Simplify the general term of a geometric sequence.The book I'm using gives me only one step-by-step example, and it's not clear enough. I have no idea what's going on in the last two lines. Please...

Simplify the general term of a geometric sequence.

The book I'm using gives me only one step-by-step example, and it's not clear enough. I have no idea what's going on in the last two lines. Please describe what's happening and why. http://tinypic.com/r/vpd8r5/6

Asked on by topderek

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embizze | High School Teacher | (Level 1) Educator Emeritus

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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...

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