Quadratic sequence with non-consecutive terms

g.h.pooles

Hello,

I am trying to figure out how to calculate the nth term for the following set of numbers with non-consecutive terms:

1st term: 0
3rd term: 2
5th term: 6
7th term: 12
9th term: 20

I now know that the answer is 1/4 n^2 - 1/4, but I'm not sure how to work it out (given that I have non-consecutive terms).

The second difference is 2, therefore the an^2 term would usually be n^2 if we had consecutive terms. But given we only have odd terms, I thought that we would just divide by 2 to give 1/2 n^2 (not 1/4 n^2). Could somebody please explain where I am going wrong and to reason me through how to work it out?

Any help much appreciated.

Thanks.

Reply 1

3pointonefour

18

Well one way of doing it is simultaneous equations, if you're good at your algebra:/

T(n) = an^2 + bn + c

The first term is 0. So if

n=1 \rightarrow T(n) = 0

a(1)^2 + b(1) + c = 0 \rightarrow a+b+c = 0

Do the same with the other terms to find out the values of a, b and c.

(edited 6 years ago)

Reply 2

g.h.pooles

OP

2

Original post by 3pointonefour

Well one way of doing it is simultaneous equations, if you're good at your algebra:/

T(n) = an^2 + bn + c

The first term is 0. So if

n=0 \rightarrow T(n) = 0

a(0)^2 + b(0) + c = 0 \rightarrow c = 0

Do the same with the other terms to find out the values of a, b and c.

Thanks for replying so quickly. Why have you put 0 as the n value in each term? If we're looking at the 1st term, shouldn't that number be 1?
I.e. a(1)^2 + b(1) + c = 0

Reply 3

3pointonefour

18

Original post by g.h.pooles

Thanks for replying so quickly. Why have you put 0 as the n value in each term? If we're looking at the 1st term, shouldn't that number be 1?
I.e. a(1)^2 + b(1) + c = 0

Yup just gave it a quick edit now. Sorry about that. :tongue:

Reply 4

g.h.pooles

OP

2

Original post by 3pointonefour

Yup just gave it a quick edit now. Sorry about that. :tongue:

I did that and I got a = 0.5; b = -1; c = 0.5

When I tested this formula, I got the right numbers for terms 1 and 2, but then for the 3rd term I got 8 instead of 6.

Can you spot any mistakes that I made? - workings below.....

Eq. 1) a + b + c = 0
Eq. 2) 9a + 3b + c = 2
Eq. 3) 25a + 5b + c = 6

Taking Eq. 1 away from Eq. 2, I got 8a + 2b = 2 (Eq. 4)
Taking Eq. 2 away from Eq. 3, I got 16a + 2b = 6 (Eq. 5)

Taking Eq. 4 away from Eq. 5, I got 8a = 4, which means that a = 0.5. Therefore, using Eq. 4, b = -1 and c = 0.5

Reply 5

g.h.pooles

OP

2

Found it!!

What a stupid mistake. I said that 6-2 = 6, instead of 4.

Thank you so much threepointonefour for all your help!!!!!

Reply 6

3pointonefour

18

Original post by g.h.pooles

I did that and I got a = 0.5; b = -1; c = 0.5

When I tested this formula, I got the right numbers for terms 1 and 2, but then for the 3rd term I got 8 instead of 6.

Can you spot any mistakes that I made? - workings below.....

Eq. 1) a + b + c = 0
Eq. 2) 9a + 3b + c = 2
Eq. 3) 25a + 5b + c = 6

Taking Eq. 1 away from Eq. 2, I got 8a + 2b = 2 (Eq. 4)
Taking Eq. 2 away from Eq. 3, I got 16a + 2b = 6 (Eq. 5)

Taking Eq. 4 away from Eq. 5, I got 8a = 4, which means that a = 0.5. Therefore, using Eq. 4, b = -1 and c = 0.5