# maths iteration homework

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#1
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.
0
6 months ago
#2
(Original post by Hannahd2003)
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.
The hard way would be to expand the RHS, rearrange so that you have a quadratic in an+1 (treating an as a constant), and solving using the formula or completing the square. The easier way would be to put a1 = 1 in the equation and solve for a2. This would give you a2 + 1 = (a2 - 1)2, which is pretty straightforward to solve.

Despite the unusual way that this is given in, it has been set up so that the sequence is quite well behaved.
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6 months ago
#3
(Original post by Hannahd2003)
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 into the form

which, if you solve, yields

You're told that .

So, hence .

Can you spot the pattern ?
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#4
(Original post by RDKGames)
You can expand into the form

which, if you solve, yields

You're told that .

So, hence .

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?
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6 months ago
#5
(Original post by Hannahd2003)
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?
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#6
(Original post by RDKGames)
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?
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6 months ago
#7
(Original post by Hannahd2003)
I just guessed, but I think that I may need to work it out and draw some sort of diagram?
Yeah working it out is usually a better idea than guessing when you're doing maths ...
2
#8
So, if a1 = 1 and a2 = 0 or 3, is it possible for a3 to equal 1 or 0?
0
6 months ago
#9
(Original post by Hannahd2003)
So, if a1 = 1 and a2 = 0 or 3, is it possible for a3 to equal 1 or 0?
Why wouldn't it be?
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#10
(Original post by RDKGames)
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.
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6 months ago
#11
(Original post by Hannahd2003)
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.
It is definitely not an arithmetic sequence. You should refer to the definition of what an arithmetic sequence is.
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6 months ago
#12
(Original post by Hannahd2003)
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.
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#13
(Original post by ghostwalker)
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.
a3 = 1, 0 and 6 but if 0 doesn't class as a positive number then it can only be 1 and 6. That helps a lot, thank you.
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#14
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.
0
6 months ago
#15
(Original post by Hannahd2003)
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.
0
6 months ago
#16
(Original post by Hannahd2003)
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
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#17
(Original post by davros)
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
That's much easier, thank you
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