Question

Order the parts of the proof for:\\ $\sum_{i=0}^{n} r^i = \frac{r^{n+1}-1}{r-1}, \forall n \ge 0$\\ $\frac{r^{k+1}-1 + r^{k+1} - r^{k+1}}{r-1} = \frac{r^{k+2}-1}{r-1}$ \\ Therefore, if $\sum_{i=0}^{k} r^i = \frac{r^{k+1}-1}{r-1}$ for some $k \ge 0$, then\\ $\sum_{i=0}^{k+1} r^i = \frac{r^{k+2}-1}{r-1}$ \\ $\sum_{i=0}^{k+1} r^i = (\sum_{i=0}^{k} r^i) + r^{k+1} = (\frac{r^{k+1}-1}{r-1}) + \frac{(r-1)r^{k+1}}{r-1}$ \\ Let n=0,\\ By definition, $\sum_{i=0}^{0} r^i = r^0 = 1$ \\ By conjecture, $\frac{r^{0+1}-1}{r-1} = \frac{r-1}{r-1} = 1$ \\ Which confirms the conjecture for n=0.\\ Assume $\sum_{i=0}^{k} r^i = \frac{r^{k+1}-1}{r-1}$ for some $k \ge 0$ \\ Therefore, $\sum_{i=0}^{n} r^i = \frac{r^{n+1}-1}{r-1}, \forall n \ge 0$

          Order the parts of the proof for:\\
$\sum_{i=0}^{n} r^i = \frac{r^{n+1}-1}{r-1}, \forall n \ge 0$\\
$\frac{r^{k+1}-1 + r^{k+1} - r^{k+1}}{r-1} = \frac{r^{k+2}-1}{r-1}$ \\
Therefore, if $\sum_{i=0}^{k} r^i = \frac{r^{k+1}-1}{r-1}$ for some $k \ge 0$, then\\
$\sum_{i=0}^{k+1} r^i = \frac{r^{k+2}-1}{r-1}$ \\
$\sum_{i=0}^{k+1} r^i = (\sum_{i=0}^{k} r^i) + r^{k+1} = (\frac{r^{k+1}-1}{r-1}) + \frac{(r-1)r^{k+1}}{r-1}$ \\
Let n=0,\\
By definition, $\sum_{i=0}^{0} r^i = r^0 = 1$ \\
By conjecture, $\frac{r^{0+1}-1}{r-1} = \frac{r-1}{r-1} = 1$ \\
Which confirms the conjecture for n=0.\\
Assume $\sum_{i=0}^{k} r^i = \frac{r^{k+1}-1}{r-1}$ for some $k \ge 0$ \\
Therefore, $\sum_{i=0}^{n} r^i = \frac{r^{n+1}-1}{r-1}, \forall n \ge 0$

Order the parts of the proof for:

∑i=0^n r^i = (r^n+1-1)/(r-1), ∀ n ≥ 0

(r^k+1-1 + r^k+1 - r^k+1)/(r-1) = (r^k+2-1)/(r-1)

Therefore, if ∑i=0^k r^i = (r^k+1-1)/(r-1) for some k ≥ 0, then

∑i=0^k+1 r^i = (r^k+2-1)/(r-1)

∑i=0^k+1 r^i = (∑i=0^k r^i) + r^k+1 = ((r^k+1-1)/(r-1)) + ((r-1)r^k+1)/(r-1)

Let n=0,

By definition, ∑i=0^0 r^i = r^0 = 1

By conjecture, (r^0+1-1)/(r-1) = (r-1)/(r-1) = 1

Which confirms the conjecture for n=0.

Assume ∑i=0^k r^i = (r^k+1-1)/(r-1) for some k ≥ 0

Therefore, ∑i=0^n r^i = (r^n+1-1)/(r-1), ∀ n ≥ 0

Added by Deborah P.

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