Posted by Raina on Sunday, December 30, 2012 at 9:29pm.
I found that the third differences were a constant of 6, so we have a cubic.
assume the function is
f(x) = ax^3 + bx^2 + cx + d
for (1,0) --> a + b + c + d = 0 ..... #1
for (2,4) --> 8a + 4b + 2c + d = 4 ....#2
for (3,18) -> 27a + 9b + 3c + d = 18 .. #3
for (4,48) -> 64a + 16b + 4c + d = 48 ... #4
#2 - #1 --> 7a + 3b + c = 4 .....#5
#3 - #2 --> 19a + 5b + c = 14 ....#6
#4 - #3 --> 37a + 7b + c = 30 ....#7
#6 - #5 ---> 12a + 2b = 10 ----> 6a + b = 5
#7 - #6 ---> 18a + 2b = 16 ----> 9a + b = 8
subtract these last two:
3a = 3
a = 1
in 6a + b = 5 ---> 6+b=5 , ---- b = -1
in #5: 7a + 3b + c = 4
7 - 3 + c = 4 ------------- c = 0
in #1:
a + b + c + d = 0
1 - 1 + 0 + d = 0
d = 0
so our general term is
x^3 - x^2
or
term(n) = n^3 - n^
= n^2 (n-1) , where n ∈ N
testing:
term(1) = 1^3 - 1^2 = 0
term(2) = 2^3 - 2^2 = 4
term(3) = 3^3 - 3^2 = 27-9 = 18
term(4) = 4^3 - 4^2 = 64-16 = 48
term(5) = 5^3 - 5^2 = 125 - 25 = 100
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