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## Answers

The Brainliest Answer!

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...

Counter example :

and is not divisible by .

and is not divisible by .