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Home » Prove the following by using the principle of mathematical induction for all n ∈ N: x^2n – y^2n is divisible by x + y
Prove the following by using the principle of mathematical induction for all n ∈ N: x^2n – y^2n is divisible by x + y
CBSE | Class 11 | Exercise 4.1 | Maths | NCERT | Principle of Mathematical Induction
Prove the following by using the principle of mathematical induction for all n ∈ N: x^2n – y^2n is divisible by x + y

. We can compose the given assertion as

 

P(n):{{x}^{2}}n-{{y}^{2}}nis distinguishable by

    \[x\text{ }+\text{ }y\]

In the event that

    \[n\text{ }=\text{ }1\]

we get

P(1)={{x}^{2}}\text{ }\!\!\times\!\!\text{ }1-{{y}^{2}}\text{ }\!\!\times\!\!\text{ }1={{x}^{2}}-{{y}^{2}}=(x+y)(x-y),

which is distinguishable by

    \[\left( x\text{ }+\text{ }y \right)\]

Which is valid.

Think about

    \[P\text{ }\left( k \right)\]

be valid for some certain whole number

    \[k\]

{{x}^{2}}k-{{y}^{2}}kis distinguishable by

    \[x\text{ }+\text{ }y\]

{{x}^{2}}k-{{y}^{2}}k=m(x+y),where

    \[m\in N\ldots \left( 1 \right)\]

Presently let us demonstrate that

    \[P\text{ }\left( k\text{ }+\text{ }1 \right)\]

is valid.

Here

{{x}^{2}}(k+1)-{{y}^{2}}(k+1)

We can compose it as

={{x}^{2}}k.{{x}^{2}}-{{y}^{2}}k.{{y}^{2}}

By adding and deducting we{{y}^{2}}k get

    \[=\text{ }x\hat{\ }2\text{ }\left( x\hat{\ }2k\text{ }\text{ }y\hat{\ }2k\text{ }+\text{ }y\hat{\ }2k \right)\text{ }\text{ }y\hat{\ }2k.\text{ }y\hat{\ }2\]

From condition

    \[\left( 1 \right)\]

we get

    \[=\text{ }x\hat{\ }2\text{ }\left\{ m\text{ }\left( x\text{ }+\text{ }y \right)\text{ }+\text{ }y\hat{\ }2k \right\}\text{ }\text{ }y\hat{\ }2k.\text{ }y\hat{\ }2\]

By duplicating the terms

    \[=\text{ }m\text{ }\left( x\text{ }+\text{ }y \right)\text{ }x\hat{\ }2\text{ }+\text{ }y\hat{\ }2k.\text{ }x\hat{\ }2\text{ }\text{ }y\hat{\ }2k.\text{ }y\hat{\ }2\]

Taking out the normal terms

    \[=\text{ }m\text{ }\left( x\text{ }+\text{ }y \right)\text{ }x\hat{\ }2\text{ }+\text{ }y\hat{\ }2k\text{ }\left( x\hat{\ }2\text{ }\text{ }y\hat{\ }2 \right)\]

Growing utilizing recipe

    \[=\text{ }m\text{ }\left( x\text{ }+\text{ }y \right)\text{ }x\hat{\ }2\text{ }+\text{ }y\hat{\ }2k\text{ }\left( x\text{ }+\text{ }y \right)\text{ }\left( x\text{ }\text{ }y \right)\]

So we get

    \[=\text{ }\left( x\text{ }+\text{ }y \right)\text{ }\left\{ mx\hat{\ }2\text{ }+\text{ }y\hat{\ }2k\text{ }\left( x\text{ }\text{ }y \right) \right\},\]

which is a factor of

    \[\left( x\text{ }+\text{ }y \right)\]

    \[P\text{ }\left( k\text{ }+\text{ }1 \right)\]

is valid at whatever point

    \[P\text{ }\left( k \right)\]

is valid.

Hence, by the rule of numerical enlistment, articulation

    \[P\text{ }\left( n \right)\]

is valid for all regular numbers for example

    \[n.\]

i

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