Question

21 .x2n-y2n is divisible by x + y.

Answer 1

The given statement is,

P( n ):( x 2n − y 2n ) is divisible by ( x+y )(1)

Here, n is a natural number.

For n=1, the statement is,

P( 1 ):( x 2 − y 2 ) is divisible by ( x+y ) P( 1 ):( x+y )( x−y ) is a multiple of 11

Hence, P( 1 ) is true.

Consider, n=k(here, k is some positive integer). Substitute n=k in the statement in equation (1).

P( k ):( x 2k − y 2k ) is divisible by ( x+y )(2)

According to the principle of mathematical induction, assume that the statement P( n ) is true for n=k. If it is also true for n=k+1, then P( n ) is true for all natural numbers.

Substitute n=k+1 in the statement in equation (1).

P( k+1 ):( x 2( k+1 ) − y 2( k+1 ) ) is divisible by ( x+y ) P( k+1 ):( x 2k ⋅ x 2 − y 2k ⋅ y 2 ) is divisible by ( x+y ) P( k+1 ):( ( x 2k + y 2k − y 2k ) x 2 − y 2k ⋅ y 2 ) is divisible by ( x+y ) P( k+1 ):( ( x 2k + y 2k − y 2k ) x 2 − y 2k ⋅ y 2 ) is divisible by ( x+y ) (3)

Further analyze from statement (3).

P( k+1 ):( ( x 2k + y 2k − y 2k ) x 2 − y 2k ⋅ y 2 ) is divisible by ( x+y ) P( k+1 ):( ( x 2k − y 2k ) x 2 + x 2 y 2k − y 2k ⋅ y 2 ) is divisible by ( x+y ) P( k+1 ):( ( x 2k − y 2k ) x 2 +( x 2 − y 2 ) y 2k ) is divisible by ( x+y ) P( k+1 ):( ( x 2k − y 2k ) x 2 +( x+y )( x−y ) y 2k ) is divisible by ( x+y ) (4)

Thus, if P( k ) is true, then P( k+1 ) is also true.

Hence, statement P( n ) is true.