Question

The expression $\frac{x^n-y^n}{x^2-y^2}$ is equal to
(where $x^2\ne y^2$ and $n$ is even natural number greater than $6$)

• A. $x^{n-1}+x^{n-3}{y}^2+...+y^{n-2}$
• B. $x^{n-2}+x^{n-4}{y}^2+...+y^{n-2}$
• C. $x^{n-1}+x^{n-2}{y}^2+...+y^{n-1}$
• D. $x^{n-2}-x^{n-4}{y}^2+...-y^{n-2}$

Answer 1

The correct option is B $x^{n-2}+x^{n-4}{y}^2+...+y^{n-2}$

Let $n=2m,m\in Z$
Then $\frac{x^n-y^n}{x^2-y^2}=\frac{(x^2{)}^m-(y^2{)}^m}{x^2-y^2}$

Let $x^2=a,y^2=b$

$=\frac{(a-b)(a^{m-1}+a^{m-2}b+...+b^{m-1})}{a-b}$
$=a^{m-1}+a^{m-2}b+...+b^{m-1}$
$=x^{n-2}+x^{n-4}{y}^2+...+x^2{y}^{n-4}+y^{n-2}$