help how to find the nth term of the sequence???

ummm, depending on the nature of the sequence, for arithmetic sequences:

Tn = a + d(n-1) where [a] is the initial term, [d] is the increment and [n] is the number of terms

for geometric sequences:

Tn = ar^(n-1) --> [a] times [r to the power of n -1]

hope this helps.

Sneaky Sequences

Prime 1,2,3,5,7,11,13,17,19,23,29,31,....

Alphabetic 5,4,8,1,9,7,6,3,2,0

Fibonacci 1,1,2,3,5,8,13,21,34,55,....,F(n)+F(n-1)

Triangular 1,3,6,10,15,21,....n(n+1)/2
(and pentagonal, sexagonal, heptagonal, octagonal well you see the possibilities (note square numbers are square sequences.)

Rap

thanks all your help

can you help me find the nth term of this sequence?
0,11,44,110,220

385

The nth term is (11/6)*(n^3 - n)

Take the difference of the terms. Then take the difference of the differences, and so on. If it is a polynomial function, then you will end up with all zeros. The order of the polynomial function is one less than the number of times you took the differences

0 11 44 110 220
11 33 66 110
22 33 44
11 11
0

Of course, you can't be certain this is the solution. There are an infinite number of sequences that start with 0, 11, 44, 110, 220.