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Prove that \(7 + 7^2 + 7^3 + \dots + 7^n= \tfrac{7}{6}(7^n-1)\) for all \(n \in \mathbb N.\)
Before we begin the proof, let us examine the propositional function involved: \[ P(n) := (7 + 7^2 + 7^3 + \dots + 7^n= \tfrac{7}{6}(7^n-1)). \] The basis step for an induction proof involves evaluating this function at a few values of \(n\). In this case \(P(1)=(7=\tfrac{7}{6}(7^1-1))=(T)\), and \(P(2)=(7+7^2=\tfrac{7}{6}(7^2-1)=56)=(T)\). Indeed, evaluate \(P(n)\) at any natural number \(n\) and you will obtain a true statement. That is the point of mathematical induction! How can we know for sure that \(P(n)\) is a true statement for all \(n\in \mathbb N\)? We certainly cannot try all values of \(n\), so we verify that \(P(n)\) is true for one or two small values of \(n\), then prove that if \(P(k)\) is true, then \(P(k+1)\) follows from that assumption; that is, \(P(k)\Rightarrow P(k+1)\). Those two steps together contstitute mathematical induction.
For this specific problem, \[P(k)=(7 + 7^2 + 7^3 + \dots + 7^k= \tfrac{7}{6}(7^k-1)) \] and \[ P(k+1)=(7 + 7^2 + 7^3 + \dots + 7^k+7^{k+1}= \tfrac{7}{6}(7^{k+1}-1)). \] How can we start with \(P(k)\) and obtain \(P(k+1)\)? The left sides of \(P(k)\) and \(P(k+1)\) tell us that adding \(7^{k+1}\) to both sides of \(P(k)\) would move us in the right direction. Let us continue with the proof to see how this works!
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Proof: We can see that and , so assume that . Continuing, we compute
Thus \(7 + 7^2 + 7^3 + \dots + 7^n= \tfrac{7}{6}(7^n-1)\) for all \(n \in \mathbb N.\)
We can also begin with the induction assumption (after the basis step of course!) and continue with valid operations.
Proof: We can see that and , so assume that . Continuing, we compute
Thus \(7 + 7^2 + 7^3 + \dots + 7^n= \tfrac{7}{6}(7^n-1)\) for all \(n \in \mathbb N.\)