Question

Find a sequence for the following terms.

2,6,12,20,30,42,56,72,90,110,132

Answer 1

when i tried to do it i found out that the differencces are

4,6,8,10,12

and the differences of those differences are 2, 2, 2, 2, 2,

I know this means something, but I simply forgot how to come up with the sequence.

its doesnt really have a common difference yet i dont think its geometric either.

Answer 2

u mean u want to find the \(n^{th}\)term

Answer 3

yea a general formula

Answer 4

you have almost worked it

Answer 5

Notice that you get a constant when you differentiate a second degree polynomial

Answer 6

so look for a quadratic function

Answer 7

yes i see a constant

from my experience that means its an exponential formula

like y=a^x where a is the constant

Answer 8

\(\large a_n = xn^2 + yn + z\)

Answer 9

but also we are taking the difference not differentiating.

Answer 10

use the terms to find the coefficients x,y,z

Answer 11

so make a system of equatoins

and solve it in a matrix

Answer 12

you're right, we are taking differences and not differeintiating. whats the difference between `nth defference` and a `nth derivative` ?

Answer 13

one is discrete, and the other is continuous. thats all right ?

Answer 14

yup pick any 3 terms since u have 3 unknonws

Answer 15

okay im going to try it.

Answer 16

you want the sequence to start at n=0 or 1 ?

Answer 17

well actually

the sequence could start at 2 or even smaller than that 0 (even though i didnt list it above)

and im trying to see where i want to start from

Answer 18

if you have the choice, make it start at 0
then you will get one coeffecient freely : z = 2

Answer 19

well i get

An=xn^2+yn+z

if it starts at zero the original sequence it then becomes

0=x(0)^2+y(0)+z

so z=0

Answer 20

is your first term 0 ?

Answer 21

\(\large a_0 = 2\) right ?

Answer 22

i just figured out that it could be zero

the first term

Answer 23

which i didnt wright in my original

Answer 24

hmm the sequence as it stands, the first term is 2

Answer 25

well it could start at 2 or start at zero and im trying to figure out which one i want to choose

Answer 26

but we can start at Ao=2 as in the original

Answer 27

okay finding the coefficients should be easy once you figure out that its a quadratic

Answer 28

but how did you know it was a quadratic and not for example a cubic function?

Answer 29

consider a linear function,
the slope is constant so the first differences AND the first derivative is constant, right ?

Answer 30

yes

Answer 31

okay i solved the matrix

and it looks as if its an infinitely many solution one

Answer 32

oh nevermind let me redo it

Answer 33

here is the general rule :
\(\large n\)th difference of a polynomial of \(\large n\)th degree is constant

Answer 34

omg that is so true

Answer 35

so the number of times you take differences until you reach a constant tell you what degree the polynomial needs to be

Answer 36

for a cubic, you will get the 3rd differences constant

Answer 37

only IF the sequence is can be represented using a polynomial

Answer 38

makes sense. but i swear i did a problem where i was taking differences of the differences and it gave the constant value which is also the base value of an exponential function y=a^x. so how would one know if by taking differences of the differences you would get a polynomical or an exponential?

Answer 39

for an exponential function, the differences are never constant... thats the reason exponential functions can be infinitely differentiable

Answer 40

the sequence i get is An=n^2+3n+2 and it looks as if this is the correct sequence!!!

Answer 41

so using an exponential function to model a polynomial data may not be a good idea

Answer 42

thank u for your help since now i got the answer and know i also know a new technique for sequences!!!

Answer 43

there is a ready made formula also

Answer 44

formual is
\(\large\tt \begin{align} \color{blue}{n^{th}term\\
=a+(n-1)d+\dfrac{(n-1)(n-2)c}{2}}\end{align}\)
where a = first term
d=difference between first two terms
c=second difference

Answer 45

wow, ive never seen that formula before. its not in my precal book.

Answer 46

thats a neat formula

Answer 47

btw, we can work this particular sequence without using differences / formulas

Answer 48

there is a nice pattern incase if you haven't noticed already

Answer 49

but it only works in case the second differences are constant

if it exceesd the second difference,then it doesnt work

Answer 50

2,6,12,20,30,42,56,72,90,110,132

1*2 = 2
2*3 = 6
3*4 = 12
4*5 = 20
...

Answer 51

lol,thats easy

Answer 52

:)

Answer 53

i would assume that if the sequence had to be subtracted 3 times to find a constant then the formula would be

An= a+ (n-1)d +(1/2)(n-1)(n-2)c + (1/3)(n-1)(n-2)(n-3)b , where b is the third difference

Answer 54

that would be an invention then

Answer 55

ahh that looks nasty, i would rather use a matrix

Answer 56

thanks again everyone