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n=1/1+2 + 1/1+2+3 + ...... + 1/1+2+3+....+2014 + 2/2015, n=?

Guest Jul 3, 2017

Best Answer 

 #6
avatar+20683 
+1

n=1/1+2 + 1/1+2+3 + ...... + 1/1+2+3+....+2014 + 2/2015, n=?

 

\(\begin{array}{|rcll|} \hline n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ \hline \end{array} \)

 

 

\(\begin{array}{rcll} n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ &=& \frac{1}{\left(\frac{1+2}{2}\right)\cdot 2} +\frac{1}{\left(\frac{1+3}{2}\right)\cdot 3} +\frac{1}{\left(\frac{1+4}{2}\right)\cdot 4} + \ldots + \frac{1}{\left(\frac{1+2013}{2}\right)\cdot 2013} + \frac{1}{\left(\frac{1+2014}{2}\right)\cdot 2014} + \frac{2}{2015} \\ &=& \frac{2}{2\cdot 3} +\frac{2}{3\cdot 4} +\frac{2}{4\cdot 5} + \ldots + \frac{2}{2013\cdot 2014} + \frac{2}{2014\cdot 2015} + \frac{2}{2015} \\ &=& 2\times \Big( \frac{1}{2015} + \frac{1}{2\cdot 3} +\frac{1}{3\cdot 4} +\frac{1}{4\cdot 5} + \ldots + \frac{1}{2013\cdot 2014} + \frac{1}{2014\cdot 2015} \Big)\\\\ && \mathbf{\frac{1}{a\cdot (a+1)} = \frac{1}{a} - \frac{1}{a+1}} \\\\ &=& 2\times \Big[ \frac{1}{2015} + \left(\frac{1}{2}-\frac{1}{3}\right) +\left(\frac{1}{3}-\frac{1}{4}\right) +\left(\frac{1}{4}-\frac{1}{5}\right) + \ldots \\ && + \left(\frac{1}{2013}-\frac{1}{2014}\right) + \left(\frac{1}{2014}-\frac{1}{2015}\right) \Big] \\\\ &=& 2\times \Big[ \frac{1}{2015} + \frac12+ \underbrace{\left(-\frac{1}{3}+\frac{1}{3}\right)}_{=0} +\underbrace{\left(-\frac{1}{4}+\frac{1}{4}\right) }_{=0} +\underbrace{\left(-\frac{1}{5}+\frac{1}{5}\right)}_{=0} + \ldots \\ && + \underbrace{\left(-\frac{1}{2013}+\frac{1}{2013}\right)}_{=0} + \underbrace{\left(-\frac{1}{2014}+\frac{1}{2014}\right)}_{=0} -\frac{1}{2015} \Big] \\\\ &=& 2\times \left(\frac{1}{2015}+ \frac12 -\frac{1}{2015} \right) \\\\ &=& 2\times \left(\frac12 \right) \\\\ &\mathbf{=}& \mathbf{1} \\ \end{array}\)

 

laugh

heureka  Jul 6, 2017
 #1
avatar
+1

What are you trying to add?? What is with 1/1?? If you are trying to add 1+2+3+.......+2015, then the answer is:

[2015 x 2016] / 2 =2,031,120

Guest Jul 3, 2017
 #2
avatar+20683 
+1

n=1/1+2 + 1/1+2+3 + ...... + 1/1+2+3+....+2014 + 2/2015, n=?

edited

 

continued fraction ?

 

\(\begin{array}{|rcll|} \hline n =\cfrac{1}{1+2+\cfrac{1}{1+2+3+\cfrac{1}{ 1+2+3+4+\ldots \cfrac{1}{1+2+3+4+\ldots + 2014 +\frac{2}{2015}}}}} \\ \hline \end{array} \)

 

 

n = 0.31606040

 

laugh

heureka  Jul 4, 2017
edited by heureka  Jul 5, 2017
 #3
avatar+94181 
0

You did not give much explanation here Heureka :)

That is not like you ://

Melody  Jul 4, 2017
 #4
avatar+27234 
0

n = 1

 

\(\sum_{r=2}^{2014}(\frac{1}{\sum_{k=1}^rk})=\frac{2013}{2015}\\ \text{ add } \frac{2}{2015} \text{ to get } 1\)

Alan  Jul 4, 2017
 #5
avatar
0

Sum of(n=1 to 2013)=[2 / (n^2 + 3n + 2)] = 2013/2015 + 2/2015 =n=1

Guest Jul 5, 2017
 #6
avatar+20683 
+1
Best Answer

n=1/1+2 + 1/1+2+3 + ...... + 1/1+2+3+....+2014 + 2/2015, n=?

 

\(\begin{array}{|rcll|} \hline n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ \hline \end{array} \)

 

 

\(\begin{array}{rcll} n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ &=& \frac{1}{\left(\frac{1+2}{2}\right)\cdot 2} +\frac{1}{\left(\frac{1+3}{2}\right)\cdot 3} +\frac{1}{\left(\frac{1+4}{2}\right)\cdot 4} + \ldots + \frac{1}{\left(\frac{1+2013}{2}\right)\cdot 2013} + \frac{1}{\left(\frac{1+2014}{2}\right)\cdot 2014} + \frac{2}{2015} \\ &=& \frac{2}{2\cdot 3} +\frac{2}{3\cdot 4} +\frac{2}{4\cdot 5} + \ldots + \frac{2}{2013\cdot 2014} + \frac{2}{2014\cdot 2015} + \frac{2}{2015} \\ &=& 2\times \Big( \frac{1}{2015} + \frac{1}{2\cdot 3} +\frac{1}{3\cdot 4} +\frac{1}{4\cdot 5} + \ldots + \frac{1}{2013\cdot 2014} + \frac{1}{2014\cdot 2015} \Big)\\\\ && \mathbf{\frac{1}{a\cdot (a+1)} = \frac{1}{a} - \frac{1}{a+1}} \\\\ &=& 2\times \Big[ \frac{1}{2015} + \left(\frac{1}{2}-\frac{1}{3}\right) +\left(\frac{1}{3}-\frac{1}{4}\right) +\left(\frac{1}{4}-\frac{1}{5}\right) + \ldots \\ && + \left(\frac{1}{2013}-\frac{1}{2014}\right) + \left(\frac{1}{2014}-\frac{1}{2015}\right) \Big] \\\\ &=& 2\times \Big[ \frac{1}{2015} + \frac12+ \underbrace{\left(-\frac{1}{3}+\frac{1}{3}\right)}_{=0} +\underbrace{\left(-\frac{1}{4}+\frac{1}{4}\right) }_{=0} +\underbrace{\left(-\frac{1}{5}+\frac{1}{5}\right)}_{=0} + \ldots \\ && + \underbrace{\left(-\frac{1}{2013}+\frac{1}{2013}\right)}_{=0} + \underbrace{\left(-\frac{1}{2014}+\frac{1}{2014}\right)}_{=0} -\frac{1}{2015} \Big] \\\\ &=& 2\times \left(\frac{1}{2015}+ \frac12 -\frac{1}{2015} \right) \\\\ &=& 2\times \left(\frac12 \right) \\\\ &\mathbf{=}& \mathbf{1} \\ \end{array}\)

 

laugh

heureka  Jul 6, 2017
 #7
avatar
0

Notice that the term in which the denominator ends in 2013 [1/1+2+3....+2013] is actually the 2012th term. Similarly, the term in which the denominator ends in 2014 [1/1+2+3.....+2014] is actually the 2013th term. Therefore, it is only necessary to sum up the terms 1 to 2013: Sum{2/(n^2 + 3n +2)}, from n=1 to 2013=2013/2015 + 2/2015 = 1.

Guest Jul 6, 2017
 #8
avatar+92856 
+1

 

Thanks, heureka, for deriving that one, step-by-step.....

 

It's certainly not intuitive that the final result will be "1".....!!!!!

 

 

cool cool cool

CPhill  Jul 6, 2017
edited by CPhill  Jul 17, 2017

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