X2 n-1-y2 n-1 is divisible by x-y for all n - N-

Let P(n) be the given statement. Now, P(n): #160;x2n 1+y2n 1 #160;is #160;divisible #160;by #160;x+y.Step1:P(1):x2 1+y2 1=x+y #160;is #160;divisib ...

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Standard X

Mathematics

Solving a Quadratic Equation by Factorization Method

x2 n-1+y2 n-1...

Question

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

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Solution

Let P(n) be the given statement.
Now,
$P (n) : x^{2 n - 1} + y^{2 n - 1} i s d i v i s i b l e b y x + y . S t e p 1 : P (1) : x^{2 - 1} + y^{2 - 1} = x + y is divisible by x + y Step 2 : Let P (m) be true . A l s o, x^{2 m - 1} + y^{2 m - 1} is divisible by x + y . Suppose : x^{2 m - 1} + y^{2 m - 1} = λ (x + y) where λ \in N . . . (1) We shall show that P (m + 1) is true whenever P (m) is true . Now, P (m + 1) = x^{2 m + 1} + y^{2 m + 1} = x^{2 m + 1} + y^{2 m + 1} - x^{2 m - 1} . y^{2} + x^{2 m - 1} . y^{2} = x^{2 m - 1} (x^{2} - y^{2}) + y^{2} (x^{2 m - 1} + y^{2 m - 1}) [From (1)] = x^{2 m - 1} (x^{2} - y^{2}) + y^{2} . λ (x + y) = (x + y) (x^{2 m - 1} (x - y) + λ y^{2}) [I t i s divisible by (x + y) .] Thus, P (m + 1) i s true . B y t h e p rinciple of m athematical induction, P (n) is true for all n \in N .$

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

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