Interest Rate for a Perpetuity

[helenli89]

Q: A perpetuity paying 1 at the beginning of the first year with each subsequent annual payment increasing by 3% has a present value of 10. If this perpetuity is exchanged for another perpetuity paying R at the beginning of every 9 years, find R to two decimal places so that the values of the two perpetuities are equal.

My attempt of solving this question is as follows:
PV = 1+ 1.03V + 1.03V^2 + 1.03V^3 +....
This is a geomatric series, therefore:
= (1-(1.03v)^n)/(1-1.03v) since it's a perpetuity, n goes to infinity. therefore v^n = 1/(1+i)^n goes to zero
= (1-0)/(1-1.03v)
So 10 = 1/(1-1.03v)
1/10 = 1-1.03/(1+i)
1.03/(1-1/10) = 1+i
i = 0.14444

Let V* = V^9
10 = RV* + R(V*)^2 + R(V*)^3 + ..... = RV* (1+v*+v*^2 + ....) =RV* (1/(1-v*)) = RV^9 /(1-v^9) = R/(1.1444^9 -1)
R = 23.67846

But the right answer is 14.1646. I know how they got this answer but I don't understand why my way of doing it does not give the right answer.
Thanks.

 

[tkhunny]

1) You paid R at the END of the first year.
2) I'm not seeing the "right" answer. I get 7.03

 

[jonah]

I'm not seeing the "right" answer

Neither could I.

A perpetuity paying 1 at the beginning of the first year with each subsequent annual payment increasing by 3% has a present value of 10.

\(\displaystyle 10 = 1 + \frac{{1.03}}{{i - 0.03}} \Leftrightarrow i = \frac{{13}}{{90}} \approx {\rm{0}}{\rm{.144444444444444}}...\)

Accordingly, the nominal rate j compounded every 9 years equivalent to an effective rate of 13/90 is easily calculated with

\(\displaystyle \left( {1 + \frac{i}{1}} \right)^1 = \left[ {1 + \frac{j}{{\left( {{\textstyle{1 \over 9}}} \right)}}} \right]^{{\textstyle{1 \over 9}}} \Leftrightarrow j = \left[ {\left( {1 + i} \right)^9 - 1} \right]\left( {{\textstyle{1 \over 9}}} \right)\)

Thus

If this perpetuity is exchanged for another perpetuity paying R at the beginning of every 9 years, find R to two decimal places so that the values of the two perpetuities are equal.

Then

\(\displaystyle 10 = R + \frac{R}{{\left[ {\frac{j}{{\left( {{\textstyle{1 \over 9}}} \right)}}} \right]}} \Leftrightarrow R \approx {\rm{7}}{\rm{.0307445477765}}...\)

Same as Sir tkhunny's.