Question

For $n\in N$, $a^{2n-1}+b^{2n-1}$ is divisible by

• A. $a+b$
• B. $(a+b{)}^2$
• C. $a^3+b^3$
• D. $a^2+b^2$

Answer 1

The correct option is A $a+b$

Consider given the expression,

$\Rightarrow a^{2n-1}+b^{2n-1},n\in N$

We know that,

$a^n+b^n=(a+b)(b^0{a}^{n-1}-b^1{a}^{n-1}+........b^{n-2}{a}^1+b^{n-1}{a}^0)$

$\therefore a^{2n-1}+b^{2n-1}=(a+b)(b^0{a}^{2n-2}-b^1{a}^{2n-3}+........b^{2n-3}{a}^1+b^{2n-2}{a}^0)$

Hence, $a^{2n-1}+b^{2n-1}$ is divisible by $(a+b)$

Hence, this is the answer.