4:54 , it doesn't mean you've gone wrong, it just shows that there isn't a bn , so the equation would be an^2+c. I have come across a question where the answer is a quadratic but is in the form of n^2+c instead of n^2+bn+c
@Davey: We have found the first part of the formula is 8n^2, so we can subtract that to leave the rest which will be linear. Looking at a simpler example: 3, 8, 15, 24,...The differences go up in 2's, so we know that the formula starts with n^2, so we subtract 1, 4, 9, 16 respectively to leave 2, 4, 6, 8 and can see that the linear formula for these numbers is 2n. So now we can put the two formulas together to get n^2 + 2n
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Got a sequence that involves n^3 and took me 4 difference of sum to get same ones... how would I convert the amount of times it takes to get a common difference to the power that I am looking for?
Maybe you will be able to work out the pattern....just try the simplest sequence of powers of 4: 1, 16, 81, 256, 625, 1296, and what is the 4th difference? Then the pattern should emerge...
You could try it yourself: what happens with the simplest example: 1, 4, 9, 16, 25,...What are the second differences? Then try multiplying up by an integer, say 10, so you get 10, 40, 90, 160, 250,...What are the second differences? Can you answer your question?
If you had to find the differences 4 times before they converge, would that mean the answer should be in the form an^4 + bn^3 + cn^2 + dn + e or should it just be an^2 + bn + c? The sequence is 1, 3 ,12, 39, 120, 316, ... which makes the differences 2, 9, 27, 81, 196 and then 7, 18, 54, 115 then 11, 36, 61 then 25, 25 If it should be the first option, do I just repeat the steps again and again or is there another way to do this?
@David: Think about the sequence of square numbers: 1, 4, 9, 16, 25,... The differences are 3, 5, 7, 9,... And the differences between the differences are 2...so we divide by 2 to get the number to multiply n^2
That's harder than this one....it will involve cubing, not just squaring. If you haven't looked at cubic nth term yet, try writing each number in your pattern as a product of two numbers and look at 2 patterns...
