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hello

in my maths coursework i have a number sequence and i dont know how to find the nth term as it's third difference that is the same, anybody know how to do these?

in my maths coursework i have a number sequence and i dont know how to find the nth term as it's third difference that is the same, anybody know how to do these?

battery99

hello

in my maths coursework i have a number sequence and i dont know how to find the nth term as it's third difference that is the same, anybody know how to do these?

in my maths coursework i have a number sequence and i dont know how to find the nth term as it's third difference that is the same, anybody know how to do these?

You're right, it's a cubic. Derive the differences for the general formula for a cubic an^3+bn^2+cn+d for n=1 to n=5.

Hope this is useful.

thanks alot for the reply!

i'm just unsure now how to put the numbers into the formula lol!

i've only been taught up to n^2 sequences where the formula is

tn = an^2 + bn + c

and that to get the values of a,b and c:

the 2nd difference = 2a

1st difference = 3a + b

nth term = a + b + c

does this still apply for cubics? and if so how do i get the value of d?

Thanks again

i'm just unsure now how to put the numbers into the formula lol!

i've only been taught up to n^2 sequences where the formula is

tn = an^2 + bn + c

and that to get the values of a,b and c:

the 2nd difference = 2a

1st difference = 3a + b

nth term = a + b + c

does this still apply for cubics? and if so how do i get the value of d?

Thanks again

Deep breath...

The method for getting the formula for a cubic sequence out of differences is very similar to that for a quadratic.

You know that if the second differences are all 2, the formula has something to do with n^2; if the second differences are all 4, the formula contains 2n^2 and so on. In other words, divide the value of the second difference by 2 to get the coefficient of n^2 in the formula.

You then compare the original sequence with the sequence of n^2 or 2n^2 or whatever is appropriate, and the differences in corresponding terms will form a linear sequence.

It's exactly the same for third differences, only you divide this value by SIX.

For instance, take a sequence that goes 2, 6, 20, 50, 102, 182.

First differences are 4, 14, 30, 52, 80.

Second differences are 10, 16, 22, 28.

Third differences are 6, 6, 6.

So the formula contains (6/6)n^3, which is just n^3.

Going along the original sequence and subtracting the relevant value of n^3 gives a new sequence (2-1), (6-8), (20-27), (50-64) etc, which works out to 1, -2, -7, -14.

Now take differences on this sequence:

First differences are -3, -5, -7

Second differences are -2, -2

So this new sequence has something to do with (-2/2)n^2, which is just -n^2.

Compare -n^2 with the second sequence: (1- -1), (-2- -4), (-7- -9)...

This turns into simply 2, 2, 2, so the linear portion of the formula is just +2.

Then go back and put it all together: nth term = n^3 - n^2 + 2

And that's it.

Jon

The method for getting the formula for a cubic sequence out of differences is very similar to that for a quadratic.

You know that if the second differences are all 2, the formula has something to do with n^2; if the second differences are all 4, the formula contains 2n^2 and so on. In other words, divide the value of the second difference by 2 to get the coefficient of n^2 in the formula.

You then compare the original sequence with the sequence of n^2 or 2n^2 or whatever is appropriate, and the differences in corresponding terms will form a linear sequence.

It's exactly the same for third differences, only you divide this value by SIX.

For instance, take a sequence that goes 2, 6, 20, 50, 102, 182.

First differences are 4, 14, 30, 52, 80.

Second differences are 10, 16, 22, 28.

Third differences are 6, 6, 6.

So the formula contains (6/6)n^3, which is just n^3.

Going along the original sequence and subtracting the relevant value of n^3 gives a new sequence (2-1), (6-8), (20-27), (50-64) etc, which works out to 1, -2, -7, -14.

Now take differences on this sequence:

First differences are -3, -5, -7

Second differences are -2, -2

So this new sequence has something to do with (-2/2)n^2, which is just -n^2.

Compare -n^2 with the second sequence: (1- -1), (-2- -4), (-7- -9)...

This turns into simply 2, 2, 2, so the linear portion of the formula is just +2.

Then go back and put it all together: nth term = n^3 - n^2 + 2

And that's it.

Jon

Ste

Just one question which is what does ^ mean?

To the power of. It is difficult to write in superscript, so you substitute ^ to indicate this. i.e., to show x squared = x^2

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- Magnesium melting point
- Proving a term is in a quadratic sequence
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