The sum of 50th powers of all the sides and of all the diagonals of a regular 100-gon inscribed in a circle of radius R is in the form of $5000\cdot \frac{\left(ab\right)!}{{\left(\left(mn\right)!\right)}^2}{R}^{50}$  then a + b + m – n is equal to----

(Where ab, mn are two digit numbers)

Let us compute the sum of 50th powers of all the sides and diagonals issued from one vertex A of a regular 100-gon inscribed in a circle of R. The problem reduces to the determination of the sum 

$S={\left(2R\sin \frac{\pi }{100}\right)}^{50}+{\left(2R\sin \frac{2\pi }{100}\right)}^{50}+\mathrm{........}+{\left(2R\sin \frac{99\pi }{100}\right)}^{50}$

We  have,

$$\begin{gathered} {\sin }^{50}\theta ={\left[\frac{(\cos \theta +i\sin \theta )-(\cos \theta -i\sin \theta )}{2i}\right]}^{50} \\ =-\frac{1}{2^{50}}{[(\cos \theta +i\sin \theta )-(\cos \theta -i\sin \theta )]}^{50} \\ \cos \theta +i\sin \theta =x, \cos \theta -i\sin \theta =\frac{1}{x} \\ \therefore {\sin }^{50}\theta =\frac{-1}{2^{50}}{(x-\frac{1}{x})}^{50} \\ \therefore {\sin }^{50}\theta =\frac{-1}{2^{50}} [x^{50}-50C_1{x}^{49}\left(\frac{1}{x}\right)+50C_2{x}^{48}{\left(\frac{1}{x}\right)}^2-\mathrm{...........}-50C_{49}x\cdot \frac{1}{x^{49}}+50C_{50}\frac{1}{x^{50}} \\ \Rightarrow {\sin }^{50}\theta =\frac{-1}{2^{50}}[(x^{50}+\frac{1}{x^{50}})-50c_1(x^{48}+\frac{1}{x^{48}})+50c_2(x^{46}+\frac{1}{x^{46}})\mathrm{.............}+50c_{24}(x^2+\frac{1}{x^2})-50c_{25}] \end{gathered}$$

$\begin{array}{l}\therefore S=-R^{50}2\{\begin{array}{l}[\cos 50\frac{\pi }{100}+\cos 50\frac{2\pi }{100}+\mathrm{..........}\cos 50\left(\frac{99\pi }{100}\right)]\\ -50C_1(\cos 48\left(\frac{\pi }{100}\right)+\cos 48\left(\frac{2\pi }{100}\right)+\mathrm{........}+\cos 48\left(\frac{99\pi }{100}\right))\mathrm{............}\end{array}\\ +50C_{24}(\cos 2\left(\frac{\pi }{100}\right)+\cos 2\left(\frac{2\pi }{100}\right)+\mathrm{............}\cos 2\left(\frac{99\pi }{100}\right))\\ -\frac{99}{2}50C_{25}]\end{array}$

$\Rightarrow S=-2R^{50}[s_1-50C_1{s}_2+50C_2{s}_3\mathrm{........}+50C_{24}{s}_{25}-\frac{99}{2}50C_{25}]$

Where letters  $s_1,s_2,s_3\mathrm{.........}{s}_{25}$ represents the sum in the brackets then from Trigonometry $s_1=s_2=s_3=\mathrm{............}{s}_{25}=-1$ 
since $\cos \alpha +\cos (\alpha +\beta )+\mathrm{..........}=\frac{\sin \left(\frac{n\beta }{2}\right)}{\sin \left(\frac{\beta }{2}\right)}\cos \left(\alpha +(n-1)\frac{\beta }{2}\right)$

$$\begin{gathered} \therefore S=-2R^{50}[-1+50C_1-50C_2+50c_3\mathrm{........}-50C_{24}-\frac{99}{2}50C_{25}] \\ S=2R^{50}[1-50C_1+50C_2-50C_3+\mathrm{........}+50C_{24}+\frac{99}{2}50C_{25}] \\ =R^{50}[2-2\cdot 50C_1+2\cdot 50C_2-2\cdot 50C_3+\mathrm{.........}+2\cdot 50C_{24}+9950C_{25}] \\ \begin{array}{l}=R^{50}[1-50C_1+50C_2-50C_3+\mathrm{........}+50C_{24}-50C_{25}+50C_{26}-50C_{27}+\mathrm{........}\\ +50C_{48}-50C_{49}+50C_{50}+100\times 50C_{25}]\end{array} \\ =R^{50}\{{(1-1)}^{50}+100\cdot 50c_{25}\} \\ \begin{array}{l}=R^{50}10050C_{25}\\ =10050C_{25}{R}^{50}\end{array} \end{gathered}$$

Hence the sum of 50th powers of all the sides and all the diagonals of the $100-gon$ is $\frac{100S}{2}=\frac{100}{2}100\cdot 50C_{25}{R}^{50}$

$=5000.\frac{50!}{{\left(\right(25)!)}^2}{R}^{50}$

So ab=50, mn = 25 are two digit numbers then a =5, b = 0, m = 5  a + b + m – n = 2