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2015-04-13T16:35:31+05:30
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...
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2015-04-13T17:31:59+05:30
Counter example : n=2

2^2-2 = 4-2 = 2 and 2 is not divisible by 6.
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