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#1 2007-02-27 02:55:10

rainbow x
Member
Registered: 2007-02-27
Posts: 3

help with quadratic sequences please.

if anyone could help me with this, it would be very much appreciated, thankyou! smile

how do i work out the nth term in the following sequence? could someone explain it to me please.

11, 44, 110, 220 <--SEQUENCE
..+33 +66 +110 <-- 1st DIFFERENCE
....+33 +44 <-- 2nd DIFFERENCE
.......+11 <-- 3rd DIFFERENCE (constant)

im stuck cos ive learnt that to work out the nth term for quadratic sequences, the second difference has to be a constant, but here, the third difference is a constant .

does that make sense? 

thnks for any help =]

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#2 2007-02-27 03:07:36

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Re: help with quadratic sequences please.

I'm pretty sure it must be a cubic, can't think of any thing else... at the moment.

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#3 2007-02-27 03:13:38

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: help with quadratic sequences please.

Yes, it’s cubic. I’ve just worked it out. smile

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#4 2007-02-27 03:23:05

rainbow x
Member
Registered: 2007-02-27
Posts: 3

Re: help with quadratic sequences please.

mhmm, makes sense that it would be cubic, im still not too sure how to go about working it out though =S

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#5 2007-02-27 06:31:41

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: help with quadratic sequences please.

Well, you just let

Since you know the values of x[sub]n[/sub] for n=1,2,3,4, substitute these into the equation and you can work out the coefficients A. B. C. D.

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#6 2007-02-27 10:18:05

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: help with quadratic sequences please.

Jane is correct. It might get a bit ugly, as you have 4 simultaneous equations with 4 unknowns, but that's the way to get the answer.

There is an alternative way of getting it by using differences, but I'm not sure whether or not it would be more efficient. As you've already found the 3rd difference to be a constant of 11 though, you might as well use that to start you off.

The difference is on the 3rd row, which means that the x³ coefficient (A, in Jane's notation) is 11/3! = 11/6.

From there, there are two ways of going forward.

1) Find B, C and D using the 4 simultaneous equations.

2) Keep using differences. Take away 11/6x³ from all the terms to get a new sequence which should have a common 2nd difference, which you use to find B (Divide the constant by 2!). Then take away Bx² from them all to get a new sequence with a common difference (that is also C), then use that to find the constant D.

I'm not sure which of those two is quicker (probably the first, actually), but they both definitely work.


Why did the vector cross the road?
It wanted to be normal.

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#7 2007-02-28 02:41:08

numen
Member
Registered: 2006-05-03
Posts: 115

Re: help with quadratic sequences please.

10+1, 40+4, 100+10, 200+20

Skip the second terms for a while and let us simplify the sequence: 1, 4, 10, 20

It's much easier now! We get: n/6(n+1)(n+2)
But that was the simplified model, let's continue:

--> 5n/3(n+1)(n+2) corresponds to the sequence 10,40,100,200. Now we need to add 1/10 of these:

5n/3(n+1)(n+2)+5n/30(n+1)(n+2) =
= 10n/6(n+1)(n+2)+1n/6(n+1)(n+2) =

= 11n/6(n+1)(n+2)

Inserting n=1,2,3,4 you'll see it works out fine for your sequence smile


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