Question 965979
let the consecutive numbers are n,(n+1),(n+2)
product = n*(n+1)*(n+2)

if n=1
P(1) = 1*2*3=6 --Even
Assume that P(n)=even then we have to prove that P(n+1)=(n+1)*(n+2)*(n+3) is also even

P(n+1)= (n+1)*(n+2)*(n+3)
      =n*(n+1)*(n+2)+ 3*(n+1)*(n+2)
we know that n*(n+1)*(n+2)is even ( assumed )
we have to show that 3*(n+1)*(n+2) is also even 
if n is odd then n+1 is even
either way n^2+3*n+2 is even
so 3*(n+1)*(n+2)=3*(n^2+3*n+2) is also even
we have shown that P(1) is even and that P(n+1)is even if P(n)is even.
hence proved.