# Math Help - Check Recurrence Relation

1. ## Check Recurrence Relation

The formula 1+r+r^2+...+r^n=(r^(n+1)-1)/(r-1) is true for all real numbers r except r=1 and for all integers n>=0. Use this fact to solve the following problem:

If n is an integer and n>=1, find a formula for the expression
2^n-2^(n-1)+2^(n-2)-2^(n-3)+...+((-1)^(n-1))(2)+(-1)^n.

For the values of r and n, I substituted r=(-2) and n=n-1 and got
((2^n)-1)/3.

Did I use the correct substitutions?

2. ## Re: Check Recurrence Relation

Originally Posted by lovesmath
The formula 1+r+r^2+...+r^n=(r^(n+1)-1)/(r-1) is true for all real numbers r except r=1 and for all integers n>=0. Use this fact to solve the following problem:

If n is an integer and n>=1, find a formula for the expression
2^n-2^(n-1)+2^(n-2)-2^(n-3)+...+((-1)^(n-1))(2)+(-1)^n.

For the values of r and n, I substituted r=(-2) and n=n-1 and got
((2^n)-1)/3.

Did I use the correct substitutions?
$2^n-2^{n-1}+...+(-1)^{n-1}2+(-1)^n=2^n\left[1-\frac{1}{2}+\frac{1}{2^2}-...+\left(-\frac{1}{2}\right)^{n-1}+\left(-\frac{1}{2}\right)^{n}\right]$

Now apply the given formula to the expression in square brackets.

CB