Binomial Theorem

Kindly consider the following figure

  $T_{r+1}{=}^{60}{C}_r{\left(x^{\frac{1}{2}}\right)}^{60-r}{\left(x^{-\frac{1}{3}}\right)}^r{\left(5^{\frac{-1}{4}}\right)}^{60-r}{\left(5^{\frac{1}{2}}\right)}^r$

for

$x^{10}\frac{60-r}{2}-\frac{r}{3}=10$    

$\Rightarrow 180-3r-2r=60$             

->r = 24

k = 3 + exponent of 5 in 

= $3+\left(\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]-\left[\frac{24}{5}\right]-\left[\frac{24}{5^2}\right]-\left[\frac{36}{5}\right]-\left[\frac{36}{5^2}\right]\right)$  

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

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

$\Rightarrow r=10$          

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

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

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

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

Let   $\frac{A_2}{A_1}=\frac{A_3}{A_2}=...=r$

$\therefore A_1{A}_3{A}_5{A}_7=\frac{1}{1296}$              

$\Rightarrow A_1{r}^3=\frac{1}{6}.......\left(i\right)$               

Again, A₂ + A₄ = $\frac{7}{36}$

$A_{1}r=\frac{7}{36}-\frac{1}{6}=\frac{1}{36}........\left(ii\right)$

$\left(i\right)\&\left(ii\right)r=\sqrt[]{6}\&A_1=\frac{1}{36\sqrt[]{6}}$

$\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$             

 $T_5{=}^n{C}_4{\left(2^{\frac{1}{4}}\right)}^{n-4}.{\left({\left(\frac{1}{3}\right)}^{\frac{1}{4}}\right)}^4{=}^n{C}_4.2^{\frac{n-4}{4}}.\frac{1}{3}$

$T_6{=}^9{C}_5{\left(2^{\frac{1}{2}}\right)}^4.{\left({\left(\frac{1}{3}\right)}^{\frac{1}{4}}\right)}^5{=}^9{C}_5.2.\frac{1}{3^{5/4}}=\frac{2}{3}.\frac{1}{3^{\frac{1}{4}}}.\frac{9*8*7*6}{4*3*2*1}$