Question

15. 12+32 52 (2n-1)2 -n(2n-I)(2n +1)

Answer 1

The given statement is,

P( n ): 1 2 + 3 2 + 5 2 +...+ ( 2n−1 ) 2 = n( 2n−1 )( 2n+1 ) 3 (1)

For n=1,

P( 1 ): 1 2 = 1( 2−1 )( 2+1 ) 3 P( 1 ):1= 3 3 P( 1 ):1=1

Thus, P( 1 ) is true.

Substitute n=k in equation (1).

P( k ): 1 2 + 3 2 + 5 2 +...+ ( 2k−1 ) 2 = k( 2k−1 )( 2k+1 ) 3 (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 equation (1).

P( k+1 ): 1 2 + 3 2 + 5 2 +...+ ( 2( k+1 )−1 ) 2 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3 (3)

Substitute the value from equation (2) in equation (3).

P( k+1 ): k( 2k−1 )( 2k+1 ) 3 + ( 2( k+1 )−1 ) 2 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3 P( k+1 ): k( 2k−1 )( 2k+1 )+3 ( 2( k+1 )−1 ) 2 3 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3 P( k+1 ): k( 2k−1 )( 2k+1 )+3 ( 2k+2−1 ) 2 3 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3 P( k+1 ): k( 2k−1 )( 2k+1 )+3 ( 2k+1 ) 2 3 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3

Further simplify,

P( k+1 ): ( 2k+1 )( k( 2k−1 )+3( 2k+1 ) ) 3 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3 P( k+1 ): ( 2k+1 )( 2 k 2 −k+6k+3 ) 3 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3 P( k+1 ): ( 2k+1 )( 2 k 2 +5k+3 ) 3 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3 P( k+1 ): ( 2k+1 )( 2 k 2 +2k+3k+3 ) 3 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3

Further simplify,

P( k+1 ): ( 2k+2−1 )( 2k+2+1 )( k+1 ) 3 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3 P( k+1 ): ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3 = ( k+1 )( 2( k+1 )−1 )( 2( k+1 )+1 ) 3

It is proved that P( k+1 ) is true whenever P( k ) is true.

Hence, statement P( n ) is true.