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.
Monday, August 8, 2016
Precalculus, Chapter 9, 9.4, Section 9.4, Problem 15
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