Prove by PMI
1) a+(a+d)+(a+2d)+ ......[a+(n-1)d] = n/2 [2a+(n-1)d], n E N.
2) n(n+1)(n+5) is divisible by 6 for all n E N.
3) 9 raised to n - 8n - 1 is a multiple of 64 for all n E N.
Assume p(k) is true
a+a+d+(a+2d)………..[a+(k-1)d]+ [a+kd ] =k/2[2a+(k-1)d]+ [a+kd ]
It is same as our assumption where k is replaced by k+1. Therefore it is true for all natural numbers