What is the Nth Term in this Sequence?

Question

The sequence is -6, -2, 4, 12, 22

I got this far in the working out and then got stuck

1st Differences 4, 6, 8, 10
2nd Differences 2, 2, 2

That gives n(squared as the start of the rule)
If you use that you get
1x1=1
2x2=4
3x3=9
4x4=16
5x5=25

The remainders are 7 for the first term
6 for the second
5 for the third
4 for the fourth
3 for the fifth
What do I do next?

Answer 1

I would write the remainders as -7, -6.. instead of positive, since you need to go smaller, not bigger.
The next step would be to see *those* differences are 1, 1, 1, etc. Thus, you have a +n term. (If they were 2, you'd have +2n, -3 would be -3n, etc).
So you have n^2 + n, and finding the constant at the end is easy (-8).

Answer 2

Here is the key:

Let t_n = nth term of the sequence



a) If you are given a sequence and the "1st differences" are constant, then it is a linear sequence. Example:
5, 7, 9, 11, 13

1st differences: 2, 2, 2, 2

The key to solve this is:
t_n = a/1! n + C

Where
a = common "1st difference"
C = is the "zero'th" term of the sequence

In the example, a = 2, and to find C, find the number that goes before the first term


b) If you are given a sequence and the "2nd differences" are a constant, then it is a quadratic sequence. Example:
2, 4, 7, 11, 16, 22

1st differences: 2, 3, 4, 5, 6
2nd differences: 1, 1, 1, 1

The key to solve this is:
t_n = a/2! n² + bn + C

Where
a = common "2nd difference"
b = constant
C = is the "zero'th" term of the sequence

In the example, a = 1, to find C, find the term that goes before the first term, and to find b, substitute t_1 = 1.



c) If the "3rd differences" are constant, then it is a cubic sequence, with
t_n = a/3! n³ + bn² + cn + C

where a is the common "3rd difference"
and C = is the "zero'th" term of the sequence




d) If the "4th differences" are constant, then it is a quartic sequence, with
t_n = a/4! n^4 + bn³ + cn² + dn + C

where a is the common "4th difference"
and C = is the "zero'th" term of the sequence




e) If the "5th differences" are constant, then it is a quintic sequence, with
t_n = a/5! n^5 + bn^4 + cn³ + dn² + en + C

where a is the common "5th difference"
and C = is the "zero'th" term of the sequence




f) If the "rth differences" are constant, then the nth term is
t_n = a/r! n^r + bn^(r - 1) + cn^(r - 2) + ... + C

where a is the common "rth difference"
and C = is the "zero'th" term of the sequence


Your problem can be solved using b). The nth term of that sequence is:
t_n = n² + n - 8

^_^

Answer 3

from what I can gather
n squared +n-8.
n is the position number
1x1+1-8=-6
2x2+2-8=-2
3x3+3-8=4
4x4+4-8=12
5x5+5-8=22

Answer 4

Use the general form of a quadratic

a n^2 +bn + c

a+b+c = -6
4a + 2b + c = -2
9a + 3b + c = 4

Solve 3 equations and 3 unknowns.

5a + b = 6
3a + b = 4

2a = 2


a = 1
b = 1
c = -8

n^2 + n - 8

Answer 5

Some of these answers appear to be over complicated.

The n'th term Un is given by n^2 +(n - 8)

Answer 6

T(n)-T(n-1) = 2n
T(n-1)-T(n-2)=2(n-1)
'
'
'
'
T(2)-T(1)=4

Add all these equations.
You get T(n)-T(1)=2n+2(n-1)+2(n-2)+........+2(2)
OR T(n)-(-6)=2n+2(n-1)+2(n-2)+...............+2(2)+2(1)-2(1)
OR T(n)+6=2(n+(n-1)+(n-2)+......+2+1)-2(1)
OR T(n)+6=n(n+1)-2
OR T(n)=n^2 +n-8

Allahuakbar
Slave of Allah

Answer 7

the second difference is 2
hence,
N is poly in n^2
N=an^2+bn+c
use terms for equations

-6=a+b+c
-2=4a+2b+c
4=9a+3b+c

1st sweep
4=3a+b
6=5a+b

2a=2>>>a=1
b=1,c= -8

if N =nth term,

for n=1,2,3,.........

N=n^2+n-8

i hope that this helps

Answer 8

Seems like you're making it harder than it is.

x(n) = x(0) + 2n + 2, where x(0) = -6.

Answer 9

Yeah what they said