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:
`2 = 1(1+1) => 2 = 1*2 => 2=2`
Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds:
`P(k): 2 + 4 + .. + 2k = k(k+1) ` holds
`P(k+1): 2 + 4 + .. + 2k + 2(k+1) = (k+1)(k+2)`
You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side, such that:
` k(k+1) + 2(k+1) = (k+1)(k+2)`
You need to notice that you can factor out k+1 to the left side, such that:
`(k+1)(k+2) = (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): 2 + 4 + 6 + ... + 2n = n(n+1)` holds for all positive integers n.
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