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" 14) "lim(n rarr oo)(1^(99)+2^(99)+3^(9...

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

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

lim_(n->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/n^(100) (1^(99)+2^(99)+3^(99)+………+n^(99)) =

lim_(x rarr oo)((x+1)^(100)+(x+2)^(100)+...+(x+50)^(100))/(x^(100)+x^(99)+......+x+1)=

(1.99)^(3)-3(1.99)+5~~…

lim_ (n rarr oo) (n ^ (3) -100n ^ (2) +1) / (100n ^ (2) + 16n)

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