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

Guest Oct 15, 2014

Best Answer 

 #3
avatar+19597 
+10

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

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

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

heureka  Oct 16, 2014
 #1
avatar+17744 
+5

What is meant by "an"?

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

geno3141  Oct 15, 2014
 #2
avatar+92748 
+10

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.    

Melody  Oct 16, 2014
 #3
avatar+19597 
+10
Best Answer

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

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

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

heureka  Oct 16, 2014
 #4
avatar+87293 
+5

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

 

 

CPhill  Oct 16, 2014
 #5
avatar+92748 
0

Thanks Chris  :)

Melody  Oct 16, 2014

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