# X2n−1 + Y2n−1 is Divisible by X + Y for All N ∈ N. - Mathematics

x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.

#### Solution

Let P(n) be the given statement.
Now,

$P(n): x^{2n - 1} + y^{2n - 1} \text{ is divisible by } x + y .$
$\text{ Step1:}$
$P(1): x^{2 - 1} + y^{2 - 1} = x + y \text{ is divisible by } x + y$
$\text{ Step2: }$
$\text{ Let P(m) be true } .$
$\text{ Also } ,$
$x^{2m - 1} + y^{2m - 1} \text{ is divisible by } x + y .$
$\text{ Suppose:}$
$x^{2m - 1} + y^{2m - 1} = \lambda\left( x + y \right) \text{ where} \lambda \in N . . . (1)$
$\text{ We shall show that } P\left( m + 1 \right) \text{ is true whenever } P\left( m \right) \text{ is true } .$
$\text{ Now } ,$
$P\left( m + 1 \right) = x^{2m + 1} + y^{2m + 1}$
$= x^{2m + 1} + y^{2m + 1} - x^{2m - 1} . y^2 + x^{2m - 1} . y^2$
$= x^{2m - 1} \left( x^2 - y^2 \right) + y^2 \left( x^{2m - 1} + y^{2m - 1} \right) \left[ \text{ From } (1) \right]$
$= x^{2m - 1} \left( x^2 - y^2 \right) + y^2 . \lambda\left( x + y \right)$
$= \left( x + y \right)\left( x^{2m - 1} \left( x - y \right) + \lambda y^2 \right) [\text{ It is divisible by } (x + y) . ]$
$\text{ Thus, } P\left( m + 1 \right) \text{ is true } .$
$\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 38 | Page 28