1-lt-a-lt-b-prove-that-b-n-k-0-n-1-k-C-n-k-a-ln-p-0-n-k-C-n-k-p-a-n-p-b-p-ln-a-

Question Number 141633 by Willson last updated on 21/May/21

1 < a < b, prove that :

b^{n} = \underset{k = 0}{\sum^{n}} {(- 1)}^{k} C_{n}^{k} a^{\frac{l n (\underset{p = 0}{\sum^{n - k}} C_{n - k}^{p} a^{n - p} b^{p})}{l n (a)}}