Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial 

General term

${=}^{15}{C}_r{\left(t^2{x}^{\frac{1}{5}}\right)}^{15-r}{\left(\frac{{\left(1-x\right)}^{\frac{1}{10}}}{t}\right)}^r$    

for term independent on t

2 (15 – r) – r = 0

$\Rightarrow r=10$          

$\therefore T_{11}{=}^{15}{C}_{10}\times \left(1-x\right)$

Maximum value of x (1 – x) occur at

x =  $\frac{1}{2}$

${\left(3x^3-2x^2+\frac{5}{x^5}\right)}^{10}$

General term $10!\frac{{\left(3x^3\right)}^{\alpha }\cdot {\left(-2^x\right)}^{\beta }\cdot {\left(5x^{-5}\right)}^{\gamma }}{\alpha !\beta !\gamma !}$

$=10!\frac{3^{\alpha }\cdot {\left(-2\right)}^{\beta }\cdot 5^{\gamma }\cdot x^{3\alpha +2\beta -5\gamma }}{\alpha !\beta !\gamma !}$

Now for constant term 3 + 2 $-5\gamma =0$ …………..(i)

$\alpha +\beta +\gamma =10$ …………(ii)

From equation (i) & (ii)

$3\alpha +2\left(10-\alpha -\gamma \right)=5\gamma$

$\alpha +20=7\gamma$

= 1, $\gamma$ = 3, = 6

Constant term $10!\cdot \frac{3^1\cdot {\left(-2\right)}^6\cdot 5^3}{3!6!}=2^9\cdot 3^2\cdot 5^4\cdot 7^1$

 ${\displaystyle \sum _{k=1}^{10}\frac{k}{k^4+k^2+1}}$

$=\frac{1}{2}{\displaystyle \sum _{k=1}^{10}\left[\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right]}$

$=\frac{1}{2}\left[1-\frac{1}{111}\right]=\frac{110}{222}=\frac{55}{111}=\frac{m}{n}$

$\therefore m+n=166$

Kindly consider the following figure

${\sum }_{r=0}^{20}?{}^{50-r}{\mathrm{C}}_6={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+{}^{30}{\mathrm{C}}_6$

$\begin{array}{rr}& ={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+\left({}^{30}{\mathrm{C}}_6+{}^{30}{\mathrm{C}}_7\right)-{}^{30}{\mathrm{C}}_7\\ & ={}^{50}{\mathrm{C}}_6+{}^{49}{\mathrm{C}}_6+{}^{48}{\mathrm{C}}_6+\dots .+\left({}^{31}{\mathrm{C}}_6+{}^{31}{\mathrm{C}}_7\right)-{}^{30}{\mathrm{C}}_7\\ & ={}^{50}{\mathrm{C}}_6+{}^{50}{\mathrm{C}}_7-{}^{30}{\mathrm{C}}_7\\ & ={}^{51}{\mathrm{C}}_7-{}^{30}{\mathrm{C}}_7\end{array}$