Mastering Mathematical Induction: Solving for the nth Term

[jackscholar]

Homework Statement

1-1/2+1/4-1/8+...nth term= [(2^n-(-1)^n)]/3(2^(n-1))]

I just can't get the nth term. I realize that the denominator should be 2^(n-1) and that, because the sign constantly changes there needs to be something like -1^n so I came up with the equation -1^n/-2^(n-1) but this doesn't satisfy n=1. Any help is highly appreciated.

 

[micromass]

Think about $(-1)^{n+1}$

 

[HallsofIvy]

Homework Statement

1-1/2+1/4-1/8+...nth term= [(2^n-(-1)^n)]/3(2^(n-1))]

I just can't get the nth term. I realize that the denominator should be 2^(n-1) and that, because the sign constantly changes there needs to be something like -1^n so I came up with the equation -1^n/-2^(n-1) but this doesn't satisfy n=1. Any help is highly appreciated.

That "3" in the denominator puzzled me also. But look what happens for different n:
n= 1 [2^1- (-1)^1]/[(3)2^{1- 1}]= (2+1)/3= 1
n= 2 [2^2- (-1)^2]/[(3)2^{2- 1}]= (4- 1)/[(3)2]= 3/(3)2= 1/2, etc.
As micromass suggests, is (-1)^n gives the wrong sign, multiply by another -1: (-1)^(n+1)- not taking it to the n-1 power as you appear to be doing. Although, because you have n-1 in the denominator, I would be inclined to divide by 1, again, not taking it to a power, to get (-1)^(n-1)/2^(n-1)= (-1/2)^{n-1}.

Are you required to have the sum starting at n= 1 ($\sum_{n=1}^\infty (-1/2)^{n-1}$)? I think the simplest thing to do is start with n= 0. $\sum_{n=0}^\infty (-1/2)^n$ will give exactly the same result.