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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 $$7 = \tfrac{7}{6}(7^1-1)$$ and $$7 + 7^2 = \tfrac{7}{6}(7^2-1)=56$$, so assume that $$7 + 7^2 + 7^3 + \dots + 7^k = \tfrac{7}{6}(7^k-1)$$ for some $$k\in \mathbb N$$. Continuing, we compute

 $$7 + 7^2 + 7^3 + \dots + 7^k+7^{k+1}$$ = $$\tfrac{7}{6}(7^k-1) +7^{k+1}$$ = $$\tfrac{7}{6}(7^k-1 +6\cdot 7^{k})$$ = $$\tfrac{7}{6}(7 \cdot 7^k-1)$$ = $$\tfrac{7}{6}( 7^{k+1}-1)$$ .

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 $$7 = \tfrac{7}{6}(7^1-1)$$ and $$7 + 7^2 = \tfrac{7}{6}(7^2-1)=56$$, so assume that $$7 + 7^2 + 7^3 + \dots + 7^k = \tfrac{7}{6}(7^k-1)$$ for some $$k\in \mathbb N$$. Continuing, we compute

 $$7 + 7^2 + 7^3 + \dots + 7^k = \tfrac{7}{6}(7^k-1) \Rightarrow$$ $$7 + 7^2 + 7^3 + \dots + 7^k + 7^{k+1} = \tfrac{7}{6}(7^k-1)+ 7^{k+1} \Rightarrow$$ $$7 + 7^2 + 7^3 + \dots + 7^k + 7^{k+1} = \tfrac{7}{6}(7^k-1 +6\cdot 7^{k}) \Rightarrow$$ $$7 + 7^2 + 7^3 + \dots + 7^k + 7^{k+1} =\tfrac{7}{6}(7 \cdot 7^k-1)\Rightarrow$$ $$7 + 7^2 + 7^3 + \dots + 7^k + 7^{k+1} = \tfrac{7}{6}( 7^{k+1}-1).$$

Thus $$7 + 7^2 + 7^3 + \dots + 7^n= \tfrac{7}{6}(7^n-1)$$ for all $$n \in \mathbb N.$$