# Calculating present value of a bond (understanding a step)

Our professor calculated the present value of a bond with $$T=10$$ years, $$FV=10,000$$€, $$C=700$$€ p.a. and an expected rate of return $$r$$. He wrote \begin{align}PV&=C\cdot\sum_{n=1}^{10}\frac{1}{(1+r)^n}+\frac{FV}{(1+r)^n}\\&=700\cdot\color{red}{\frac{1-\frac{1}{(1+r)^n}}{(1+r)^n}}+\frac{FV}{(1+r)^n}\end{align} I don't understand the red step (I guess he transformed the sum).

In our lectures we have to different formulas for calculating the present value (with cash flows $$C_n$$): $$PV=\sum_{n=1}^T\frac{C_n}{(1+r)^n} \tag{1}$$ and annuity (value of $$C$$ received each year for $$T$$ years): $$PV = \frac{C}{r}\left(1-\frac{1}{(1+r)^T}\right).\tag{2}$$ I understand that if in $$(1)$$ $$C_n=C\ \forall n=1,...,T:$$ $$PV=C\cdot\sum_{n=1}^T\left(\frac{1}{(1+r)}\right)^n$$ Maybe it's connected to the geometric sum? Thanks for every help!

• Professor may have made a mistake in the denominator of the red portion. Need $r$ there and not $(1+r)^n$ Jan 21 at 18:14
• May be. Then it would look more like $(1)$ and $(2)$. But how do I simplify the sum to the red fraction?
– Uhmm
Jan 21 at 18:16

Not sure what the issue is here, but $$S_n=\frac{a(x^n-1)}{x-1}$$ for the sum of the first $$n$$ terms of a series with first term $$a$$ and common ratio $$x$$. In your case $$a=x=\frac{1}{1+r}$$ so $$\sum_{n=1}^{10} \frac{1}{(1+r)^n}=\frac{1}{r}\left( 1-\frac{1}{(1+r)^{10}}\right)$$ which as @nbbo2 has pointed out is your red term with the denominator corrected to be $$r$$ instead of $$(1+r)^n$$.