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

प्रश्न

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

उत्तर

Let P(n) be the given statement.
Now,

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

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Principle of Mathematical Induction

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 38 Q 37 Q 39

पाठ 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २८]

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पाठ 12 Mathematical Induction
Exercise 12.2 | Q 38 | पृष्ठ २८

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

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