1+4+7+---------+(3n-2)=1 n(3n-1) 

                                         2

For n=1
 (3*1-2) = 1( 1(3*1-1)

1 = 1

for n = k.

 1+4+7+---------+(3k-2) = 1/2 k(3k-1) 

for n= k +1

1+4+7+---------+(3k-2) +(3(k+1)-2)

= 1/2 k(3k-1) + 3k + 3 - 2

= 1/2 3k² - k/2 + 3k + 1

= [3k² - k + 6k + 2]/2

= [3k² + 5k + 2]/2

= [ 3k² + 3k + 2k + 2]/2

= [ 3k(k+1) + 2(k+1)]/2

= [(3k+2)(k+1)]/2

= (k+1)[3(k+1)-1]/2

= R.H.S.