Question

For all $n\in N,x^{2n}-y^{2n}$ is divisible by

• A. $(x+y)$
• B. $(x+y{)}^3$
• C. $(x+2y)$
• D. $(2x+y)$

Answer 1

Answer: (A)

Let the statement $P(n)$ be defined as
$P(n):x^{2n}-y^{2n}$ is divisible by $x+y$.
Step I For $n=1,P(1):x^2-y^2=(x-y)(x+y)$
which is divisible by $(x+y)$, that is true.
Step II Let it is true for $n=k$,
i.e., $x^{2k}-y^{2k}=\lambda (x+y)....$(i)
Step III For $n=k+1$,
$x^{2(k+1)}-y^{2(k+1)}=x^{2k+2}-y^{2k+2}$
$=x^{2k}{x}^2-y^{2k}{y}^2$
$=\left[\lambda (x+y)+y^{2k}\right]x^2-y^{2k}{y}^2\text{ [using Eq. (i)] }$
$=\lambda (x+y)x^2+y^{2k}{x}^2-y^{2k}{y}^2$
$=\lambda (x+y)x^2+y^{2k}\left(x^2-y^2\right)$
$=\lambda (x+y)x^2+y^{2k}(x-y)(x+y)$
$=(x+y)\left[\lambda x^2+y^{2k}(x-y)\right]$
which is a multiple of $(x+y)$ i.e., divisible by $(x+y)$.
Therefore, $P(k+1)$ is true when $P(k)$ is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers $n$.