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  • For any natural number n, xn -yn is divisible by x -y, where x integers with x ≠ y.

Prove by the Principle of Mathematical Induction: For any natural number n, x^n -y^n is divisible by x-y where x and y are integers with x \neq y.

Answers (1)

P(n)= x^n -y^n is divisible by x -y, where x integers withx \neq y. ........(given)

 

Now, we’ll substitute different values for n,

P(0) = x^0 -y^0 = 0, is divisible by x-y

P(1) = x -y, is divisible by x-y

P(2) = x^2 -y^2 = (x+y)(x-y), is divisible by x-y

Now, let us consider,

P(k) = x^k -y^k, is divisible by x-y

Thus,x^k -y^k= a(x-y)

We also get that

P(k+1) = x^{k+1} -y^{k+1}

= x^k(x-y) + y(x^k-y^k)

= x^k(x-y) + y a(x-y), is divisible by x-y

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n) = x^n - y^n is divisible by x-y, x integers with x≠ y   
 

Posted by

infoexpert21

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