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

प्रश्न

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

उत्तर

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

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

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Principle of Mathematical Induction

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Q 38 Q 37 Q 39

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

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आर.डी. शर्मा Mathematics [English] Class 11

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

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Principle of Mathematical Induction

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