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Given the following sequence of numbers

1, 3, 7, 9, 13, 15, 21, 25, 27, 31, 33, 37, 43, 45, 49, 51, 55, 57, 63, 67, 69, 73, 75, 79, 85, 91, 93, 97, 99

Can anyone show me the calulation for the nth term?

The sequence of first differences can be seen to repeat 2 sets of numbers

2, 4, 2, 4, 2, 6, 4, 2, 4, 2, 4, 6, 2, 4, 2, 4, 2, 6, 4, 2, 4, 2, 4, 6, 2, 4, 2, 4, 2

I'm mucking about with codes for my lower level students just for fun - then realised I couldn't actually $n$th sequence the code I'd created - oops!

Thanks!

  • 0
    You made an error, I think. You have two adjacent first differences of 6: $85 - 79 = 91 - 85 = 6$.2011-09-16

1 Answers 1

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Call the sequence $a(n)$ with $n$ starting from $0$. Then$a(12n)=42n+1$ $a(12n+1)=42n+3$$a(12n+2)=42n+7$ $...$$a(12n+11)=42n+37$