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The problem:
A lottery winner is given two payment options: Receive 131 million dollars in 25 yearly installments of equal size, the first payable immediately, or receive a single immediate payment of 70.3 million dollars. Assuming that these plans are of equal value to the state lottery system, what interest rate is the state getting on its investments?

My progress so far:
Using the formula $c = \frac{(1+r)^nPr}{{(1+r)^n-1}}$ (where $c$ = monthly payment, $P$ = principal, $r$ = monthly interest rate, and $n$ = number of payment periods), I have set up the equation $$25\cdot\frac{(1+r)^{25}\cdot70300000r}{(1+r)^{25}-1}=131000000,$$ which I have simplified to $$(262-3515r)(1+r)^{25}=262.$$ I do not know how I can solve this equation. Please help!

J. W. Tanner
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ERROR 404
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  • What you can do is let $r$ be $1+ the interest rate$. – GSmith Jun 26 '24 at 23:38
  • Normally the annuity problems where you find the interest rate require numerical methods. Financial calculators do this pretty easily (I like the BAII Plus), but any numerical solver can do it. – J. Chapman Jun 26 '24 at 23:54
  • @GSmith With $x=1+r,$ we have $$(262-3515(x-1))x^{25}=262,$$ which simplifies to $$3515x^{26}-3777x^{25}-262=0,$$ which, I assume, is pretty difficult to solve algebraically. – ERROR 404 Jun 27 '24 at 00:09
  • https://math.stackexchange.com/questions/1536653/approximating-the-compond-interest-for-a-loan – Claude Leibovici Jun 27 '24 at 04:16

1 Answers1

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If you look for an approximation of $r$ solution of $$a=\frac{r\,(1+r)^n}{{(1+r)^n-1}}\quad \quad\text{where} \quad a=\frac c P$$ beside what I wrote here, you can use the $[k,k]$ Padé approximant $P_k$ of the rhs, such as $$P_1=\frac 1 n\,\,\frac{1+\frac{1}{3} (n+2) r }{1-\frac{1}{6} (n-1) r }$$ $$P_2=\frac 1 n\,\,\frac{1+\frac{2}{5} (n+3) r+\frac{1}{20} (n+2) (n+3)r^2 } {1-\frac{1}{10} (n-7) r+\frac{1}{60} (n-2) (n-1) r^2 }$$

For your case, $n=25$ and $a=\frac{262}{3515}$, they give respectively $0.0524769$ and $0.0550283$ while the solution is $0.0549840$.

If you use $P_3$ (have a look here for its generation), at the price of a cubic equation, you would obtain $0.0549836$.

If the link is broken, the Wolfram Alpha input is

PadeApproximant[(r*(1+r)^n)/((1+r)^n-1),{r,0,{3,3}}]

Claude Leibovici
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  • I'm not convinced. OP states "installments of equal size, the first payable immediately" while your solution is for paying installments after each period. Thus the interest rate is $\approx 6,11680$%. BTW, this is the annual interest rate, the formula shown in OP is described with $c =$ monthly payment and $r =$ monthly interest rate but by implication used for annual account. – m-stgt Jun 28 '24 at 10:43
  • @m-stgt. I just played with the formula. If the equation is wrong, tell it. – Claude Leibovici Jun 28 '24 at 13:28
  • The equation is wrong. Pls compare with Appendix D, Formulas Used of HP12C Usr Gde.. A virtual 12C for free is still available. – m-stgt Jun 28 '24 at 14:15