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Home » Prove For any natural number n, x^n – y^n is divisible by x – y, where x integers with x ≠ y.
Prove For any natural number n, x^n – y^n is divisible by x – y, where x integers with x ≠ y.
CBSE | Class 11 | NCERT Exemplar | Principle of Mathematical Induction
Prove For any natural number n, x^n – y^n is divisible by x – y, where x integers with x ≠ y.

As indicated by the inquiry,

    \[P\left( n \right)\text{ }=\text{ }xn\text{ }\text{ }yn\]

is detachable by

    \[x\text{ }\text{ }y,\text{ }x\]

whole numbers with

    \[x\text{ }\ne \text{ }y\]

.

In this way, subbing various qualities for n, we get,

    \[P\left( 0 \right)\text{ }=\text{ }x0\text{ }\text{ }y0\text{ }=\text{ }0\]

Which is separable by

    \[x-y.\]

    \[P\left( 1 \right)\text{ }=\text{ }x\text{ }-\text{ }y\]

Which is separable by

    \[x-y.\]

    \[P\left( 2 \right)\text{ }=\text{ }x2\text{ }\text{ }y2\]

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

Which is separable by

    \[x-y.\]

    \[P\left( 3 \right)\text{ }=\text{ }x3\text{ }\text{ }y3\]

    \[=\text{ }\left( x-y \right)\left( x2+xy+y2 \right)\]

Which is separable by

    \[x-y.\]

Let

    \[P\left( k \right)\text{ }=\text{ }xk\text{ }\text{ }yk\]

be separable by

    \[x-y.\]

Thus, we get,

    \[\Rightarrow xkyk\text{ }=\text{ }a\left( x-y \right).\]

Presently, we additionally get that,

    \[\Rightarrow P\left( k+1 \right)\text{ }=\text{ }xk+1yk+1\]

    \[=\text{ }xk\left( x-y \right)\text{ }+\text{ }y\left( xk-yk \right)\]

    \[=\text{ }xk\left( x-y \right)\text{ }+y\text{ }a\left( x-y \right)\]

Which is separable by

    \[x\text{ }-\text{ }y.\]

    \[\Rightarrow P\left( k+1 \right)\]

is valid when P(k) is valid.

Along these lines, by Mathematical Induction,

    \[P\left( n \right)\text{ }xn\text{ }\text{ }yn\]

is distinct by

    \[x\text{ }\text{ }y,\]

where x numbers with x ≠ y which is valid for any normal number n.

i

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