Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.

Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
  $$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$

Along my proof I am stuck at the above section where it would be shown that:

$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$   is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.

let f(n)=3n-2
let s(n)=[f(1)+f(2)+f(3)+...+f(n-1)+f(n)]

s(n)    =(3(1)-2)+(3(2)-2)+(3(3)-2)+...+(3(n-1)-2)+(3(n)-2)
s(n)    =(3(1)   + 3(2)   + 3(3)   +...+ 3(n-1)   + 3(n)  )-2n
s(n)    = 3(1    +   2    +   3    +...+  (n-1)   +   n   )-2n
s(n)    =3(sum of all #s from 1 to n)       -2n
s(n)    =3(sum of all #s from 1 to n-1)+3(n)-2n
s(n)    =3(sum of all #s from 1 to n-1)+n

Now we reverse the order:
s(n)    =3((n-1)+(n-2)+...3+2+1+(n-n))+n
s(n)    =3(n^2)-3(sum of all #s from 1 to n)+n
s(n)    =3(n^2)-3(sum of all #s from 1 to n)+n+n-n
s(n)    =3(n^2)-3(sum of all #s from 1 to n)+2n-n
s(n)    =3(n^2)-3(sum of all #s from 1 to n)-(-2n)-n
s(n)    =3(n^2)-[3(sum of all #s from 1 to n)-2n]-n

We've got another value for s(n) on the right side now.
s(n)    =3(n^2)-[s(n)]-n
2[s(n)] =3(n^2)-n
2[s(n)] =n(3n)-n(1)
2[s(n)] =n(3n-1)
s(n)    =n(3n-1)/2

