Question

n(4n2 +6n-1)

Answer 1

The given statement is,

P( n ):1⋅3+3⋅5+5⋅7...+( 2n−1 )( 2n+1 )= n( 4 n 2 +6n−1 ) 3 (1)

For n=1,

P( 1 ):1⋅3= 1( 4⋅ 1 2 +6⋅1−1 ) 3 P( 1 ):1⋅3= 4+6−1 3 P( 1 ):3= 9 3 P( 1 ):3=3

Thus, P( 1 ) is true.

Substitute n=k in equation (1).

P( k ):1⋅3+3⋅5+5⋅7...+( 2k−1 )( 2k+1 )= k( 4 k 2 +6k−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⋅3+3⋅5+5⋅7...+( 2( k+1 )−1 )( 2( k+1 )+1 )= ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 P( k+1 ):1⋅3+3⋅5+5⋅7...+( 2k+1 )( 2k+3 )= ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 (3)

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

P( k+1 ): k( 4 k 2 +6k−1 ) 3 +( 2k+1 )( 2k+3 )= ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 P( k+1 ): k( 4 k 2 +6k−1 )+3( 2k+1 )( 2k+3 ) 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 P( k+1 ): k( 4 k 2 +6k−1 )+3( 4 k 2 +8k+3 ) 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 P( k+1 ): 4 k 3 +6 k 2 −k+12 k 2 +24k+9 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3

Further simplify.

P( k+1 ): 4 k 3 +18 k 2 +23k+9 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 P( k+1 ): 4 k 3 +14 k 2 +4 k 2 +14k+9k+9 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 P( k+1 ): k( 4 k 2 +14k+9 )+1( 4 k 2 +14k+9 ) 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 P( k+1 ): ( k+1 )( 4 k 2 +14k+9 ) 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3

Further simplify.

P( k+1 ): ( k+1 )( 4 k 2 +8k+6k+4+6−1 ) 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 P( k+1 ): ( k+1 )( 4( k 2 +2k+1 )+6( k+1 )−1 ) 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 P( k+1 ): ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3 = ( k+1 )( 4 ( k+1 ) 2 +6( k+1 )−1 ) 3

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

Hence, statement P( n ) is true.