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1. Hi all

We've just started a sequences topic, and Ive never been good at sequences (finding the nth term) at GCSE and now Im struggling to find the nth terms again.

Once Im given the rule its easy, its just finding it

Does anyone have any tips on the best ways to improve myself at finding the nth terms? We have also now started Sigma notation (using r) so any ideas how to help me?

I would appreciate all help, as I really need to brush up my skills, but my text book is utter rubbish at providing advice

thanks!
2. Also, I can do simple ones, e.g. 5,9,13,17...

Its ones that involve fractions and negatives for example:

2/3 + 5/9 + 8/27 + 11/81 + 14/243 + 17/729

and

1 x 4 - 3 x 7 + 5 x 10.......+ 29 x46

Help? lol

3. What you want to do is look for a pattern with these problems. Ie is the next term just the previous one multiplied by a number, or is it some combination of addition and multiplication etc. If you can predict the next term in the sequence, the general formula will follow.
For the 2nd series you gave, you might have spotted that the top part of each fraction is the top part of the previous one, with 3 added on. And that the bottom part of the next term is the bottom of the previous term multiplied by 3. So the next term after 17/729 will be (17+3)/(3*729). Now you already know how to get the expressions for simple series which are of the form a(n) = k*n+d, or a(n)=d+k^n, and the thing you need to realise is that this is a combination of both, and is hence a(n)=(3*n-1)/(3^2)

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Updated: January 28, 2010
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