Using the sum of G.P., prove that `a^(n)+b^(n)(a,binN)` is divisble by a+b for odd natural numbers n. Hence prove that `1^(99)+2^(99)+….100^(99)` is divisble by 10100

Explanation

`(a^(n)+b^(n))/(a+b)=(a^(n)(1-(-b/a)^(n)))/(a(1-(-b/a)))`, if n is odd
`=a^(n-1)((1-(-b/a)^(n))/(1-(-b/a)))`
`=a^(n-1)(1-b/a+(b/a)^(2)-(b/a)^(3)+...+(-1)^(n-1)(b/a)^(n-1))`
`=a^(n-1)-bxxa^(n-2)+b^(2)xxa^(n-3)….+(-1)^(n-1)xxb^(n-1)`, which is an integer
Thus, `a^(n)+b^(n)` is divisble by a+b for odd positive natural numbers. Now, `1^(99)+2^(99)+...+100^(99)=(1^(99)+100^(99))+(2^(99)+99^(99))+...+(50^(99)+51^(99))`
Each bracket is divisble by 101 and hence, the sum is divisbleby 101.
Also, `1^(99)+2^(99)+...+100^(99)=(1^(99)+99^(99))+(2^(99)+98^(99)+...+(49^(99)+51^(99))+50^(99)+100(99)`
Here, each bracket and `50^(99)` and `100^(99)` are divisble by 100.
Hence, the sum is divisble by 100.
Hence, sum is divisble by `101xx100`=10100.