an=(n+1)/(2n+3) find the " n " in this que

an=(n+1)/(2n+3) find the " n " in this que
an-an-1=1/99

$$a_n=\dfrac{n+1}{2n+3} \qquad \text{ and } \qquad a_n-a_{n-1}=\dfrac{1}{99}$$

$$\begin{array}{rcl} \frac{n+1}{2n+3}-\frac{(n-1)+1}{2(n-1)+3} &=& \frac{1}{99} \\\\ \frac{n+1}{2n+3}-\frac{n}{2n+1} &=& \frac{1}{99} \\\\ \frac{(2n+1)(n+1)-n(2n+3)}{(2n+3)(2n+1)} &=& \frac{1}{99} \\\\ (2n+3)(2n+1) &=& 99[(2n+1)(n+1)-n(2n+3)]\\\\ (2n+3)(2n+1) &=& 99(2n+1)(n+1)-99n(2n+3)]\\\\ 4n^2+2n+6n+3 &=& 99(2n^2+2n+n+1)-99(2n^2+3n)\\\\ 4n^2+8n+3 &=& 99(2n^2+3n) +99 -99(2n^2+3n)\\\\ 4n^2+8n+3 &=& 99 \\\\ 4n^2+8n-96 &=& 0 \quad | \quad :4 \\\\ n^2 +2n-24 &=& 0 \\\\ n_{1,2}=\frac{-2\pm\sqrt{4-4(-24)} } {2} \\\\ n_{1,2}=\frac{-2\pm\sqrt{100}} {2} \\\\ n &=& \frac{-2+10}{2} \\\\ n &=& \frac{8}{2} \\\\ n &=& 4 \end{array}$$

What is meant by "an"?

What is "an-an-1=1/99"?

Maybe

Chris is teaching me the science of forensic mathematics  - How did I do Chris?     LOL

$$\\a_n=(n+1)/(2n+3)\\\\ a_n-a_{n-1}=1/99\\\\ so\\\\ a_{n-1}=(n-1+1)/(2(n-1)+3)\\\\ a_{n-1}=n/(2n+1)\\\\ so\\\\ a_n-a_{n-1}=1/99\\\\ \frac{n+1}{2n+3}-\frac{n}{2n+1}=\frac{1}{99}\\\\ 99(n+1)(2n+1)-99n(2n+3)=(2n+3)(2n+1)\\\\ etc$$

If you want me to continue, just ask.    

Good job by Melody and heureka....points for both........maybe some of my detective work is rubbing off on Melody ..... LOL!!!

Thanks Chris  :)