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    Use the method of finite differences ...

    Here's a page that explains it reasonably well:


    Just so you know, because the page uses the notation 'n!' Here is a note explaining this:

    n! = n(n - 1)(n - 2)...(3)(2)(1)

    So 4! = 4 * 3 * 2 * 1 = 24

    10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3,628,800

    There's probably a button on your calculator that can do calculations of the form 'n!' ...

    Hope this helps!


    for the formula where the first difference stays the same, eg 8 16 24 32, multiply n by the common difference and then add or take off the difference to give you the answer/

    eg. 7 15 23 31

    common difference = 8 therefore 8n

    8(1)+c = 7
    therefore c = -1

    Give me a minute on the quadratic ones...

    Simba, didn't know that method of finite differences existed, that's a good method to know. Many thanks.

    Hi Alex.

    Basically, find common differences between the numbers in the sequence. This means take each number and subtract it from its adjacent number, e.g. if the sequence was 6, 9, 12, 15, ..., then the common difference would be 3. Then, under each pair of adjacent numbers, write the common difference down, so in this case, write 3 3 3. Because we have a constant number ('3'), the formula will *involve* 3n. Now the next step is to check if this works. I.e. if n=1, we have the answer as 6, n=2 the answer is 9 etc. Well no it doesn't work for our formula for n - but notice if we did 3n for n=1, 2, 3 and 4, we get the sequence: 3, 6, 9, 12. Now you ask yourself: "What do I do to get from this sequence to the one I had originally". Answer: Add 3. So the formula now reads 3n+3. Then, check to see if this is right, and indeed it is.

    If you don't have a common difference, proceed as follows:
    Example: 2 6 14 26 42
    The differences are 4 8 12 16.

    We then go one stage further and take the differences of *these*. The differences of 4 8 12 and 16 are: 4 4 and 4. Now these are constant so we stop here. Because we had to go to *two* lot of differences, our formula will involve n^2. Usually, the coefficient of n^2 will always be *half* of your common difference, in our case, the common difference was 4 so the coefficient of n^2 is 2. (Remember: Coefficient is a posh word for the number that goes in front of the letter, i.e. the coefficient of 3n is 3, the coefficient of 7p is 7 etc)

    Right, now we have a formula involving 2n^2. We then check using n values of 1, 2, 3, 4 and 5 as in my first example above. Plugging these values of n in we get:
    2 8 18 32 50

    Again, we compare this sequence with the sequence we started with:

    2 8 18 32 50 with:
    2 6 14 26 42

    To get from the 1st equation to the 2nd equation we have to subtract different numbers. From the first one we subtracted 0 (2-2=0), the 2nd we subtracted 2 (8-6), then 4 (18-14) then 6 (32-26) then 8 (50-42).
    There is a pattern here. We are just subtracting '2n-2' from our sequence. [It can be tricky to see where I got the 2n-2 from. To see it, write out the sequence 0, 2, 4, 6, 8 and then you will see the sequence is 2n-2...by using my rules explained above].

    So, our sequence now reads: 2n^2 - (2n - 2) which is 2n^2-2n+2 which factorises again to 2(n^2-n+1). Does this work? Plug values of n=1...5 back in and check. Yes, we do get 2, 6, 14, 26 and 42 back!

    I hope you understood some of that. It might take some reading and practise with some examples. Remember the method though, always take differences until they are the same. Then use the rules I have outlined above. Obviously it's easier to help face to face but I hope this internet explanation is good enough. Let me know if you need more help!


    (Original post by alexsmithson)
    I am so totally confused :confused: . Some of the ones at the end are extremely hard and I just can't do them. It seems so easy for some of you excellent mathematicians *flutters eye lashes*, please could someone have a go for me in the file and repost it? It would only take some of you genii (whats the plural for genius again?) minutes I'm sure. lol

    Look forward to hearing from you soon, Alex.
    Personally, I'm happy to help you and teach you how to do sequences questions but I'm not happy giving you the answers. That way you don't learn anything, and you get the easy way out. I've been through GCSE maths and A-level maths and I didn't get through A-level Further Maths by cheating.

    Take a read of my post, *do some examples of your own* or just write down my example and take it line by line slowly (you won't understand it by just reading it) and you'll see sequences are actually quite simple.

    Use your time wisely to learn rather than use your time playing with your mates and spending 5 minutes of an evening posting a message basically saying "give me the answers".

    (Original post by alexsmithson)
    Sorry if I enraged you there. I understood your explanation, but only upto a certain point. This isn't my GCSE coursework, but questions leading up to it. Maybe if you showed me how to do some of the harder ones instead of the easy ones I may get it?

    Sorry, Alex.
    Let me know what this "certain point" is. Also, if you type on here one of the sequences you class as "harder ones"...I'll solve it for you, with explanations. If it just involves methods I've already taught you, then there's not much point.

    (Original post by alexsmithson)
    -----17, 29, 42, 56-----

    Formula: ½(n²+21n+12)

    How about that one?

    Thanks, Alex.
    Right, that falls into the category I explained earlier about going to the second difference and it being constant (in the example I gave it was 4, in this one it is 1). Therefore the formula will have 1/2 n^2 involved (which it does). As for the remainder (10.5 n + 6) that can be found out exactly as I have shown in the previous message. It took forever for me to type it all out, so don't make me type it all out again. Perhaps someone else could help him? I'm off to bed...
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Updated: December 20, 2005

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