2022/029) Show that $\frac{1}{29} + \frac{1}{31} + ... \frac{1}{55}$ when the sum is put in the form $\frac{p}{q}$ then p is divisible by 83

Using $\frac{1}{n} + \frac{1}{a-n} = \frac{a}{n(a-n)}$
We get $\frac{1}{n} =  \frac{a}{n(a-n)}-\frac{1}{a-n}$
Putting a= 83 we get
$\frac{1}{n} =  \frac{83}{n(83-n)}-\frac{1}{83-n}$

Running n from 29 upto 55 in increment of 2 we have

$\sum_{n=14}^{27} \frac{1}{2n+1}=\sum_{n=14}^{27} (\frac{83}{(2n+1)(82-2n)}-\frac{1}{82-2n})$

Or $\sum_{n=14}^{27} \frac{1}{2n+1}=83\sum_{n=14}^{27}\frac{1}{(2n+1)(82-2n)}- \sum_{n=14}^{27}\frac{1}{82-2n}$

As $\sum_{n=14}^{27}\frac{1}{(2n+1)(82-2n)}$ when we take the sum  we shall get it of the form $\frac{p}{q}$ where q does not have a divisor 83. as 83 does not divide any denominator

So  $\sum_{n=14}^{27} \frac{1}{2n+1}=\frac{83p}{q}- \sum_{n=14}^{27}\frac{1}{82-2n}\cdots(1)$

Further $82-2n = 2k => n = 41-k$

So $\sum_{n=14}^{27}\frac{1}{82-2n} = \sum_{n=14}^{27}\frac{1}{2n}\cdots(2)$

 Now the given expression

= $\sum_{n=0}^{27} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n}$

= $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n} + \sum_{n=14}^{27} \frac{1}{2n+1 }$

= $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n} + \frac{83p}{q}- \sum_{n=14}^{27}\frac{1}{82-2n}$

= $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n} + \frac{83p}{q}- \sum_{n=14}^{27}\frac{1}{2n}$

=  $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} + \frac{83p}{q}- 2 \sum_{n=14}^{27}\frac{1}{2n}$

=  $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} + \frac{83p}{q}-  \sum_{n=14}^{27}\frac{1}{n}$

=  $\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} + \frac{83p}{q}-   \sum_{n=7}^{13}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{2n+1}$

=  $(\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=7} ^{13} \frac{1}{2n+1}) + \frac{83p}{q}-   \sum_{n=1}^{6}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{2}{2n}$

=  $\sum_{n=0}^{6} \frac{1}{2n+1 } + \frac{83p}{q}-   \sum_{n=1}^{6}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{n}$

= $\frac{83p}{q} + \sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} -   \sum_{n=7}^{13}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{2n+1}$
= $\frac{83p}{q} + \sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=7}^{13} \frac{1}{2n+1} -   \sum_{n=1}^{13}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{2n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{2n}-  \sum_{n=7}^{13}\frac{1}{2n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{n}$
= $\frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - (\sum_{n=3}^{6}\frac{1}{2n+1} + \sum_{n=4}^{6}\frac{1}{2n})$
= $\frac{83p}{q} + \sum_{n=0}^{2} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=4}^{6}\frac{1}{2n}$
= $\frac{83p}{q} + \sum_{n=0}^{2} \frac{1}{2n+1 } - \sum_{n=1}^{3}\frac{1}{2n} - \sum_{n=4}^{6}\frac{1}{2n} -  \sum_{n=4}^{6}\frac{1}{2n}$
= $\frac{83p}{q} + \sum_{n=0}^{2} \frac{1}{2n+1 } - \sum_{n=1}^{3}\frac{1}{2n} - \sum_{n=4}^{6}\frac{1}{n}$
= $\frac{83p}{q} + 1 + \frac{1}{3} +  \frac{1}{5} - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6}$
$=\frac{83p}{q}$

Numerator is divisible by 83 and denominator is not and 83 is prime . hence proved