lim-n-k-1-n-1-k-1-k-n-1-C-k-1-n-C-r-n-r-n-r-

Question Number 136854 by Raxreedoroid last updated on 26/Mar/21

\lim_{n \to \infty} \underset{k = 1}{\sum^{n}} ({(- 1)}^{k + 1} k \times^{n - 1} C_{k - 1}) = ?

^{n} C_{r} = \frac{n!}{r! (n - r)!}

Commented by mr W last updated on 27/Mar/21

\underset{k = 1}{\sum^{n}} {(- 1)}^{k + 1} {k C}_{k - 1}^{n - 1} = 0 f o r a n y n ⩾ 3

Answered by mr W last updated on 27/Mar/21

{(1 - x)}^{n} = \underset{k = 0}{\sum^{n}} C_{k}^{n} {(- x)}^{k}

{(1 - x)}^{n} = \underset{k = 0}{\sum^{n}} {(- 1)}^{k} \frac{n}{k} C_{k - 1}^{n - 1} x^{k}

- n {(1 - x)}^{n - 1} = \underset{k = 0}{\sum^{n}} {(- 1)}^{k} {n C}_{k - 1}^{n - 1} x^{k - 1}

{(1 - x)}^{n - 1} = \underset{k = 0}{\sum^{n}} {(- 1)}^{k + 1} C_{k - 1}^{n - 1} x^{k - 1}

x {(1 - x)}^{n - 1} = \underset{k = 0}{\sum^{n}} {(- 1)}^{k + 1} C_{k - 1}^{n - 1} x^{k}

{(1 - x)}^{n - 1} - x (n - 1) {(1 - x)}^{n - 2} = \underset{k = 0}{\sum^{n}} {(- 1)}^{k + 1} {k C}_{k - 1}^{n - 1} x^{k - 1}

{(1 - x)}^{n - 1} - x (n - 1) {(1 - x)}^{n - 2} = \underset{k = 1}{\sum^{n}} {(- 1)}^{k + 1} {k C}_{k - 1}^{n - 1} x^{k - 1}

l e t x = 1

0 = \underset{k = 1}{\sum^{n}} {(- 1)}^{k + 1} {k C}_{k - 1}^{n - 1}

\lim_{n \to \infty} \underset{k = 1}{\sum^{n}} {(- 1)}^{k + 1} {k C}_{k - 1}^{n - 1} = 0

Facebook Tweet Pin