The annuity formula, proved

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

1. 1Fix a rate $r > -1$ with $r \ne 0$, and write $v = \frac{1}{1 + r}$, so that $rv = 1 - v$.
2. 2Add the next discount factor, $v^{k+1} = v \cdot v^k$, to both sides. On the right, write it as $\frac{rv \cdot v^k}{r}$.
3. 3Since $rv = 1 - v$, the right side becomes $\frac{1 - v^k + (1 - v)v^k}{r}$, which is $\frac{1 - v^{k+1}}{r}$.
4. 4Base case: for $n = 1$ the sum is $v$, and the formula gives $\frac{1 - v}{r} = \frac{rv}{r} = v$.
5. 5That is the formula for $n = k + 1$. By induction it holds for every $n \ge 1$, and multiplying by $C$ gives the claim.
6. 6Inductive step: assume that $v + v^2 + \cdots + v^k = \frac{1 - v^k}{r}$ for some $k \ge 1$.

The proof by this route, with its key steps left out. Type each missing piece in $r$ and $k$, as you would a quick answer, such as 1/(1+r) or -k, then check.

This proof runs the wrong way. One line draws a conclusion that the lines before it do not support. Select it.

1. 1Base case: for $n = 1$ both sides are $v$, since $\frac{1 - v}{r} = \frac{rv}{r} = v$.
2. 2Inductive step: assume the formula for $n = k$, and suppose that $v + \cdots + v^{k+1} = \frac{1 - v^{k+1}}{r}$.
3. 3Subtract $v^{k+1}$ from both sides. The left side becomes $v + \cdots + v^k$.
4. 4The right side becomes $\frac{1 - v^{k+1} - rv^{k+1}}{r} = \frac{1 - v^k}{r}$, because $v^{k+1}(1 + r) = v^k$.
5. 5This is the formula for $n = k$, which we assumed. So the supposition is true, and the formula holds for $n = k + 1$.
6. 6By induction, the formula holds for every $n \ge 1$.

Prove by induction that $B_n = C \cdot \frac{(1 + r)^n - 1}{r}$ for every $n \ge 1$, when $B_1 = C$, $B_{n+1} = (1 + r)B_n + C$ and $r \ne 0$. Write the base case, the inductive step and the conclusion, in sentences.

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