Define a_{n} by a_{0} = α, a_{n+1} = 2a_{n} - n^{2}. For which α are all a_{n} positive?

**Solution**

Answer: α ≥ 3.

The trick is that we can solve the recurrence relation. A particular solution is obviously a polynomial with leading term n^{2}, so we soon find a_{n} = n^{2} + 2n + 3. The general solution is then a_{n} = n^{2} + 2n + 3 + k2^{n}. The initial condition a_{0} = α gives that k = α - 3. A necessary and sufficient condition for all a_{n} to be positive is evidently k ≥ 0, or α ≥ 3.

© John Scholes

jscholes@kalva.demon.co.uk

10 Dec 1999

Last corrected/updated 21 Nov 02