+0  
 
0
93
8
avatar

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

Guest Jul 3, 2017

Best Answer 

 #6
avatar+18369 
+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
Sort: 

8+0 Answers

 #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+18369 
+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+89804 
0

You did not give much explanation here Heureka :)

That is not like you ://

Melody  Jul 4, 2017
 #4
avatar+25995 
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+18369 
+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+75333 
+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

8 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details