Could anyone solve this challenge problem on series, very similar to the Fibonacci Sequence? an = an-1 + an-2 - an-3.

Question

It is given that a₁ = $0.2$, a₂ = $0.3$, a₃ = $0.5$ and a_n = a_n-1 + a_n-2 - a_n-3

P = a₂₀₁₃

Q = a₁ + a₂ + a₃ + ... + a₂₀₁₃

R is the minimum value of i such that a_i > 2013.

S is the number of possible values for i such that ai is an integer for i $\leqslant$ i $\leqslant$ 2013.

Find the values of P, Q, R and S.

(It should be able to be solved without the use of a calculator.)

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• I got the answer 301.8 for P, by realising that the value of the terms increases by 0.1 and 0.2 in an alternating way.

P = $(2013-1) / 2 * (0.1+0.2)$ = $1006 * 0.3$ = $301.8$

Answer 1

Your method is pretty close to optimal, I'd say.

If you want another approach, note that the characteristic polynomial is $$x^3-x^2-x+1=(x-1)^2(x+1)$$ so the general solution to the recursion is $$A+B (-1)^n+Cn$$

Using your given initial conditions we see that $$a_n=\frac 1{40}-\frac 1{40}(-1)^n+\frac 3{20}n$$ And you can now sum and solve analytically.

Note, however, that the simplicity of this solution really is a lucky accident (or a contrivance of the problem). Most cubics have unpleasant solutions so this analytic method would return a very complex formula.