Yet Another Cycling Forum
General Category => The Knowledge => OT Knowledge => Topic started by: a lower gear on 18 January, 2013, 07:44:54 pm
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The smallest gear's maths homework has me
struggling utterly failing to remember how to express the nth number in a sequence. Here are the three sequences I'm grappling with:
2,4,6,8,10,12,16... (constant interval of 2) is the answer 2n+2?
1,4,9,16,25,36,49... (interval grows in sequence 3,5,7,9,11,13...)
4,9,14,19,24,29,34... (constant interval of 5)
B*ggered if I can recall how to express the nth term in each series - can anyone in this parish remind me how its done?
40 years is a long time...
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When the difference is fixed; its (difference x n) +- fiddle factor.
The first one is 2n (assuming a typo in there?)
Third is 5n-1 (as 5n would give 5, 10, 15, 20)
Second is just n^2. Finding the nth term of a quadratic sequence is National Curriculum Level 7. Do you want a rundown on how to work it out?
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You basically have to think about what you have to do to the original 1,2,3,4,5...n to get to the answers in each sequence. So:
Sequence 1: 2,4,6,8,10 etc
To get from the original 1,2,3,4,5 you have to multiple each number in the original by 2 i.e. 2n
Sequence 2: 1,4,9,16,25 etc
To get from the original 1,2,3,4,5 you have to square each number in the original I.e. n2
Sequence 3: 4,9,14,19,24
To get from the original 1,2,3,4,5 you have to multiple each number in the original by 5 and subtract 1 I.e. 5n-1
So rather than thinking about the intervals you need to think what you do to each number in 1,2,3,4,5etc to get to the answers in the sequence.
Hope that makes sense...(and is right!).
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I was composing a reply, but icenutter and Pippa covered pretty much what I had in there, so I abandoned it in favour of this observation...
1,4,9,16,25,36,49... (interval grows in sequence 3,5,7,9,11,13...)
As already noted, this is just n2, however if you want to see where your differences in the subsequent terms being 2n-1 arises from then see below:--
difference between nth and (n-1)th term = n2 - (n-1)2
= n2 - (n-1)(n-1)
= n2 - (n2 - 2n + 1)
= n2 - n2 + 2n - 1
= 2n -1
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Thank you ever so much - its amazing how much one forgets over the years from lack of use - and I cannot recall ever having done this in my schooldays, though presumably I must have. :facepalm:
Pippa - your explanation of the underlying fundamental primciple was wonderfully clear. I had become side-tracked by the interval, leading me astray.
My son read this thread before bedtime and said he was very impressed at how friendly and helpful everyone was. Take a communal bow! ;)
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Handy hint to find icenutter's 'fiddle factor'. Calculate and extra term at the beginning, whatever would come before the first term if you were counting backwards. Call it the 'zeroth term', that's your fiddle factor.