Question-113309

Question Number 113309 by Hassen_Timol last updated on 12/Sep/20

Commented by Hassen_Timol last updated on 12/Sep/20

Can you please help me \dots ?

Commented by mathmax by abdo last updated on 12/Sep/20

S_{n} = \sum_{k = 0}^{n} n^{n - k} a^{k} \Rightarrow S_{n} = n^{n} \sum_{k = 0}^{n} {(\frac{a}{n})}^{k} = n^{n} \times \frac{1 - {(\frac{a}{n})}^{n + 1}}{1 - \frac{a}{n}}

= n^{n + 1} \times \frac{n^{n + 1} - a^{n + 1}}{n^{n + 1} (n - a)} = \frac{n^{n + 1} - a^{n + 1}}{n - a} if a \neq n

if a = n S_{n} = \sum_{k = 0}^{n} n^{k} n^{n - k} = n^{n} \sum_{k = 0}^{n} (1) = (n + 1) n^{n} .

Commented by Hassen_Timol last updated on 12/Sep/20

Thank you so much Sir

Commented by abdomsup last updated on 12/Sep/20

y o u a r e w e l c o m e

Answered by Dwaipayan Shikari last updated on 12/Sep/20

\underset{k = 1}{\sum^{n}} k . k! = 1.1! + 2.2! + \dots .

1.1! = (2 - 1) 1! = 2! - 1!

2.2! = 3.2! - 2! = 3! - 2!

\dots .

\underset{n = 1}{\sum^{n}} k . k! = 2! - 1! + 3! - 2! + 4! - 3! + \dots . . n! - (n - 1)! + (n + 1)! - n!

= (n + 1)! - 1

Answered by Dwaipayan Shikari last updated on 12/Sep/20

\underset{k = 0}{\sum^{n}} (\frac{a^{k}}{n^{k}}) n^{n} = n^{n} \underset{k = 0}{\sum^{n}} {(\frac{a}{n})}^{k} = n^{n} (\frac{{(\frac{a}{n})}^{n + 1} - 1}{\frac{a}{n} - 1})

= n^{n + 1} (\frac{a^{n + 1} - n^{n + 1}}{n^{n + 1} (a - n)}) = (\frac{a^{n + 1} - n^{n + 1}}{a - n})

Commented by Hassen_Timol last updated on 12/Sep/20

Thank you so much