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Mathematics 21 Online
OpenStudy (anonymous):

Can somebody explain to me a logical way of working out the nth term of a quadratic sequence? I'm tired of remembering formulas. Hope this isn't asking too much ;)

OpenStudy (anonymous):

any example of such quadratic sequence??

OpenStudy (anonymous):

-1; 0; 3; 8 ;15 for example

OpenStudy (anonymous):

The difference of the terms is 1, 3, 5, 7 Which, in turn, have a difference of 2. This means that there is 1 n^2 term involved. Subtracting n^2 from all gives: -2, -4, -6, -8, -10 This has a difference of -2, so -2x is involved. Subtracting -2x from all gives 0,0,0,0,0 so you're done. Nth term = n^2-2n

OpenStudy (anonymous):

Ahh that's really cool I think I get what you did. I'd need to try some examples myself.

OpenStudy (anonymous):

If you know that the equation is quadratic, the difference of differences will be constant and will be twice your coefficient of x^2. Subtracting this kx^2 from the original sequence gives a linear series with a constant difference. This is the coefficient of x so subtract nx from the linear series. This would leave every term being equal, and that's the constant (d) to add. The series can then be written as kx^2+nx+d

OpenStudy (anonymous):

Haha that was hard to wrap my head around but I gotcha. Thanks :)

OpenStudy (anonymous):

"the difference of differences will be constant and will be twice your coefficient of x^2." I wonder why this is so... It seems each term is made up of: an^2 + bn + c so... Tn = an^2 + bn + c And then we found a which is half the second difference for reasons unknown an minused it from the term so we only had the linear part left to deal with Tn - an^2 = bn + c And the difference between each term became b which makes sense since that's how you'd normally do a linear series and then we minused that part Tn - an^2 - bn = c And we were left with the constant that must be added to each term... So the only part I don't get as why a must be half the 2nd difference

OpenStudy (anonymous):

hey friend i had a query let us assume a sequence -6, 5, 22, 45, 74, 109 how to get the nth term of this

OpenStudy (anonymous):

I'll have to see if I can do it haha

OpenStudy (anonymous):

Ok so I get a second difference of 6 so I have to minus 3n^2 from each term... then you get the linear sequence -9; -7; -5; -3 which has a difference of 2 so 2n... -11 so I guess that's c so it's 3n^2 + 2n - 11

OpenStudy (anonymous):

how did u found out 3n^2

OpenStudy (anonymous):

the differences between the terms are 11; 17; 23; 29 etc and the differences between those (or the 2nd difference) is always 6... and so you have to use half of that 2nd difference to go in front of n^2... but I don't really understand why you use half :?

OpenStudy (anonymous):

k got this point wt next

OpenStudy (anonymous):

ok so now take the first term.. -6 and minus 3(1)^2 then the second - 3(2)^2 and so on to get -9; -7; -5; -3

OpenStudy (anonymous):

hmmm then i had got this point then

OpenStudy (anonymous):

now the difference in this new series is 2... so you minus 2n from each of those terms and you should get a constant -9 -2(1) = -11 -7 -2(2) = -11 so that's the final part of the equation

OpenStudy (anonymous):

thanks alot buddy.......

OpenStudy (anonymous):

Or, if you can, you can just see that this linear patter is in fact 2n - 11

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