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 #4
avatar+18566 
0

Algebra

Compute the sum \(\mathbf{\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots}\)

 

\(\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ ? } } \\\\ \begin{array}{|lcll|} \hline s_n = \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} \\ \hline \end{array} \\ \end{array}\\\)

 

Formula:

\(\begin{array}{|lcll|} \hline \text{in general}:\ \frac{1}{n(n+d)} = \frac{1}{d}\left(\frac{1}{n}- \frac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

 

we rearrange:

\(\begin{array}{|rcll|} \hline \dfrac{2}{n \cdot (n+1) \cdot (n+2)} \\\\ &=& \dfrac{2}{n}\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \dfrac{2}{n}\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{2}{n}\times \dfrac{1}{n+1} - \dfrac{2}{n}\times \dfrac{1}{n+2} \\\\ &=& 2\times \left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)- 2\times \dfrac{1}{2} \times \left(\dfrac{1}{n} -\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{2}{n} - \dfrac{2}{n+1} -\dfrac{1}{n} + \dfrac{1}{n+2} \\\\ \mathbf{\dfrac{2}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ \dfrac{1}{n} - \dfrac{2}{n+1} + \dfrac{1}{n+2} } \\ \hline \end{array}\)

 

telescoping series

\(\begin{array}{|rcll|} \hline s_n &=& \mathbf{\dfrac{1}{1}} &\mathbf{-}& \mathbf{\dfrac{2}{2}} &\color{red}+& \color{red}\dfrac{1}{3} \\\\ &\mathbf{+}& \mathbf{\dfrac{1}{2}} &\color{red}-& \color{red}\dfrac{2}{3} &\color{blue}+& \color{blue}\dfrac{1}{4} \\\\ &\color{red}+& \color{red}\dfrac{1}{3} &\color{blue}-& \color{blue}\dfrac{2}{4} &\color{red}+& \color{red}\dfrac{1}{5} \\\\ &\color{blue}+& \color{blue}\dfrac{1}{4} &\color{red}-& \color{red}\dfrac{2}{5} &\color{green}+& \color{green}\dfrac{1}{6} \\\\ && \ldots \\\\ &+\color{red}& \color{red}\dfrac{1}{n-2} &\color{green}-& \color{green}\dfrac{2}{n-1} &\color{red}+& \color{red}\dfrac{1}{n} \\\\ &\color{green}+& \color{green}\dfrac{1}{n-1} &\color{red}-& \color{red}\dfrac{2}{n} &\mathbf{+}& \mathbf{\dfrac{1}{n+1}} \\\\ &\color{red}+& \color{red}\dfrac{1}{n} &\mathbf{-}& \mathbf{\dfrac{2}{n+1}} &\mathbf{+}& \mathbf{\dfrac{1}{n+2}} \\ \hline \end{array}\)

 

The part of each term cancelling with part of the next two diagonal terms:

Example:

\(\begin{array}{|lcll|} \hline \frac{1}{3}-\frac{2}{3}+\frac{1}{3} = 0 \\ \frac{1}{4}-\frac{2}{4}+\frac{1}{4} = 0 \\ \frac{1}{5}-\frac{2}{5}+\frac{1}{5} = 0 \\ \ldots \\ \frac{1}{n}-\frac{2}{n} + \frac{1}{n} = 0 \\ \hline \end{array}\)

 

So \(s_n\) is, we have all black terms left :

\(\begin{array}{|rcll|} \hline s_n &=& \dfrac{1}{1}-\dfrac{2}{2}+\dfrac{1}{2} + \dfrac{1}{n+1} - \dfrac{2}{n+1} + \dfrac{1}{n+2} \\\\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{1}{2} - \dfrac{1}{n+1} + \dfrac{1}{n+2}} \\ \hline \end{array} \)

 

 \(\lim \limits_{n\to \infty} { \dfrac{1}{n+1}} = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{1}{n+2} } = 0 \)

 

\( \begin{array}{|rcll|} \hline \lim \limits_{n\to \infty} s_n &=& \dfrac{1}{2} - 0 + 0 \\ &=& \dfrac{1}{2} \\ \hline \end{array} \)

 

\(\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ \dfrac{1}{2} } } \\ \end{array}\\\)

 

 

laugh

heureka 3 hours ago
 #3
avatar+18566 
+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
 #2
avatar+18566 
+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