If a_(n+1) = a_(n) / a_(n-1), find a_2017
We are given; a1 = 2, a2 = 6 an+1 = an / an-1 And we want to find a2017. The question is hard because if we tried simply applying the formula to get an answer, we would have to do it thousands of times. We want an+1 to depend on other terms of the sequence as little as possible. Right now it depends on two of them. But if we apply the recursive formula to an, we get this: an+1 = an / an-1 = (an-1 / an-2) / an-1 = 1/an-2 (only works if n ≥ 3) Equivalently: an+3 = an (for all n ≥ 1) This is perfect! It means that if you only look at every third term of the sequence, the terms just keep being "flipped". a1 = 2 a3+1 = 1/2 a6+1 = 2 a9+1 = 1/2 ... a2016+1 = 2 This works out because 2016 is an even multiple of 3, so we know that the 2 has been inverted an even amount of times - that's the same as not flipping it at all. Awesome right?
Related MAT University answers
All answers ▸Let r and s be integers. Then ( 6^(r+s) x 12^(r-s) ) / ( 8^(r) x 9^(r+2s) ) is an integer when: (a) r+s <= 0, (b) s <= 0, (c) r <= 0, (d) r >= s.
How many distinct real roots does the equation x^3 − 30x^2 + 108x − 104 = 0 have?
Circle the correct letter: The equation x^3 - 30x^2 + 108x - 104 = 0 has a) No real roots; b) Exactly one real root; c) Three distinct real roots; d) A repeated root.
(Note this is the kind of exercise I would ask someone who is doing further maths and especially someone MAT/STEP) Sketch the graph of y=sin(1/x)