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