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Not 100 percent sure, but here are my workings for this question. For the third column of my workings (with the characteristic equation), it is noteworthy to know that the golden ratio for the Fibonacci sequence (1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...) with recurrence relation a (n + 2) = a (n + 1) + a n is obtained using very similar workings.
Date Posted:
5 years ago
Also, I don't recall this question being taught in JC during my JC days more than 10 years ago (and in fact I already forgot most of my JC and uni stuffs!), so I had to use my uni knowledge (with a little modification by introducing lg on both sides first) to do this question.
I am not so sure what you have learnt in differential equations so far, because I only learned this technique during university (my A Level syllabus ten years ago had no such level of differential equations).
Below I have provided a general idea to solve the recurrence relations. It’s not apparently clear how they found out such general solutions, because I would have never guessed it. I suspect it was due to years of analysis of equations that the older generations of mathematicians undertook.
In fact, I noticed that the numerators follow a pattern, which is simply the previous term multiplied by 4 and then either a plus 1 or -1, where the +-1 is done alternately. Hence, it cannot take up a single ratio r1, but because there are two repeating but related patterns, it could be made of two ratios r1 and r2.
Of course, to talk about the powers only rather than the entire value of 10^power, I used lg to kill off the base 10.
This question was given as a "question of the week" by my teacher !