We make what is called a "difference table": Write the numbers in a row: 9 73 241 561 1081 1849 Under that row, subtract each adjacent pair of numbers, placing the difference between and below them. (73-9=64, so write 64 between and below the pair 9 73, 241-73=168, so write 168 between and below the pair 73 241, etc.) Like this: 9 73 241 561 1081 1849 64 168 320 520 768 That row is called the row of 1st differences. Under that row, again subtract each adjacent pair of numbers, placing the difference between and below them, like this: 9 73 241 561 1081 1849 64 168 320 520 768 104 152 200 248 That row is called the row of 2nd differences. Under that row, again subtract each adjacent pair of numbers, placing the difference between and below them, like this: 9 73 241 561 1081 1849 64 168 320 520 768 104 152 200 248 48 48 48 That row is called the row of 3rd differences. Now notice that they are all the same, so we don't need to make any more rows of differences. So we assume that the next 3rd difference, if we had the next number after 1849, would also be a 48, so we write another 48 on the bottom row. (I'll color it red): 9 73 241 561 1081 1849 64 168 320 520 768 104 152 200 248 48 48 48 48 Then we work backwards by adding. Add the 48 to the 248, getting 296, Write that to the right of the 248. (I'll color it red too). 9 73 241 561 1081 1849 64 168 320 520 768 104 152 200 248 296 48 48 48 48 Still working backward, Add the 296 to the 768, getting 1064, Write that to the right of the 768. (I'll color it red too). 9 73 241 561 1081 1849 64 168 320 520 768 1064 104 152 200 248 296 48 48 48 48 One more step working backwards. Add 1064 to the 1849, getting 2913, write it at the end of the first row, and there is your answer!: 9 73 241 561 1081 1849 2913 64 168 320 520 768 1064 104 152 200 248 296 48 48 48 48 Answer: The next term is 2913 Edwin
It can be any number you want.
The solution arises from the fact that anything times zero is just zero and also adding zero to something doesn't change it.
Now granted the formula will get a bit lengthy but it is a valid solution and it can be any number you want.
Lets start with a simple sequence of three random numbers between (1-10) from random.org I got 6,9,2,_ whats next, Ok back to random.org its came up with 5.
So our problem is to come up with a formula f(n) that produces f(1)=6, f(2)=9, f(3)=2, and f(4)=5
f(1)=6
f(2)=9
f(3)=2
f(4)=5
Lets initially choose
f(n) = 6
f(1) = 6 check
f(2) = 6 fail
so lets refine it
f(n) = 6 + (n-1)(3/1)
f(1) = 6 check
f(2) = 9 check
f(3) = 12 fail
so we refine further
f(n) = 6 + (n-1)(3/1) + (n-1)(n-2)(-10/2)
f(1) = 6 check
f(2) = 9 check
f(3) = 2 check
f(4) = -15 fail
so we refine further
f(n) = 6 + (n-1)(3/1) + (n-1)(n-2)(-10/2) + (n-1)(n-2)(n-3)(10/3)
f(1) = 6 check
f(2) = 9 check
f(3) = 2 check
f(4) = 5 check
So to answer the original question using the same methods the formula looks like this...
f(7) = 0
simplified...