We start with the general form of a geometric sequence:
`t_n=ar^(n-1)`
Apparently you know that the first term is `1/2` and that the common ratio `r=-4` . Thus the sequence in question is `1/2,-2,8,-32,...`
Plugging into the general form we get:
`t_n=(1/2)(-4)^(n-1)`
The negative base is a bother. What is really happening is you have a geometric sequence whose terms alternate in sign. This can be written as `(-1)^n` or `(-1)^(n-1)` times the underlying sequence; whether it is n or n-1 depends on whether the first term or second term is negative.
Thus we rewrite the negative 4 as (-1)(4). But we notice that we are multiplying by `1/2` . Since `1/2` and 4 are both powers of 2, we write them that way:
`t_n=(1/2)[(-1)(4)]^(n-1)`
`=(2^(-1))[(-1)2^2]^(n-1)`
We use the product of powers rule `(ab)^m=a^mb^m`
`=(2^(-1))[(-1)^(n-1)(2^2)^(n-1)]`
We use the power to a power rule `(a^m)^n=a^(mn)`
`=(2^(-1))[(-1)^(n-1)(2)^(2n-2)]`
We are multiplying three terms, and multiplication is commutative, so we multiply the first and last terms using the product of powers rule `a^m*a^n=a^(m+n)`
`=(-1)^(n-1)(2)^(2n-2-1)`
`=(-1)^(n-1)2^(2n-3)`
which is the form they gave.
For n=1 you get `(-1)^0*2^(-1)=1/2`
For n=2 you get `(-1)^1*2^1=-2`
For n=3 you get `(-1)^2*2^3=-8` etc...
See eNotes Ad-Free
Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.
Already a member? Log in here.