Question

for any positive integer n.prove that n^2-n is divisible by 6.

Answer 1

For S(n) = if n=1  then n2-n/6 = 0 . Which is divisible by 6

there for S(1) true.

We can assume that n=k
S(k) = k^2 - k/ 6 = m (a number).
then k^2 - k = 6m
k^2= 6m+k --------------(eqn 1)
there for S(k) is true 

now we have to prove n =k+1
=> (k+1)^2 -(k+1)
=> (k^2+2k+1)-(k+1)
=>k^2+k+0---------->(eqn 2)

subtitute eqn 1 in eqn 2

=>6m+2k
=>2(3m+k)

Sorry i could solve this far...