`lim_(n->oo)(1^(99)+2^(99)+3^(99)+.......n^(99))/(n^(100))=`

Lt_(n rarr oo)[(1^(99)+2^(99)+3^(99)+...+n^(99))/(n^(100))]

Iim_(n to oo) (1^(99) + 2^(99) + …..+ n^(99))/(n^(100)) equals :

lim_ (n rarr oo) (1 ^ (99) + 2 ^ (99) + 3 ^ (99) + ...... n ^ (99)) / (n ^ (100)) =

lim_(n rarr oo) 1/n^(100) (1^(99)+2^(99)+3^(99)+………+n^(99)) =

(1.99)^(3)-3(1.99)+5~~…

Using binomial theorem show that 1^(99) + 2^(99) +3^(99) + 4^(99) + 5^(99) is divisible by 5

The value quad ^(^^)99C_(0)+^(99)C_(1)+^(99)C_(2)+.......+^(99)C_(49)=

Using the binomial theorem show that 1^(99) + 2^(99) +3^(99) + 4^(99) + 5^(99) is divisible by 3 and 5 so that it is actually divisible by 15.