!} = e - 1 + 3 e + 2 e = 6 e - 1$ $= {ae}^{b} + c \Rightarrow a = 6, b = 1, c = - 1 . Therefore,$ $a + b + c = 6$
	Commented by OlafThorendsen last updated on 22/Jul/20

	$The sum starts at 1, not at 0 .$ $Finally the good result is 6 e - 1 sir .$
	Commented by 1549442205PVT last updated on 22/Jul/20

	$Thank you Sir . I mistaked and$ $corrected .$
	Answered by OlafThorendsen last updated on 22/Jul/20

	$S = \underset{k = 1}{\sum^{\infty}} \frac{(k + 1) (k + 2) - 1}{k!}$ $S = \underset{k = 1}{\sum^{\infty}} \frac{k^{2} + 3 k + 1}{k!}$ $S = \underset{k = 1}{\sum^{\infty}} \frac{k}{(k - 1)!} + 3 \underset{k = 1}{\sum^{\infty}} \frac{1}{(k - 1)!} + \underset{k = 1}{\sum^{\infty}} \frac{1}{k!}$ $S = \underset{k = 0}{\sum^{\infty}} \frac{k + 1}{k!} + 3 \underset{k = 0}{\sum^{\infty}} \frac{1}{k!} + \underset{k = 1}{\sum^{\infty}} \frac{1}{k!}$ $S = \underset{k = 1}{\sum^{\infty}} \frac{1}{(k - 1)!} + 4 \underset{k = 0}{\sum^{\infty}} \frac{1}{k!} + \underset{k = 1}{\sum^{\infty}} \frac{1}{k!}$ $S = \underset{k = 0}{\sum^{\infty}} \frac{1}{k!} + 4 \underset{k = 0}{\sum^{\infty}} \frac{1}{k!} + (\underset{k = 0}{\sum^{\infty}} \frac{1}{k!} - 1)$ $S = 6 \underset{k = 0}{\sum^{\infty}} \frac{1}{k!} - 1$ $S = 6 e - 1$
	Commented by Dwaipayan Shikari last updated on 22/Jul/20
	��
	Commented by Dwaipayan Shikari last updated on 22/Jul/20

	$G r e a t s o l u t i o n$
	Commented by mr W last updated on 22/Jul/20

	$i a g r e e w i t h M J S s i r . i n f a c t t h e$ $s u m o f L H S i s n o t u n i q u e, s i n c e t h e$ $a_{n} t e r m i s n o t d e f i n e d!$
	Commented by 1549442205PVT last updated on 22/Jul/20

	$we can find out the rule to define$ $general term follows as :$ $a_{1} = 5, a_{2} = 11, a_{3} = 19, a_{4} = 29, a_{5} = 41 . So,$ $\| \begin{matrix} a_{2} - a_{1} & a_{3} - a_{2} & a_{4} - a_{3} & a_{5} - a_{4} \\ 6 & 8 & 10 & 12 \end{matrix} \|$ $Since 12 - 10 = 10 - 8 = 8 - 6, it follows$ $that the sequence 6, 8, 10, 12 . . . form$ $an arithmetic progression which has$ $the difference equal to 2 . Hence its$ $general term is 2 n + 4 . So, by above result$ $a_{n + 1} - a_{n} = 2 n + 4 . It follows that$ $\underset{k = 1}{\sum^{n}} (a_{k + 1} - a_{k}) = \underset{k = 1}{\overset{n}{Σ}} (2 k + 4)$ $\Rightarrow a_{n + 1} - a_{1} = 2 \underset{k = 1}{\overset{n}{Σ}} k + 4 n = n (n + 1) + 4 n$ $\Rightarrow a_{n + 1} = n^{2} + 5 n + 5 \Rightarrow a_{n} = n^{2} + 3 n + 1$
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