The geometric series, proved

Here is the induction proof, with its lines shuffled. Put them in order with the arrows, or drag a line by its handle.

1. 1Its top simplifies to $1 - r^{k+1}$, so the right side is the formula for $n = k + 1$.
2. 2Fix a number $r \ne 1$. Base case: for $n = 1$ the left side is $1$, and the right side is $\dfrac{1 - r}{1 - r} = 1$.
3. 3Both steps hold, so by induction the formula holds for every $n \ge 1$.
4. 4Over a common denominator, the right side becomes $\dfrac{1 - r^k + r^k(1 - r)}{1 - r}$.
5. 5Add the next term, $r^k$, to both sides. The left side becomes $1 + r + \cdots + r^k$.
6. 6Inductive step: assume the formula holds for some $n = k \ge 1$.

The same proof, with its key steps left out. Type each missing piece as you would a quick answer, such as r^k or (1-r)/(2+r), then check.

A different proof, by subtraction, as in lesson 1.5.2. One line claims more than the lines before it have shown. Select it.

1. 1Let $S = 1 + r + r^2 + \cdots + r^{n-1}$, for any number $r$.
2. 2Multiply both sides by $r$: $rS = r + r^2 + \cdots + r^n$.
3. 3Subtract the second equation from the first. Every term but $1$ and $r^n$ cancels, so $S - rS = 1 - r^n$.
4. 4Factor the left side: $S(1 - r) = 1 - r^n$.
5. 5Divide both sides by $1 - r$: $S = \dfrac{1 - r^n}{1 - r}$ for every number $r$.

Prove by induction that $1 + 3 + 5 + \cdots + (2n - 1) = n^2$ for every $n \ge 1$. Write the base case, the inductive step and the conclusion, in sentences.

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