+0

# Series Question

0
103
2
+26

Determine the value of $$\frac{\frac{2016}{1} + \frac{2015}{2} + \frac{2014}{3} + \dots + \frac{1}{2016}}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2017}}.$$

Any help is appreciated!

Aug 30, 2019

#1
+23273
+3

Determine the value of  $$\dfrac{2016}{1} + \dfrac{2015}{2} + \dfrac{2014}{3} + \dots + \dfrac{1}{2016} \above 1pt \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2017}$$

$$\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{2016}{1} + \dfrac{2015}{2} + \dfrac{2014}{3} + \dots + \dfrac{1}{2016} \above 1pt \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2017} } \\\\ &=& \dfrac{\dfrac{2017-1}{1} + \dfrac{2017-2}{2} + \dfrac{2017-3}{3} + \dots + \dfrac{2017-2016}{2016}} {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+ \dfrac{1}{2017}} \\\\ &=& \dfrac{\dfrac{2017}{1} + \dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016}-2016\cdot 1} {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+ \dfrac{1}{2017}} \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016}+2017-2016 } {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+ \dfrac{1}{2017}} \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016}+1 } {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+ \dfrac{1}{2017}} \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016} +1 } {\left( \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{2016}+\dfrac{1}{2017} \right)\times \dfrac{2017}{2017} } \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016} +1 } {\left( \dfrac{2017}{2} + \dfrac{2017}{3} + \dfrac{2017}{4} + \dots + \dfrac{2017}{2016}+\dfrac{2017}{2017} \right)\times \dfrac{1}{2017} } \\\\ &=& \dfrac{\dfrac{2017}{2} + \dfrac{2017}{3} + \dots + \dfrac{2017}{2016} +1 } {\left( \dfrac{2017}{2} + \dfrac{2017}{3} + \dfrac{2017}{4} + \dots + \dfrac{2017}{2016}+1 \right)\times \dfrac{1}{2017} } \\\\ &=& \dfrac{\left(\dfrac{2017}{2} + \dfrac{2017}{3}+ \dfrac{2017}{4} + \dots + \dfrac{2017}{2016} +1\right) } {\left( \dfrac{2017}{2} + \dfrac{2017}{3} + \dfrac{2017}{4} + \dots + \dfrac{2017}{2016}+1 \right) } \times 2017 \\\\ &=& \mathbf{2017} \\ \hline \end{array}$$

Aug 30, 2019
#2
+26
0

Oh wow! I used the same approach and was off by one. Just a careless mistake.

Aug 30, 2019