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 ;)
any example of such quadratic sequence??
-1; 0; 3; 8 ;15 for example
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
Ahh that's really cool I think I get what you did. I'd need to try some examples myself.
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
Haha that was hard to wrap my head around but I gotcha. Thanks :)
"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
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
I'll have to see if I can do it haha
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
how did u found out 3n^2
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 :?
k got this point wt next
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
hmmm then i had got this point then
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
thanks alot buddy.......
Or, if you can, you can just see that this linear patter is in fact 2n - 11
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