`3 + 6 + 9 + 13 + .... 3n = (3n)/2 (n + 1)` Use mathematical induction to prove the formula for every positive integer n.

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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:

`3 = 3*1*(1+1)/2 => 3 = 3*1 => 3=3`

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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:

`3 = 3*1*(1+1)/2 => 3 = 3*1 => 3=3`

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

`P(k): 3+ 6 + .. + 3k = (3k(k+1))/2 ` holds

`P(k+1):  3+ 6 + .. + 3k + 3(k+1) =  (3(k+1)(k+2))/2`

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

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

`3k(k+1) + 6(k+1) = 3(k+1)(k+2)`

Factor out (k+1) to the left side:

`(k+1)(3k+6) = 3(k+1)(k+2)`

Factor out 3 to the left side:

`(k+1)3*(k+2) = 3(k+1)(k+2)`

`3(k+1)(k+2) = 3(k+1)(k+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): 3+6+9+...+3n = (3n(n+1))/2 ` holds for all positive integers n.

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