Problem-15-Find-the-sum-of-S-3-1-2-3-4-2-3-4-5-3-4-5-2016-2014-2015-2016-

Question Number 8554 by Sopheak last updated on 16/Oct/16

P r o b l e m . 15

F i n d t h e s u m o f

S = \frac{3}{1! + 2! + 3!} + \frac{4}{2! + 3! + 4!} + \frac{5}{3! + 4! + 5!} + \dots + \frac{2016}{2014! + 2015! + 2016!}

Commented by Yozzias last updated on 16/Oct/16

\frac{3}{1! (1 + 2 + 2 \times 3)}, \frac{4}{2! (1 + 3 + 3 \times 4)}, \frac{5}{3! (1 + 4 + 4 \times 5)}, \dots

\frac{n + 2}{n! (1 + n + 1 + (n + 1) (n + 2))}

= \frac{n + 2}{n! (2 + n + n^{2} + 3 n + 2)}

= \frac{n + 2}{n! (4 + 4 n + n^{2})}

= \frac{n + 2}{n! {(n + 2)}^{2}}

= \frac{1}{n! (n + 2)}

S = \underset{r = 1}{\sum^{n}} \frac{1}{r! (r + 2)} = \underset{r = 1}{\sum^{n}} \frac{r + 1}{(r + 2)!} = \underset{r = 1}{\sum^{n + 1}} \frac{r}{(r + 1)!} - \frac{1}{(1 + 1)!} = 1 - \frac{1}{(n + 2)!} - \frac{1}{2}

S = \frac{1}{2} - \frac{1}{(n + 2)!}

n = 2014 \Rightarrow S = \frac{1}{2} - \frac{1}{2016!}

\underset{r = 1}{\sum^{2014}} \frac{r + 2}{r! + (r + 1)! + (r + 2)!} = \frac{1}{2} - \frac{1}{2016!}

Answered by Yozzias last updated on 16/Oct/16

Check answer in comments .

Commented by Sopheak last updated on 16/Oct/16

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