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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

 

$P(n):{{x}^{2}}n-{{y}^{2}}n$is 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 \[k\]

${{x}^{2}}k-{{y}^{2}}k$is 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 \[n.\]

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