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 = 3^1 - 1=> 2 = 2`
Step 2: Inductive step: Show that if P(k) holds, then also P(k + 1) holds:
`P(k): 2(1 + 3 + 3^2 + .. +3^(k-1)) = 3^k - 1` holds
`P(k+1): 2(1 + 3 + 3^2 + .. + 3^(k-1) + 3^k) = 3^(k+1) - 1`
You need to use induction hypothesis that P(k) holds, hence, you need to re-write the left side, such that:
`3^k - 1 + 2*3^k = 3^(k+1) - 1`
Reduce like terms, such that:
`3^k + 2*3^k = 3^(k+1)`
`3*3^k = 3^(k+1) `
Use the rule of exponents:
`3^(k+1) = 3^(k+1)`
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(1 + 3 + 3^2 + .. +3^(n-1)) = 3^n - 1` holds for all positive integers n.
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