Posted by Raina on .
0,4,18,48,100.....
How can I find the general term for this ?

math 
Reiny,
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 (n1) , where n ∈ N
testing:
term(1) = 1^3  1^2 = 0
term(2) = 2^3  2^2 = 4
term(3) = 3^3  3^2 = 279 = 18
term(4) = 4^3  4^2 = 6416 = 48
term(5) = 5^3  5^2 = 125  25 = 100