Assume that the interest rate is r, so that capital of k becomes k(1 + r)n after n years. How much do we need to invest to be able to withdraw 1 at the end of year 1, 4 at the end of year 2, 9 at the end of year 3, 16 at the end of year 4 and so on (in perpetuity)?
Solution
To meet the withdrawal at the end of year n we need to invest n2/(1+r)n. Thus the total investment required is 1/(1+r) + 4/(1+r)2 + 9/(1+r)3 + 16/(1+r)4 + ... . Now we know that 1 + x + x2 + ... = 1/(1 - x). Differentiating, we get: x/(1 - x)2 = x + 2x2 + 3x3 + ... . Differentiating again: x(1 + x)/(1 - x)3 = 12x + 22x2 + 32x3 + ... . Substituting x = 1/(1+r) gives (1 + r)(2 + r)/r3 as the required investment.
© John Scholes
jscholes@kalva.demon.co.uk
25 Feb 2002