AP EAMCET Engineering 18 Sep 2020 Shift 1 Paper

For any odd positive integer ' $n$ ' $a^{n}+b^{n}$ is divisible by $(a+b)$.To prove this identity, we can use the factorization:$a^n + b^n = (a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - ... + b^{n-1})$which can be derived using the sum of powers formula for $a^n$ and $b^n$:$a^n + b^n = (a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - ... + ab^{n-2} + b^{n-1})$Since $n$ is odd, the expression inside the parentheses is a sum of an odd number of terms, so it can be written as:$a^{n-1} - a^{n-2}b + a^{n-3}b^2 - ... + ab^{n-2} + b^{n-1} = (a-b)(a^{n-2} + a^{n-4}b^2 + ... + b^{n-2})$Now we can substitute this factorization back into the original expression:$a^n + b^n = (a+b)(a-b)(a^{n-2} + a^{n-4}b^2 + ... + b^{n-2})$Since $n$ is odd, $a+b$ is a factor of $a^n+b^n$ by the sum of powers formula. Therefore, we can conclude that $a^n+b^n$ is divisible by $a+b$ for any odd positive integer $n$.