Question-159479

Question Number 159479 by quvonnn last updated on 17/Nov/21

Answered by Tokugami last updated on 17/Nov/21

\frac{2}{4 \times 2!} + \frac{2}{5 \times 3!} + \frac{2}{6 \times 4!} + \frac{2}{7 \times 5!} + \frac{2}{8 \times 6!} + \dots

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

= 2 \underset{n = 2}{\sum^{\infty}} \frac{1 (n + 1)}{(n + 2) (n + 1) n!}

= 2 \underset{n = 2}{\sum^{\infty}} \frac{n + 1}{(n + 2)!}

n + 2 \to n

= 2 \underset{n = 4}{\sum^{\infty}} \frac{n - 1}{n!}

= 2 \underset{n = 4}{\sum^{\infty}} (\frac{n}{n!} - \frac{1}{n!})

= 2 \underset{n = 4}{\sum^{\infty}} \frac{1}{(n - 1)!} - 2 \underset{n = 4}{\sum^{\infty}} \frac{1}{n!}

\underset{n = 4}{\sum^{\infty}} \frac{1}{(n - 1)!}

n - 1 \to n

= 2 (\underset{n = 3}{\sum^{\infty}} \frac{1}{n!} - \underset{n = 4}{\sum^{\infty}} \frac{1}{n!})

= 2 ((\underset{n = 0}{\sum^{\infty}} \frac{1}{n!} - \underset{n = 0}{\sum^{2}} \frac{1}{n!}) - (\underset{n = 0}{\sum^{\infty}} \frac{1}{n!} - \underset{n = 0}{\sum^{3}} \frac{1}{n!}))

= 2 (- (\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!}) + (\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!}))

= \frac{2}{3!} = \frac{2}{6}

= \frac{1}{3}

Commented by quvonnn last updated on 17/Nov/21

t h a n h y o u

t h a n h y o u

Facebook Tweet Pin