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 #1
avatar+18610 
+1
heureka 16.10.2017
 #5
avatar+18610 
+1

6 × 6! + 7 × 7! + 8 × 8! + 9 × 9! + 10 × 10! + 11 × 11! + 12 × 12! + 13 × 13! + 14 × 14! = a! - b!

a + b= ?

 

\(\begin{array}{|rcll|} \hline && n\times n! \\ &=& (n+1-1)\times n! \\ &=& [(n+1)-1]\times n! \\ &=& (n+1)\times n! -1\times n! \\ &=& n!\times (n+1) -n! \\ &=& (n+1)! -n! \\\\ &&\mathbf{ n\times n! = (n+1)! -n! } \\ \hline \end{array}\)

 

\(\begin{array}{|rcrcr|} \hline 6\times 6! &=& 7!- 6! \\ 7\times 7! &=& 8!- 7! \\ 8\times 8! &=& 9!- 8! \\ 9\times 9! &=& 10!- 9! \\ 10\times 10! &=& 11!-10! \\ 11\times 11! &=& 12!-11! \\ 12\times 12! &=& 13!-12! \\ 13\times 13! &=& 14!-13! \\ 14\times 14! &=& 15!-14! \\ \hline \text{sum} &=& 7! &\mathbf{-}& \mathbf{6!} \\ &+& 8! &-& 7! \\ &+& 9! &-& 8! \\ &+& 10! &-& 9! \\ &+& 11! &-& 10! \\ &+& 12! &-& 11! \\ &+& 13! &-& 12! \\ &+& 14! &-& 13! \\ &+& 15! &-& 14! \\\\ &=& \not{ {\color{green}7!}} && -6! \\ && {\color{red}{-}}\not{ {\color{red}7!}} && {\color{green}{+}} \not{ {\color{green}8!}} \\ && {\color{green}{+}}\not{ {\color{green}9!}} && {\color{red}{-}}\not{ {\color{red}8!}} \\ && {\color{red}{-}}\not{ {\color{red}9!}} && {\color{green}{+}} \not{ {\color{green}10!}} \\ && {\color{green}{+}} \not{ {\color{green}11!}} && {\color{red}{-}}\not{ {\color{red}10!}} \\ && {\color{red}{-}}\not{ {\color{red}11!}} && {\color{green}{+}} \not{ {\color{green}12!}} \\ && {\color{green}{+}} \not{ {\color{green}13!}} && {\color{red}{-}}\not{ {\color{red}12!}} \\ && {\color{red}{-}}\not{ {\color{red}13!}} && {\color{green}{+}} \not{ {\color{green}14!}} \\ && \mathbf{+15!} && {\color{red}{-}}\not{ {\color{red}14!}} \\\\ &\mathbf{=}& \mathbf{15! - 6!} \\ \hline \end{array} \)

 

so a = 15 and b = 6

a+b = 15+6 = 21

 

laugh

heureka 10.10.2017
 #4
avatar+18610 
+2

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 25.09.2017