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 #3
avatar+18548 
+1

What is the value of the sum 1/1*3 + 1/3*5 + 1/5*7+ 1/7*9+...+1/199*201?

Express your answer as a fraction in simplest form.

 

\(\begin{array}{rcll} && \frac{1}{1*3} + \frac{1}{3*5} + \frac{1}{5*7}+ \frac{1}{7*9}+\ldots+\frac{1}{199*201} \\ &=& \frac{1}{1*3} + \frac{1}{3*5} + \frac{1}{5*7}+ \frac{1}{7*9}+\ldots+\frac{1}{(2n-1)(2n+1)} \\ \hline && \frac{1}{(2n-1)(2n+1)} = \frac12\left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \\ && \frac{1}{1*3} = \frac12\left( \frac{1}{1} - \frac{1}{3} \right) \\ && \frac{1}{3*5} = \frac12\left( \frac{1}{3} - \frac{1}{5} \right) \\ && \frac{1}{5*7} = \frac12\left( \frac{1}{5} - \frac{1}{7} \right) \\ && \frac{1}{7*9} = \frac12\left( \frac{1}{7} - \frac{1}{9} \right) \\ && \ldots \\ && \frac{1}{199*201} = \frac12\left( \frac{1}{199} - \frac{1}{201} \right) \\ \hline &=& \frac12\left( \frac{1}{1} - \frac{1}{3} \right) + \frac12\left( \frac{1}{3} - \frac{1}{5} \right) + \frac12\left( \frac{1}{5} - \frac{1}{7} \right) + \frac12\left( \frac{1}{7} - \frac{1}{9} \right)+\ldots+\frac12\left( \frac{1}{199} - \frac{1}{201} \right) \\ &=& \frac12\left( \frac{1}{1} - \underbrace{\frac{1}{3} + \frac{1}{3}}_{=0} - \underbrace{\frac{1}{5}+\frac{1}{5}}_{=0} - \underbrace{\frac{1}{7} + \frac{1}{7}}_{=0} - \underbrace{\frac{1}{9}+ \frac{1}{9}}_{=0} +\ldots- \underbrace{\frac{1}{199}+\frac{1}{199}}_{=0} - \frac{1}{201} \right) \\ &=& \frac12\left( \frac{1}{1} - \frac{1}{201} \right) \\ &=& \frac12\left( 1 - \frac{1}{201} \right) \\ &=& \frac12\left( \frac{201-1}{201} \right) \\ &=& \frac12\left( \frac{200}{201} \right) \\ &=& \frac{100}{201} \\ \end{array} \)

 

laugh

heureka Sep 20, 2017 8:09:57 AM
 #2
avatar+18548 
+1

Compute 1*1/2+2*1/4+3*1/8+...+n*1/2n+...

 

\(\begin{array}{|rcll|} \hline S_n &=& 1*\frac{1}{2}+2*\frac{1}{4}+3*\frac{1}{8}+4*\frac{1}{16}+\ldots+n*\left(\frac{1}{2}\right)^n+\ldots \\ \hline \end{array} \)

 

(i)

\(\text{Multiply } S_n \text{ by } \tfrac12 \text{ the common ratio of the geometric series} \)

\(\small{ \begin{array}{rcccccccccccccl} S_n &=& 1*\frac{1}{2} &+& 2*\frac{1}{4} &+& 3*\frac{1}{8} &+& 4*\frac{1}{16} &+& \ldots &+& n*\left(\frac{1}{2}\right)^n && \\ \frac12 S_n &=& & & 1*\frac{1}{4} &+& 2*\frac{1}{8} &+& 3*\frac{1}{16} &+& \ldots &+& (n-1)*\left(\frac{1}{2}\right)^n &+& n*\left(\frac{1}{2}\right)^{n+1} \\ \hline (1-\frac12)S_n &=& [~ 1*\frac{1}{2} &+& 1*\frac{1}{4} &+& 1*\frac{1}{8} &+& 1*\frac{1}{16} &+& \ldots &+& 1*\left(\frac{1}{2}\right)^n ~] &-& n*\left(\frac{1}{2}\right)^{n+1} \quad (\text{subtract})\\ \end{array} } \)

 

The series in the square brackets is a geometric series with \(a = \frac12\), \(r = \frac12\) and \(n\) terms,
Thus, \(S_n\) for this series =\( \dfrac{a(1-r^{n})}{1-r} = \dfrac{\frac12(1-(\frac12)^n)}{1-\frac12}=1-(\frac12)^n\)

\(\begin{array}{rcll} (1-\frac12)S_n &=& [~ 1*\frac{1}{2} + 1*\frac{1}{4} + 1*\frac{1}{8} + 1*\frac{1}{16} + \ldots + 1*\left(\frac{1}{2}\right)^n ~] - n*\left(\frac{1}{2}\right)^{n+1} \\ (1-\frac12)S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ \frac12S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ \end{array} \)

 

(ii)

\(\text{Because } \left|\frac12\right| < 1 \text{, then } \lim \limits_{n\to \infty} { \left(\frac12 \right)^n } = 0 \text{ and } \lim \limits_{n\to \infty} { \left(\frac12 \right)^{n+1} } = 0\)

\(\begin{array}{rcll} \frac12S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ S_n &=& 2 \left( 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \right) \\ \lim \limits_{n\to \infty} {S_n} &=& 2 \left( 1-0 - n*0 \right) \\ &=& 2\cdot( 1 ) \\ &=& 2 \\ \end{array}\)

 

laugh

heureka Sep 20, 2017 7:38:32 AM
 #1
avatar+18548 
+1

[Hilfe]Vereinfachen Sie folgende beiden Ausdrücke:

 

 

5.1

\(\begin{array}{|rcll|} \hline && \dfrac{1}{1-\frac{1}{p}} - \dfrac{1}{p-1} \\\\ &=& \dfrac{1}{\frac{p}{p}-\frac{1}{p}} - \dfrac{1}{p-1} \\\\ &=& \dfrac{1}{\frac{p-1}{p}} - \dfrac{1}{p-1} \\\\ &=& \dfrac{p}{p-1} - \dfrac{1}{p-1} \\\\ &=& \dfrac{p-1}{p-1} \\\\ &=& 1 \\ \hline \end{array}\)

 

 

5.2

\(\begin{array}{|rcll|} \hline && \dfrac{ \left(-\frac{1}{2}\right)^{-3} \left(\frac{3}{4}\right)^2 x^2 y } { \left(\frac{2}{3}\right)^{-2} \left(-\frac{1}{2}\right) \left(x y\right)^{-2} } \\\\ &=& \dfrac{ \left(\frac{3}{4}\right)^2\left(\frac{2}{3}\right)^{2} x^2 y \left(x y\right)^{2} } { \left(-\frac{1}{2}\right)^{3} \left(-\frac{1}{2}\right) } \\\\ &=& \dfrac{ \left(\frac{3}{4}\right)^2\left(\frac{2}{3}\right)^{2} x^2 y x^2y^2 } { \left(-\frac{1}{2}\right)^{3} \left(-\frac{1}{2}\right)^{1} } \\\\ &=& \dfrac{ \left(\frac{3}{4}\right)^2\left(\frac{2}{3}\right)^{2} x^{2+2} y^{1+2} } { \left(-\frac{1}{2}\right)^{3+1} } \\\\ &=& \dfrac{ \left(\frac{3}{4}\right)^2\left(\frac{2}{3}\right)^{2} x^{4} y^{3} } { \left(-\frac{1}{2}\right)^{4} } \\\\ &=& \dfrac{ \left(\frac{3}{4}\right)^2\left(\frac{2}{3}\right)^{2} x^{4} y^{3} } { \left(-1 \right)^4 \left(\frac{1}{2}\right)^{4} } \\\\ &=& \dfrac{ \left(\frac{3}{4}\right)^2\left(\frac{2}{3}\right)^{2} x^{4} y^{3} } { 1\cdot \left(\frac{1}{2}\right)^{4} } \\\\ &=& \dfrac{ \frac{3^2}{4^2} \cdot \frac{2^2}{3^2} \cdot x^{4} y^{3} } { \left(\frac{1}{2}\right)^{4} } \\\\ &=& \dfrac{ \frac{2^2}{4^2} x^{4} y^{3} } { \left(\frac{1}{2}\right)^{4} } \\\\ &=& \dfrac{ \left(\frac{2}{4}\right)^2 x^{4} y^{3} } { \left(\frac{1}{2}\right)^{4} } \\\\ &=& \dfrac{ \left(\frac{1}{2}\right)^2 x^{4} y^{3} } { \left(\frac{1}{2}\right)^{4} } \\\\ &=& \dfrac{ x^{4} y^{3} } { \left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{-2} } \\\\ &=& \dfrac{ x^{4} y^{3} } { \left(\frac{1}{2}\right)^{4-2} } \\\\ &=& \dfrac{ x^{4} y^{3} } { \left(\frac{1}{2}\right)^{2} } \\\\ &=& \dfrac{ x^{4} y^{3} } { \frac{1}{2^2}} \\\\ &=& 2^2x^{4} y^{3} \\\\ &=& 4 x^{4} y^{3} \\ \hline \end{array}\)

 

laugh

heureka Sep 20, 2017 6:19:07 AM