Let p(n) = n3 - n
i)If n=1 , p(1 ) = 1^3 - 1 [ 1^3 means 1 cube ]
= 1 - 1
= 0 [ 0 divisible by 3]
If n =1 , p(1) is true--------------------( 1)
ii) Assume that p( m ) is true.--------------------------(2 )
iii) we have to prove that p(m+ 1) is true.
p(m + 1) = (m + 1) ^3 - (m + 1)
= (m + 1 ) [ (m + 1)^2 -1] [ (m + 1) is the common factor ]
= ( m + 1) [ m 2 + 2m + 1 - 1 ] [ expansion of (m + 1 ) ^2 ]
= ( m + 1 ) [ m2 + 2m ]
= ( m + 1 ) m ( m + 2 ) [ m is the common factor ]
{ divisibility rule of 3 is If sum of the terms divisible by 3 then it is divisible by 3}
sum = m + 1 + m + m + 2
= 3m + 3
= 3 ( m + 1)
it is divisible by 3.
Therefore ( m+1) m (m+2 ) is divisible by 3
p(m+1) is divisible by 3
p(m+1) is true---------------------------(3)
From (1), (2) and (3)
for all values of n , p(n) = n3 - 3 is divisible by 3