mathwannabe
Junior Member
- Joined
- Feb 20, 2012
- Messages
- 122
Hello everyone 
I have never worked on this type of problem before. I got the same answer that the book provides, but that book is proven to be evil and I don't trust it one bit x)
Here is the problem and how I did it (or not LOL):
1) \(\displaystyle \dfrac{a^{2n+1}-a^{2n-1}}{a^{n-1}+a^n}=\)
\(\displaystyle =\dfrac{a^{2n}a-a^{2n}a^{-1}}{a^na^{-1}+a^n}=\)
\(\displaystyle =\dfrac{a^{2n}(a-a^{-1})}{a^n(a^{-1}+1)}=\)
\(\displaystyle =\dfrac{a^n(a-a^{-1})}{a^{-1}+1}=\)
\(\displaystyle =\dfrac{a^n(a-\frac{1}{a})}{\frac{1}{a}+1}=\)
\(\displaystyle =\dfrac{a^n(\frac{a^2-1}{a})}{\frac{a+1}{a}}=\)
\(\displaystyle =a^n\dfrac{(a-1)(a+1)}{a}\times\dfrac{a}{a+1}=a^n(a-1)\)
I have never worked on this type of problem before. I got the same answer that the book provides, but that book is proven to be evil and I don't trust it one bit x)
Here is the problem and how I did it (or not LOL):
1) \(\displaystyle \dfrac{a^{2n+1}-a^{2n-1}}{a^{n-1}+a^n}=\)
\(\displaystyle =\dfrac{a^{2n}a-a^{2n}a^{-1}}{a^na^{-1}+a^n}=\)
\(\displaystyle =\dfrac{a^{2n}(a-a^{-1})}{a^n(a^{-1}+1)}=\)
\(\displaystyle =\dfrac{a^n(a-a^{-1})}{a^{-1}+1}=\)
\(\displaystyle =\dfrac{a^n(a-\frac{1}{a})}{\frac{1}{a}+1}=\)
\(\displaystyle =\dfrac{a^n(\frac{a^2-1}{a})}{\frac{a+1}{a}}=\)
\(\displaystyle =a^n\dfrac{(a-1)(a+1)}{a}\times\dfrac{a}{a+1}=a^n(a-1)\)