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

To prove that \( a^n + b^n \) is divisible by \( a + b \) for any odd natural number \( n \), we can follow these steps:

### Step 1: Factor the expression
We start with the expression \( a^n + b^n \). Since \( n \) is odd, we can factor this expression using the identity:
\[ a^n + b^n = (a + b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - \ldots + b^{n-1}) \]
This shows that \( a^n + b^n \) is divisible by \( a + b \).
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