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Question Number 90251 by Hanumantha Rao DAMARAJU last updated on 22/Apr/20

\frac{1}{1!} + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{2}}{3!} + \dots \dots . = ? ? ?

Commented by MJS last updated on 22/Apr/20

\frac{17}{6} e

but I cannot prove it

Commented by mathmax by abdo last updated on 22/Apr/20

S = \sum_{n = 1}^{\infty} \frac{\sum_{k = 1}^{n} k^{2}}{n!} w e h a v e \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6} \Rightarrow

S = \sum_{n = 1}^{\infty} \frac{n (n + 1) (2 n + 1)}{6 n!} = \sum_{n = 1}^{\infty} \frac{(n + 1) (2 n + 1)}{6 (n - 1)!}

=_{n - 1 = p} \sum_{p = 0}^{\infty} \frac{(p + 2) (2 p + 3)}{6 p!} = \sum_{p = 0}^{\infty} \frac{2 p^{2} + 3 p + 4 p + 6}{6 p!}

= \frac{1}{3} \sum_{p = 1}^{\infty} \frac{p}{(p - 1)!} + \frac{7}{6} \sum_{p = 1}^{\infty} \frac{1}{(p - 1)!} + \sum_{p = 0}^{\infty} \frac{1}{p!}

\sum_{p = 0}^{\infty} \frac{1}{p!} = e

\sum_{p = 1}^{\infty} \frac{1}{(p - 1)!} = \sum_{p = 0}^{\infty} \frac{1}{p!} = e

\sum_{p = 1}^{\infty} \frac{p}{(p - 1)!} =_{p - 1 = k} \sum_{k = 0}^{\infty} \frac{k + 1}{k!} = \sum_{k = 1}^{\infty} \frac{1}{(k - 1)!} + \sum_{k = 0}^{\infty} \frac{1}{k!}

= 2 e \Rightarrow S = \frac{2}{3} e + \frac{7}{6} e + e = (\frac{2}{3} + \frac{7}{6} + 1) e

= \frac{4 + 7 + 6}{6} e = \frac{17}{6} e

Answered by TANMAY PANACEA. last updated on 22/Apr/20

T_{r} = \frac{1^{2} + 2^{2} + 3^{2} + . . + r^{2}}{r!} = \frac{r (r + 1) (2 r + 1)}{6 r!}

T_{r} = \frac{1}{6} [\frac{(r + 1) (2 r + 1)}{(r - 1)!}] = \frac{1}{6} [\frac{2 r^{2} + 3 r + 1}{(r - 1)!}]

= \frac{1}{6} [\frac{2 {(r - 1)}^{2} + 7 r - 7 + 6}{(r - 1)!}]

= \frac{1}{6} [\frac{2 (r - 1)}{(r - 2)!} + \frac{7}{(r - 2)!} + \frac{6}{(r - 1)!}]

= \frac{1}{6} [\frac{2 r - 4 + 2}{(r - 2)!} + \frac{7}{(r - 2)!} + \frac{6}{(r - 1)!}]

= \frac{1}{6} [\frac{2}{(r - 3)!} + \frac{9}{(r - 2)!} + \frac{6}{(r - 1)!}]

T_{4} = \frac{1}{6} [\frac{2}{1!} + \frac{9}{2!} + \frac{6}{3!}]

T_{5} = \frac{1}{6} [\frac{2}{2!} + \frac{9}{3!} + \frac{6}{4!}]

T_{6} = \frac{1}{6} [\frac{2}{3!} + \frac{9}{4!} + \frac{6}{5!}]

\dots

\dots

T_{r} = \frac{1}{6} [\frac{2}{(r - 3)!} + \frac{9}{(r - 2)!} + \frac{6}{(r - 1)!}]

\dots

\dots

a d d t h e m

S = \underset{r = 4}{\sum^{\infty}} T_{r} = \frac{1}{6} [2 \times e + 9 (e - \frac{1}{1!}) + 6 (e - \frac{1}{1!} - \frac{1}{2!})]

= \frac{1}{6} [17 e - 9 - 6 - 3] = \frac{17 e}{6} - \frac{18}{6}

h e n c e r e q u i r e d a n s w e r i s

T_{1} + T_{2} + T_{3} + \underset{r = 4}{\sum^{\infty}} T_{r}

= \frac{1}{1!} + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{2}}{3!} + \frac{17 e}{6} - 3

= 1 + \frac{5}{2} + \frac{14}{6} - 3 + \frac{17 e}{6}

= \frac{6 + 15 + 14 - 18}{6} + \frac{17 e}{6}

= \frac{17}{6} + \frac{17}{6} e = \frac{17}{6} (1 + e)

p l s c h e c k

Commented by mathmax by abdo last updated on 22/Apr/20

s m a l l e r r o r s i r t a n m a y \dots

s m a l l e r r o r s i r t a n m a y \dots