Please illustrate that a bond with maturity N years that has coupon equal to its yield is associated with the conversion factor of 1.
I do this by writing out $$\frac1{100} \left( \sum_{t=1}^N \left[ \frac{100 (0.06)}{1.06^t} \right]+\frac{100}{1.06^N} \right)$$ but I do not get that this = 1.
I use the formula:
$$\sum_{k=m}^n a^k = \begin{cases}\frac{a^{n+1} - a^m}{a-1}, \quad &a \neq 1\\n-m+1, \quad &a=1\end{cases}$$
We are given a bond with Coupon = Yield = $6 \%$ and Maturity $N$. We want to check that the conversion factor = 1, in other words that $$\frac1{100} \left( \frac{100}{1.06^N} + \sum_{t=1}^N \frac{100 \cdot 0.06}{1.06^t} \right) = 1 $$ or equivalently $$ \frac{1}{1.06^N} + \sum_{t=1}^N \frac{0.06}{1.06^t} = 1. $$
The first term is $\frac{1}{1.06^N} = \left( \frac{1}{1.06} \right)^N.$ The second term can be simplified using the formula for geometric sums $$\sum_{t=1}^N \frac{0.06}{1.06^t} = 1 - \left( \frac{1}{1.06} \right)^N.$$ Now we add the first term and the second term and see that $$\left( \frac{1}{1.06} \right)^N \quad + \quad 1 - \left( \frac{1}{1.06} \right)^N = 1. $$
