maths iteration homework

Consider sequences a1, a2, a3, . . . of positive real numbers with a1 = 1 and such that an+1 + an = (an+1 − an)^2 for each positive integer n. How many possible values can a2017 take?

I am really confused on this question, I have tried to find out what a2 is, but I got (a2)^2-2a2. I then found out what was being added to each term of the sequence ( I got (a2)^2-2a2-1 ) Finally i substituted it into the sequence to get a2017 = 2016(a2)^2 - 4032a2 - 2015. However, because I can't find a value for a2, I can't find out what a2017 is. If someone could please tell me what I have done wrong or how I can find other values, it would be great.

Thank you.

Consider sequences a1, a2, a3, . . . of positive real numbers with a1 = 1 and such that an+1 + an = (an+1 − an)^2 for each positive integer n. How many possible values can a2017 take?

I am really confused on this question, I have tried to find out what a2 is, but I got (a2)^2-2a2. I then found out what was being added to each term of the sequence ( I got (a2)^2-2a2-1 ) Finally i substituted it into the sequence to get a2017 = 2016(a2)^2 - 4032a2 - 2015. However, because I can't find a value for a2, I can't find out what a2017 is. If someone could please tell me what I have done wrong or how I can find other values, it would be great.

Thank you.

You can expand $a_{n+1} + a_n = (a_{n+1} - a_n)^2$ into the form

$a_{n+1}^2 - (2a_n + 1)a_{n+1} + (a_n^2 - a_n) = 0$

which, if you solve, yields

$a_{n+1} = \dfrac{2a_n + 1 \pm \sqrt{8a_n + 1}}{2}$

You're told that $a_1 = 1$.

So, $a_2 = \dfrac{3 \pm 3}{ 2 }$ hence $a_2 \in \{ 0, 3\}$.

What about $a_3$ ?

Can you spot the pattern ?

So, if a1 has 1 value, a2 has 2 values, a3 has 4 values... So the number of values equals 2n-2, so if n=2017, the number of values is 4032?

Did you calculate a3 values explicitly, or are you guessing that it has 4?

I just guessed, but I think that I may need to work it out and draw some sort of diagram?

So, if a1 = 1 and a2 = 0 or 3, is it possible for a3 to equal 1 or 0?

Why wouldn't it be?

Because an arithmetic sequence goes up/down by the same number each time, so if the sequence goes 1,0... It goes down by one each time so the third term couldn't be 0. Just a thought.

So, if a1 = 1 and a2 = 0 or 3, is it possible for a3 to equal 1 or 0?

Perhaps worth commenting that the sequence is of positive real numbers, so 0 is not acceptable. Also, if 1 and 0 are your only choices for a3, then you're gone wrong somewhere.

So I have now realised that all of the numbers are triangular, so I have used gauss' formula to find out the 2017th triangular number (which was 2035153) and then I divided it by 2017 to get the number of values as 1009.

So I have now realised that all of the numbers are triangular, so I have used gauss' formula to find out the 2017th triangular number (which was 2035153) and then I divided it by 2017 to get the number of values as 1009.

Just to point out now you've got the answer...you didn't need to work out that large triangular number - since the formula is (n/2)(n + 1) and you're dividing by n again (in this case 2017) you just needed to work out (1/2) (2018) = 1009