You need to use mathematical induction to prove the formula for every positive integer n, hence, you need to perform the two steps of the method, such that:

Step 1: Basis: Show that the statement P(n) hold for n = 1, such that:

`1 = 1/2*(3*1-1) => 1 =2/2 => 1=1`

Step 2:...

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You need to use mathematical induction to prove the formula for every positive integer n, hence, you need to perform the two steps of the method, such that:

Step 1: Basis: Show that the statement P(n) hold for n = 1, such that:

`1 = 1/2*(3*1-1) => 1 =2/2 => 1=1`

Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds:

`P(k): 1 + 4 + 7 + .. + (3k-2) = (k(3k-1))/2 ` holds

`P(k+1): 1 + 4 + 7 + .. + (3k-2) + (3k+1) = ((k+1)(3k+2))/2`

You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side, such that:

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

`3k^2 - k + 6k + 2 = 3k^2 + 2k + 3k + 2`

You need to add the like terms, such that:

`3k^2 + 5k + 2 = 3k^2 + 5k + 2`

Notice that P(k+1) holds.

**Hence, since both the basis and the inductive step have been verified, by mathematical induction, the statement `P(n): 1 + 4 + 7 + .. + (3n-2) = (n(3n-1))/2` holds for all positive integers n.**