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Letf(x) = \frac{x^2}{x^2 - 1}.Find the largest n so that f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98.

difficulty advanced
Guest Jan 25, 2015

Best Answer 

 #1
avatar+19652 
+5

Let 

f(x) = \frac{x^2}{x^2 - 1}.

Find the largest n so that

$$$ f(2)\cdot f(3)\cdot f(4)\cdots f(n-1)\cdot f(n) < 1.98. $$$

$$\boxed{ \small{\text{ $ \prod \limits_{i=2}^{n} \frac{i^2}{i^2-1} < 1.98 $ }} } \qquad \dfrac{i^2}{i^2-1} = \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right)
$\\ \\ \\$
\small{\text{
$ \prod \limits_{i=2}^{n} \dfrac{i^2}{i^2-1} =
\prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right) = \prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \prod\limits_{i=2}^{n} \left(\dfrac{i}{i+1} \right)
$
}} $\\\\$
\small{\text{
$ =
\left( \frac{2}{2-1} \cdot
\frac{3}{3-1} \cdot \frac{4}{4-1} \cdot \frac{5}{5-1} \dots \frac{n}{n-1}
\right)
\cdot
\left( \frac{2}{2+1} \cdot
\frac{3}{3+1} \cdot \frac{4}{4+1} \cdot \frac{5}{5+1} \dots \frac{n-1}{n} \cdot \frac{n}{n+1}
\right)
$
}} $\\\\$
\small{\text{
$ =
\left( \frac{2}{1} \cdot
\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \dots \frac{n}{n-1}
\right)
\cdot
\left( \frac{2}{3} \cdot
\frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \dots \frac{n-1}{n} \cdot \frac{n}{n+1}
\right)
$
}} $\\\\$
\small{\text{
$ =
\left(
\frac{2}{1} \cdot
\frac{\textcolor[rgb]{1,0,0}{\not{3}} }{ \textcolor[rgb]{0,0,1}{\not{2}} } \cdot
\frac{\textcolor[rgb]{1,0,0}{\not{4}} }{ \textcolor[rgb]{0,0,1}{\not{3}} } \cdot
\frac{\textcolor[rgb]{1,0,0}{\not{5}} }{ \textcolor[rgb]{0,0,1}{\not{4}} } \dots
\frac{\textcolor[rgb]{1,0,0}{\not{n}} }{ \textcolor[rgb]{0,0,1}{\not{n-1}} } \right)
\cdot
\left(
\frac{ \textcolor[rgb]{0,0,1}{\not{2}} }{\textcolor[rgb]{1,0,0}{\not{3}}} \cdot
\frac{ \textcolor[rgb]{0,0,1}{\not{3}} }{\textcolor[rgb]{1,0,0}{\not{4}}} \cdot
\frac{ \textcolor[rgb]{0,0,1}{\not{4}} }{\textcolor[rgb]{1,0,0}{\not{5}}} \cdot
\frac{ \textcolor[rgb]{0,0,1}{\not{5}} }{\textcolor[rgb]{1,0,0}{\not{6}}} \dots
\frac{ \textcolor[rgb]{0,0,1}{\not{n-1}} }{\textcolor[rgb]{1,0,0}{\not{n}} }
\cdot \frac{n}{n+1}
\right)
$
}} $\\\\$
\small{\text{
$ =
\left(
\frac{2}{1} \right)
\cdot
\left(
\frac{n}{n+1}
\right)
$
}} $\\\\$
\boxed{
\small{\text{
$
\left(
\frac{2}{1} \right)
\cdot
\left(
\frac{n}{n+1}
\right) < 1.98
$
}}
} \\\\
\small{\text{$ \dfrac{n}{n+1} < \dfrac{1.98}{2} $ }$ \\\\$
\small{\text{ $ \dfrac{n}{n+1} < 0.99 $ } $\\\\$
\small{\text{ $ n < 0.99 *(n+1) $ }} $\\$
\small{\text{ $ n < 0.99 *n+ 0.99 $ }} $\\$
\small{\text{ $ n - 0.99 *n < 0.99 $ }} $\\$
\small{\text{ $ n *(1 - 0.99) < 0.99 $ }} $\\$$$

$$\\
\small{\text{ $ n * 0.01 < 0.99 $ }} $\\$
\small{\text{ $ n < \frac{ 0.99 } { 0.01 } $ }} $\\$
\small{\text{ $ n < 99 $ }} \\
\boxed{ \small{\text{ $ n = 98 $ }} }$$

heureka  Jan 26, 2015
 #1
avatar+19652 
+5
Best Answer

Let 

f(x) = \frac{x^2}{x^2 - 1}.

Find the largest n so that

$$$ f(2)\cdot f(3)\cdot f(4)\cdots f(n-1)\cdot f(n) < 1.98. $$$

$$\boxed{ \small{\text{ $ \prod \limits_{i=2}^{n} \frac{i^2}{i^2-1} < 1.98 $ }} } \qquad \dfrac{i^2}{i^2-1} = \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right)
$\\ \\ \\$
\small{\text{
$ \prod \limits_{i=2}^{n} \dfrac{i^2}{i^2-1} =
\prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \left(\dfrac{i}{i+1} \right) = \prod\limits_{i=2}^{n} \left(\dfrac{i}{i-1} \right) * \prod\limits_{i=2}^{n} \left(\dfrac{i}{i+1} \right)
$
}} $\\\\$
\small{\text{
$ =
\left( \frac{2}{2-1} \cdot
\frac{3}{3-1} \cdot \frac{4}{4-1} \cdot \frac{5}{5-1} \dots \frac{n}{n-1}
\right)
\cdot
\left( \frac{2}{2+1} \cdot
\frac{3}{3+1} \cdot \frac{4}{4+1} \cdot \frac{5}{5+1} \dots \frac{n-1}{n} \cdot \frac{n}{n+1}
\right)
$
}} $\\\\$
\small{\text{
$ =
\left( \frac{2}{1} \cdot
\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \dots \frac{n}{n-1}
\right)
\cdot
\left( \frac{2}{3} \cdot
\frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \dots \frac{n-1}{n} \cdot \frac{n}{n+1}
\right)
$
}} $\\\\$
\small{\text{
$ =
\left(
\frac{2}{1} \cdot
\frac{\textcolor[rgb]{1,0,0}{\not{3}} }{ \textcolor[rgb]{0,0,1}{\not{2}} } \cdot
\frac{\textcolor[rgb]{1,0,0}{\not{4}} }{ \textcolor[rgb]{0,0,1}{\not{3}} } \cdot
\frac{\textcolor[rgb]{1,0,0}{\not{5}} }{ \textcolor[rgb]{0,0,1}{\not{4}} } \dots
\frac{\textcolor[rgb]{1,0,0}{\not{n}} }{ \textcolor[rgb]{0,0,1}{\not{n-1}} } \right)
\cdot
\left(
\frac{ \textcolor[rgb]{0,0,1}{\not{2}} }{\textcolor[rgb]{1,0,0}{\not{3}}} \cdot
\frac{ \textcolor[rgb]{0,0,1}{\not{3}} }{\textcolor[rgb]{1,0,0}{\not{4}}} \cdot
\frac{ \textcolor[rgb]{0,0,1}{\not{4}} }{\textcolor[rgb]{1,0,0}{\not{5}}} \cdot
\frac{ \textcolor[rgb]{0,0,1}{\not{5}} }{\textcolor[rgb]{1,0,0}{\not{6}}} \dots
\frac{ \textcolor[rgb]{0,0,1}{\not{n-1}} }{\textcolor[rgb]{1,0,0}{\not{n}} }
\cdot \frac{n}{n+1}
\right)
$
}} $\\\\$
\small{\text{
$ =
\left(
\frac{2}{1} \right)
\cdot
\left(
\frac{n}{n+1}
\right)
$
}} $\\\\$
\boxed{
\small{\text{
$
\left(
\frac{2}{1} \right)
\cdot
\left(
\frac{n}{n+1}
\right) < 1.98
$
}}
} \\\\
\small{\text{$ \dfrac{n}{n+1} < \dfrac{1.98}{2} $ }$ \\\\$
\small{\text{ $ \dfrac{n}{n+1} < 0.99 $ } $\\\\$
\small{\text{ $ n < 0.99 *(n+1) $ }} $\\$
\small{\text{ $ n < 0.99 *n+ 0.99 $ }} $\\$
\small{\text{ $ n - 0.99 *n < 0.99 $ }} $\\$
\small{\text{ $ n *(1 - 0.99) < 0.99 $ }} $\\$$$

$$\\
\small{\text{ $ n * 0.01 < 0.99 $ }} $\\$
\small{\text{ $ n < \frac{ 0.99 } { 0.01 } $ }} $\\$
\small{\text{ $ n < 99 $ }} \\
\boxed{ \small{\text{ $ n = 98 $ }} }$$

heureka  Jan 26, 2015

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