write a formula for the general term of the infinite sequence -1, 4, -16, 64, -256, 1024...?

Answer 1

Here's the formula for the general term, a (sub n):

a (sub n) = -1(-4)^(n -1).

For n = 1, a (sub 1) = -1(-4)^(1-1) = -1(-4)^0 = -1(1) = -1.
For n = 2, a (sub 2) = -1(-4)^(2-1) = -1(-4)^1 = -1(-4) = 4.
For n = 3, a (sub 3) = -1(-4)^(3-1) = -1(-4)^2 = -1(16) = -16.
For n = 4, a (sub 4) = -1(-4)^(4-1) = -1(-4)^3 = -1(-64) = 64, and so forth.

My formula is correct! Look what happens if we use the formula of answer #1. If x = 3 or the third term, then f(x) = -4(3) = -12, which isn't even in the sequence.

Answer 2

f(x)=-4(x)

Answer 3

This is a Geometric Progression..
and the general term of any GP is equal to:
a(r raised to the power of n-1)
but I dunno abt this one..

Answer 4

1,132,096