x-1-3-n-x-1-x-2-x-n-x-n-n-N-n-1-Find-n-1-1-n-1-x-n-

Question Number 160008 by HongKing last updated on 23/Nov/21

x_{1} = 3; n (x_{1} + x_{2} + \dots + x_{n}) = x_{n}; n \in N; n ⩾ 1

Find :

Ω = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} x_{n}

Commented by mr W last updated on 23/Nov/21

Ω = 6 e ?

Commented by HongKing last updated on 23/Nov/21

Yes my dear Ser but how please

Answered by Tokugami last updated on 23/Nov/21

n (x_{1} + x_{2} + \dots + x_{n}) = x_{n}

n \underset{k = 1}{\sum^{n}} x_{k} = x_{n}

{n x}_{n} + n \underset{k = 1}{\sum^{n - 1}} x_{k} = x_{n}

n \underset{k = 1}{\sum^{n - 1}} x_{k} = x_{n} - {n x}_{n}

\frac{n}{1 - n} \underset{k = 1}{\sum^{n - 1}} x_{k} = x_{n}

\frac{n + 1}{- n} \underset{k = 1}{\sum^{n}} x_{k} = x_{n + 1}

\underset{k = 1}{\sum^{n}} x_{k} - \underset{k = 1}{\sum^{n - 1}} x_{k} = \frac{- n}{n + 1} x_{n + 1} - \frac{1 - n}{n} x_{n}

x_{n} = \frac{- n}{n + 1} x_{n + 1} - \frac{1 - n}{n} x_{n}

(1 + \frac{1 - n}{n}) x_{n} = - \frac{n}{n + 1} x_{n + 1}

\frac{1}{n} x_{n} = - \frac{n}{n + 1} x_{n + 1}

- \frac{n + 1}{n^{2}} x_{n} = x_{n + 1}

(- \frac{1 + 1}{1^{2}}) x_{1} = x_{2}

(- \frac{2 + 1}{2^{2}}) x_{2} = (- \frac{2 + 1}{2^{2}}) (- \frac{1 + 1}{1^{2}}) x_{1} = x_{3}

x_{n} = {(- 1)}^{n - 1} (\frac{n!}{{((n - 1)!)}^{2}}) x_{1}

\frac{n!}{(n - 1)!} = n

x_{n} = {(- 1)}^{n - 1} (\frac{3 n}{(n - 1)!})

Ω = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} x_{n}

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

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

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

= 6 e

Commented by HongKing last updated on 23/Nov/21

perfect my dear Ser, thank you so much

Answered by mr W last updated on 23/Nov/21

f i n d \underset{n = 1}{\sum^{\infty}} \frac{n^{2}}{n!} = ?

\underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} = e^{x}

\underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n!} = e^{x} - 1

\underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n - 1}}{n!} = \frac{d}{d x} (e^{x} - 1) = e^{x}

\underset{n = 1}{\sum^{\infty}} \frac{{n x}^{n}}{n!} = {x e}^{x}

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

l e t x = 1,

\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n^{2}}{n!} = 2 e

x_{1} + x_{2} + x_{3} + \dots + x_{n - 1} + x_{n} = \frac{x_{n}}{n}

x_{1} + x_{2} + x_{3} + \dots + x_{n - 1} = \frac{x_{n - 1}}{n - 1}

\Rightarrow x_{n} = \frac{x_{n}}{n} - \frac{x_{n - 1}}{n - 1}

\Rightarrow \frac{x_{n} (n - 1)}{n} = - \frac{x_{n - 1}}{n - 1}

\Rightarrow \frac{x_{n}}{x_{n - 1}} = - \frac{n}{{(n - 1)}^{2}}

\Rightarrow \frac{x_{n - 1}}{x_{n - 2}} = - \frac{n - 1}{{(n - 2)}^{2}}

\dots

\Rightarrow \frac{x_{2}}{x_{1}} = - \frac{2}{1^{2}}

\Rightarrow \frac{x_{n}}{x_{1}} = {(- 1)}^{n - 1} \frac{n!}{{((n - 1)!)}^{2}}

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

\Rightarrow x_{n} = {(- 1)}^{n - 1} \frac{n^{2} x_{1}}{n!}

w i t h x_{1} = 3,

\Rightarrow x_{n} = {(- 1)}^{n - 1} \frac{3 n^{2}}{n!}

\Rightarrow {(- 1)}^{n + 1} x_{n} = {(- 1)}^{2 n} \frac{3 n^{2}}{n!} = \frac{3 n^{2}}{n!}

Ω = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} x_{n} = 3 \underset{n = 1}{\sum^{\infty}} \frac{n^{2}}{n!} = 3 \times 2 e = 6 e

Commented by HongKing last updated on 23/Nov/21

perfect my dear Ser, thank you so much