ce a-maths

Let P ( n ) be the proposition

1/1 x 5+1/3 x 7+1/5 x 9+‧‧‧+1 /(2n – 1)(2n + 3)=1/3 - (n+1)/(2n+1)(2n+3)

When n = 1,

L.H.S. = 1 / ( 2 – 1 )( 2 + 3 ) = 1 / 5

R.H.S. = 1 / 3 – ( 1 + 1 ) / ( 2 + 1 )( 2 + 3 ) = 1 / 5 = L.H.S.

Therefore P ( 1 ) is true.

Assume P ( k ) is true for some positive integers k.

i.e. 1/1x5+1/3x7+1/5x9+‧‧‧+1/(2k-1)(2k+3)=1/3- (k+1)/(2k+1)(2k+3)

When n = k + 1 ,

L.H.S. = 1/1x5+1/3x7+1/5x9+‧‧‧+1/(2k-1)(2k+3) +1 /( 2k + 2 – 1 )( 2k + 2 + 3 )

= 1/3- (k+1)/(2k+1)(2k+3) + 1 / ( 2k + 1 )( 2k + 5 )

= 1 / 3 + 1 / ( 2k + 1 )( 2k + 5 ) - (k+1)/(2k+1)(2k+3)

= 1 / 3 + {( 2k + 3 ) – ( k + 1 )( 2k + 5 )} / ( 2k + 1 )( 2k + 5 )( 2k + 3 )

= 1 / 3 – ( 2k2 + 5k + 2 ) / ( 2k + 1 )( 2k + 5 )( 2k + 3 )

= 1 / 3 – ( 2k + 1 )( k + 2 ) / ( 2k + 1 )( 2k + 5 )( 2k + 3 )

= 1 / 3 – ( k + 2 ) / ( 2k + 3 )( 2k + 5 )

R.H.S. = 1 / 3 – ( k + 1 + 1 ) / ( 2k + 2 + 1 )( 2k + 2 + 3 )

= 1 / 3 – ( k + 2 ) / ( 2k + 3 )( 2k + 5 )

= L.H.S.

Therefore P ( k + 1 ) is true.

By the principle of mathematical induction, P ( n ) is true for all positive integers n.