Maths M.C.Q (1 Marks)

$\sum_{k=0}^{10}\left(2^{2}+3 k\right){ }^{10} C _{ k }=\alpha .3^{10}+\beta .2^{10}$

$LHS =4 \sum_{ k =0}^{10}{ }^{10} C _{ k }+3 \sum_{ k =0}^{10} k \cdot \frac{10}{ k } \cdot{ }^{9} C _{ k -1}$

$=4.2^{10}+3.10 .2^{9}$

$=19.2^{10}=\alpha .3^{10}+\beta .2^{10}$

$\Rightarrow \alpha=0, \beta=19 \Rightarrow \alpha+\beta=19$

$=\sum_{r=1}^{15}(-1)^{r} \cdot r^{15} C_{r}+\left({ }^{14} C_{1}+{ }^{14} C_{3}+\ldots+{ }^{14} C_{11}+{ }^{14} C_{13}\right)-{ }^{14} C_{3}$

$=\sum_{r=1}^{15}(-1)^{r} 15 \cdot{ }^{14} C_{r-1}+2^{13}-14$

$=15\left(-14 C_{0}+{ }^{14} C_{1} \ldots \ldots . .-14 C_{14}\right)+2^{13}-14$

$=2^{13}-14$

$=30\left({ }^{30} C _{30}\right)+29\left({ }^{30} C _{29}\right)+\ldots \ldots+2\left({ }^{30} C _{2}\right)+1\left({ }^{30} C _{1}\right)$

$=\sum_{ r =1}^{30} r \left({ }^{30} C _{ r }\right)$

$=\sum_{ r =1}^{30} r \left(\frac{30}{ r }\right)\left({ }^{29} C _{ r -1}\right)$

$=30 \sum_{ r =1}^{30}{ }^{29} C _{ r -1}$

$=30\left({ }^{29} C _{0}+{ }^{29} C _{1}+{ }^{29} C _{2}+\ldots+{ }^{29} C _{29}\right)$

$=30\left(2^{29}\right)=15(2)^{30}= n (2)^{ m }$

$\therefore n =15, m =30$

$n + m =45$

$\sum(4(r-1)+r) \cdot{ }^{20} C_{r}$

$\sum r(r-1) \cdot \frac{20 \times 19}{r(r-1)} \cdot{ }^{18} C_{r}+r \cdot \frac{20}{r} \cdot \sum^{19} C_{r-1}$

$\Rightarrow 20 \times 19.2^{18}+20.2^{19}$

$\Rightarrow 420 \times 2^{18}$

sum of suffix is const. so summation will be ${ }^{40} \mathrm{C}_{20}$

$y=(1-x)\left(x^{3}-1\right)^{100}$

$y=\left(x^{3}-1\right)^{100}-x\left(x^{3}-1\right)^{100}$

Coff. Of $x^{256}$ in $y=-$ coff of $x^{255}$ in $\left(x^{3}-1\right)^{100}$

$={ }^{-100} \mathrm{C}_{85}(-1)^{15}={ }^{100} \mathrm{C}_{15}$

$=\frac{1}{a} \sum_{r=1}^{n}\left(1-\frac{r b}{a}\right)^{-1}$ $=\frac{1}{a} \sum_{r=1}^{n}\left\{\left(1+\frac{r b}{a}+\frac{r^{2} b^{2}}{a^{2}}\right)+(\right.$ terms to be neglected $\left.)\right\}$

$=\frac{1}{a}\left[n+\frac{n(n+1)}{2} \cdot \frac{b}{a}+\frac{n(n+1)(2 n+1)}{6} \cdot \frac{b^{2}}{a^{2}}\right]$

$=\frac{1}{a}\left[n^{3}\left(\frac{b^{2}}{3 a^{2}}\right)+\ldots\right]$

So $\gamma=\frac{\mathrm{b}^{2}}{3 \mathrm{a}^{2}}$

$\Rightarrow \frac{25^{x-1}+7}{5^{(x-1)}+1}=4$

$\Rightarrow \frac{t^{2}+7}{t+1}=4$

$\Rightarrow t=1,3=5^{x-1}$

$\Rightarrow x-1=0 \text { (one of the possible value) }$

$\Rightarrow x=1$

Let $x=10^{100}$

$\Rightarrow P=\left(1+\frac{1}{x}\right)^{x}$

$\Rightarrow P=1+(x)\left(\frac{1}{x}\right)+\frac{(x)(x-1)}{\lfloor 2} \cdot \frac{1}{x^{2}}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+\frac{(x)(x-1)(x-2)}{\lfloor 3} \cdot \frac{1}{x^{3}}+\ldots$

(upto $10^{100}+1$ terms $)$

$\Rightarrow P=1+1+\left(\frac{1}{\lfloor2}-\frac{1}{\lfloor2 x^{2}}\right)+\left(\frac{1}{\lfloor3}-\ldots\right)+\ldots \text { so on }$

$\Rightarrow P=2+\left(\text { Positive value less than } \frac{1}{\lfloor 2}+\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}+\ldots\right)$

$\Rightarrow P \in(2,3)$

Also $e=1+\frac{1}{\lfloor1}+\frac{1}{\lfloor2}+\frac{1}{\lfloor3}+\frac{1}{\lfloor4}+\ldots$

$\Rightarrow \frac{1}{\lfloor2}+\frac{1}{\lfloor 3}+\frac{1}{\lfloor 4}+\ldots=\mathrm{e}-2$

$\Rightarrow \mathrm{P}=2+($ Positive value less than $\mathrm{e}-2)$

$\Rightarrow \mathrm{P} \in(2,3)$

$\Rightarrow$ least integer value of $\mathrm{P}$ is $3$

$\frac{10 !}{\alpha ! \beta ! \gamma !} \mathrm{a}^{\alpha+\gamma} \cdot \mathrm{b}^{\beta+\gamma} \cdot 2^{\beta} \cdot 4^{\gamma}$

$\alpha+\beta+\gamma=10....(1)$

$\alpha+\gamma=7....(2)$

$\beta+\gamma=8....(3)$

$\text { (2) }+(3)-(1) \Rightarrow \gamma=5$

$\alpha=2$

$\beta=3$

$\text { so coefficients }=\frac{10 !}{2 ! 3 ! 5 !} 2^{3} \cdot 2^{10}$

$=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{2 \times 3 \times 2 \times 5 !} \times 2^{13}$

$=315 \times 2^{16} \Rightarrow \mathrm{k}=315$

${ }^{60} C _{ r }\left(3^{1 / 4}\right)^{60- r } \cdot\left(5^{1 / 8}\right)^{ r }$

${ }^{60} C _{ r }(3)^{\frac{60- r }{4}} .5^{\frac{ r }{8}}$

For rational terms.

$\frac{r}{8}=k ; \quad 0 \leq r \leq 60$

$0 \leq 8 k \leq 60$

$0 \leq k \leq \frac{60}{8}$

$0 \leq k \leq 7.5$

$k =0,1,2,3,4,5,6,7$

$\frac{60-8 k }{4}$ is always divisible by $4$ for all value of $k$

Total rational terms $=8$; Total terms $=61$; irrational terms $=53$ ;$n -1=53-1=52$

$52$ is divisible by $26$

$=17 k _{2}+2^{3762}($ as $2023=17 \times 17 \times 9)$

$=17 k _{2}+4 \times 16^{940}$

$=17 k _{2}+4 \times(17-1)^{940}$

$=17 k _{2}+4\left(17 k _{3}+1\right)$

$=17 k +4 \Rightarrow$ remainder $=4$

$\therefore(2020+ x )^{2022}=(2020+4 k +3)^{2022}$

$=(4(505+ k )+3)^{2022}$

$=(4 \lambda+3)^{2022}=\left(16 \lambda^{2}+24 \lambda+9\right)^{1011}$

$=\left(8\left(2 \lambda^{2}+3 \lambda+1\right)+1\right)^{1011}$

$=(8 p +1)^{1011}$

$\therefore$ Remainder when divided by $8=1$

$=-39+18 . \mathrm{I}$

$=(54-39)+18(\mathrm{I}-3)$

$\Rightarrow \quad=15+18 \mathrm{I}_{1}$

$\Rightarrow$ Remainder$=15.$

$\left(1-x+x^{3}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2} \ldots \ldots .+a_{3 n} x^{3 n}$

$\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}=\operatorname{Sum}$ of $a_{0}+a_{2}+a_{4} \ldots \ldots . .$

$\sum_{j=0}^{\left[\frac{3 n -1}{2}\right]} a_{2 j+1}  =$ Sum of $a_{1}+a_{3}+a_{5} \ldots \ldots$

put $x=1$

$1= a _{0}+ a _{1}+ a _{2}+ a _{3} \ldots \ldots \ldots+ a _{3 n } \quad \ldots \ldots$

Put $x=-1$

$1= a _{0}- a _{1}+ a _{2}- a _{3} \ldots \ldots \ldots+(-1)^{3 n } a _{3 n } \ldots \ldots$

Solving (A) and (B)

$a_{0}+a_{2}+a_{4} \ldots . .=1$

$a_{1}+a_{3}+a_{5} \ldots \ldots .=0$

$\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right]} a_{2 j+1}=1$

Now $\frac{{ }^{n} C_{r-1}}{{ }^{n} C_{r}}=\frac{2}{5}$

$\Rightarrow 7 r=2 n+2$

$\frac{{ }^{n} C_{r}}{{ }^{n} C_{r+1}}=\frac{5}{12}$

$\Rightarrow 17 r =5 n -12$

On solving (1)$\&(2)$

$\Rightarrow n =118$

$T _{ r +1}={ }^{9} C _{ r }\left(\frac{3}{2}\right)^{9- r }\left(-\frac{1}{3}\right)^{ r } x ^{18-3 r }$

For independent of x

$18-3 r=0, r=6$

$\therefore \quad T _{7}={ }^{9} C _{6}\left(\frac{3}{2}\right)^{3}\left(-\frac{1}{3}\right)^{6}=\frac{21}{54}= k$

$\therefore \quad 18 k =\frac{21}{54} \times 18=7$

Clearly r should be a multiple of 8 .

$\because$ there are exactly 33 integral terms

Possible values of $r$ can be

$0,8,16, \ldots \ldots \ldots, 32 \times 8$

$\therefore$ least value of $n =256$

$T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r}\left(\frac{-k}{x^{2}}\right)^{r}$

$T_{r+1}={ }^{10} C_{r} \cdot x^{\frac{10-r}{2}} \cdot(-k)^{r} \cdot x^{-2 r}$

$T_{r+1}={ }^{10} C_{r} x^{\frac{10-5 r}{2}}(-k)^{r}$

Constant term $: \frac{10-5 r}{2}=0 \Rightarrow r=2$

$T_{3}={ }^{10} C_{2} \cdot(-k)^{2}=405$

$k^{2}=\frac{405}{45}=9$

$k=\pm 3 \Rightarrow|k|=3$

$=(1+x)^{6}\left(1+x^{2}\right)^{6}$

$=\sum_{ r =0}^{6}{ }^{6} C _{ r } x ^{ r } \sum_{ r =0}^{6}{ }^{6} C _{ t } x ^{2 t }$

$=\sum_{ r =0}^{6} \sum_{ t =0}^{6}{ }^{6} C _{ r }^{6} C _{ t } x ^{ r +2 t }$

For coefficient of $x^{4} \Rightarrow r+2 t=4$

$\begin{array}{|c|c|} \hline r & t \\ \hline 0 & 2 \\ \hline 2 & 1 \\ \hline 4 & 0 \\ \hline \end{array}$

Coefficient of $x^{4}$ $={ }^{6} C _{0}{ }^{6} C _{2}+{ }^{6} C _{2}{ }^{6} C _{1}+{ }^{6} C _{4}{ }^{6} C _{0}$

$=120$

$^{10} \mathrm{C}_{7}+^{9} \mathrm{C}_{6}+^{5} \mathrm{C}_{5}+\ldots+^{4} \mathrm{C}_{1}+^{3} \mathrm{C}_{0}$

$\underbrace{^{4} C_{0}+^{4} C_{1}}_{^{5} C_{1}}+^{5} C_{2}+\ldots+^{10} C_{7}=^{11} C_{7}=330$

$\alpha=-96 \;and\; \beta=36$

$\therefore \alpha-\beta=-132$

$=^{16} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{16-2 \mathrm{r}} \times \frac{1}{(\cos \theta)^{16-\mathrm{r}}(\sin \theta)^{\mathrm{r}}}$

For independent of $x ; 16-2 r=0 \Rightarrow r=8$

$\Rightarrow \mathrm{T}_{9}=^{16}\mathrm{C}_{8} \frac{1}{\cos ^{8} \theta \sin ^{8} \theta}$

$=^{16} \mathrm{C}_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}$

for $\theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right] \ell_{1}$ is least for $\theta_{1}=\frac{\pi}{4}$

for $\theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right] \ell_{2}$ is least for $\theta_{2}=\frac{\pi}{8}$

$\frac{\ell_{2}}{\ell_{1}}=\frac{\left(\sin 2 \theta_{1}\right)^{8}}{\left(\sin 2 \theta_{2}\right)^{8}}=(\sqrt{2})^{8}=\frac{16}{1}$

$=^{10} \mathrm{C}_{0}+^{10} \mathrm{C}_{1} \mathrm{x}(1+\mathrm{x})+^{10} \mathrm{C}_{2} \mathrm{x}^{2}(1+\mathrm{x})^{2}$

$+^{10} \mathrm{C}_{3} \mathrm{x}^{3}(1+\mathrm{x})^{3}+^{10} \mathrm{C}_{4} \mathrm{x}^{4}(1+\mathrm{x})^{4}+\ldots \ldots$

Coeff. of $\mathrm{x}^{4}=^{10} \mathrm{C}_{2}+^{10} \mathrm{C}_{3} \times^{3} \mathrm{C}_{1}+^{10} \mathrm{C}_{4}=615$

$\mathrm{r}+1 \mathrm{th}$ term of $\left(\alpha \mathrm{x}^{\frac{1}{9}}+\beta \mathrm{x}^{-\frac{1}{6}}\right)^{10}$

$t_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}} \alpha^{10-\mathrm{r}}(\mathrm{x})^{\frac{10-\mathrm{r}}{9}} \cdot \beta^{\mathrm{r}} \mathrm{x}^{-\frac{\mathrm{r}}{6}}$

$={ }^{10} \mathrm{C}_{\mathrm{r}} \alpha^{10-\mathrm{r}} \beta^{\mathrm{r}}(\mathrm{x})^{\frac{10-\mathrm{r}}{9}-\frac{\mathrm{r}}{6}}$

If $t_{\mathrm{r}}+1$ is independent of $\mathrm{x}$

$\frac{10-r}{9}-\frac{r}{6}=0 \Rightarrow r=4$

maximum value of $\mathrm{t}_{5}$ is $10 \mathrm{K}$ (given)

$\Rightarrow{ }^{10} \mathrm{C}_{4} \alpha^{6} \beta^{4}$ is maximum

By $\mathrm{AM} \geq \mathrm{GM}$ (for positive numbers)

$\frac{\frac{\alpha^{3}}{2}+\frac{\alpha^{3}}{2}+\frac{\beta^{2}}{2}+\frac{\beta^{2}}{2}}{4} \geq\left(\frac{\alpha^{6} \beta^{4}}{16}\right)^{\frac{1}{4}}$

$\Rightarrow \alpha^{6} \beta^{4} \leq 16$

So, $10 \mathrm{K}={ }^{10} \mathrm{C}_{4} 16$

$\Rightarrow \mathrm{K}=336$

$N _{ C _{ r -1}}: N _{ C _{ r }}: N _{ C _{ r +1}}=5: 10: 14$

$\Rightarrow \frac{ N _{ C _{r}}}{ N _{ C _{ r -1}}}=\frac{ N +1- r }{ r }=2$

$\frac{N_{C_{r+1}}}{N_{C_{r}}}=\frac{N-r}{r+1}=\frac{7}{5}$

$\Rightarrow \quad r=4, N=11$

$\Rightarrow \quad(1+x)^{11}$

Largest coefficient $={ }^{11} C _{6}=462$

$=22 C _{ r } x$

$\because 22 C _{3}=22 C _{19}=1540$

$\therefore r =3$ or 19

$Z 2 m - mr -2 r =1$

$m =\frac{2 r +1}{22-5}$

$r =3, \quad m =\frac{7}{19} \notin N$

$r =19, m =\frac{38+1}{22-19}=\frac{39}{3}=13$

$m =13$

$=a_{0}+a_{1} x_{+} a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+\ldots+a_{4 n} x^{4 n}$

$\mathrm{So}$

$a_{0}+a_{1}+a_{2}+\ldots+a_{4 n}=2 n+1$

$a_{0}-a_{1}+a_{2}-a_{3} \ldots+a_{4 n}=2 n+1$

$\Rightarrow a_{0}+a_{2}+a_{4}+\ldots+a_{4 n}=2 n+1$

$\Rightarrow 2 \mathrm{n}+1=61 \quad \Rightarrow \mathrm{n}=30$

$\mathrm{S}=101^{25} \mathrm{C}_{25}+97^{25} \mathrm{C}_{1}+\ldots \ldots \ldots .+1^{25} \mathrm{C}_{25}$

$2 \mathrm{S}=(102)\left(2^{25}\right)$

$\mathrm{S}=51\left(2^{25}\right)$

replace x by $\frac{2}{x}$ in above identity :-

$\frac{2^{10}\left(2 x ^{2}+3 x +4\right)^{10}}{ x ^{20}}=\sum_{ r =0}^{20} \frac{ a _{ r } 2^{ r }}{ x ^{ r }}$

$\Rightarrow 2^{10} \sum_{ r =0}^{20} a _{ r } x ^{ r }=\sum_{ r =0}^{20} a _{ r } 2^{ r } x ^{(20- r )}($ from

now, comparing coefficient of $x^{7}$ from both sides

(take $r=7$ in L.H.S. $\& r=13$ in $R .$ H.S. $)$

$2^{10} a_{7}=a_{13} 2^{13} \Rightarrow \frac{a_{7}}{a_{13}}=2^{3}=8$

$=\frac{(49)^{126}-1}{49-1}=(49^{63}+1) \frac{\left(49^{63}-1\right)}{(48)}$

So greatest value of $\mathrm{k}=63$

$=\left\{\frac{(1+8)^{100}}{8}\right\}$

$=\left\{\frac{1+{ }^{100} C _{ T } 8+{ }^{100} C _{2} \cdot 8^{2}+\ldots+{ }^{100} C _{100} 8^{100}}{8}\right\}$

$=\left\{\frac{1+8 m }{8}\right\}$

$=\frac{1}{8}$

$\left(1-t^{18}-3 t^{6}+3 t^{12}\right)(1-t)^{-3}$

$\Rightarrow$ coefficient of $t^{4}$ in $(1-t)^{-3}$ is

$^{3+1-1} C_{4}=^{6} C_{2}=15$

third term say $\mathrm{T}_{3}=^{5} \mathrm{C}_{2}\left(\mathrm{x}^{\log _{2} \mathrm{x}}\right)^{2}=2560$

$\Rightarrow\left(x^{\log x}\right)^{2}=256$

taking lograthium to the base $2$ on both sides

$\Rightarrow 2\left(\log _{2} x\right)^{2}=8 \Rightarrow\left(\log _{2} x\right)=\pm 2$

$\Rightarrow x=4, \frac{1}{4}$

Here $x=\frac{1}{4}$

Consider constant term

$^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{x}})^{10-\mathrm{r}}\left(\frac{\lambda}{\mathrm{x}^{2}}\right)^{r}$

$\frac{10-r}{2}-2 r=0$

$10-5 r=0$

$r=2$

$\Rightarrow^{10} \mathrm{C}_{2} \times \lambda^{2}=720$

$ \Rightarrow \lambda=4$

$t_{4+1}=^{8} C_{4}\left(\frac{x^{3}}{3}\right)^{4}\left(\frac{3}{x}\right)^{4}=5670$

$=^{8} \mathrm{C}_{4} \cdot \mathrm{x}^{8}=5670$

$=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2} \mathrm{x}^{8}=5670$

$=x^{8}=\frac{567}{7}=81$

$=x^{8}-81=0$

$\Rightarrow$ Real value of $x=\pm \sqrt{3}$

$a_{0}=\left(10^{50}\right)(2)$

$a_{2}=^{50} C_{2}(10)^{48}(2)$

$\frac{a_{2}}{a_{0}}=\frac{^{50} C_{2}(10)^{48}(2)}{10^{52}(2)}=12.25$

$=\frac{2^{2} 2^{-2} 3^{-4 / 3}}{2^{4} 2^{(4 / 3)-6} 3^{-2}}=3^{2 / 3} \cdot 2^{8 / 3}$

$=4 \cdot(36)^{1 / 3}$

for rational term, $\mathrm{r}=0,10,20,30,40,50,60$

$\Rightarrow$ number of rational terms $=7$

$\therefore $ number of irrational terms $=54$

$x^{\frac{1}{4}- \frac{3}{2}(1+\operatorname{log} x)}=10$

$\Rightarrow \frac{1}{4}-\frac{3}{2}\left(1+\log _{10} x\right) \cdot \log _{10} x=1$

$\Rightarrow 6 t^{2}+5 t+4=0, t=\log _{10} x$

$D<0$

So no real solution

All options are incorrect

$20 \times 8^{7}=\frac{160}{x^{3}} \cdot x^{3000} x$

$8^{6}=x^{\log _{2} x}-3$

$2^{18}=x^{\log _{2} x-3}$

$\Rightarrow 18=\left(\log _{2} x-3\right)\left(\log _{2} x\right)$

Let $\log _{2} x=t$

$\Rightarrow t^{2}-3 t-18=0$

$\Rightarrow(t-6)(t+3)=0$

$\Rightarrow \mathrm{t}=6,-3$

$\log _{2} x=6$

$ \Rightarrow x=2^{6}=8^{2}$

$\log _{2} x=-3$

$\Rightarrow x=2^{-3}=8^{-1}$

$\Rightarrow \frac{r}{n-r+1}=\frac{2}{15}$

$\Rightarrow 15 r=2 n-2 r+2$

$\Rightarrow 17 r=2 n+2.........(1)$

also given $\frac{^{n} C_{r}}{^{n} C_{r+1}}=\frac{15}{70}$

$ \Rightarrow \frac{r+1}{n-r}=\frac{3}{14}$

$\Rightarrow 3 n-3 r=14 r+14.........(2)$

$\Rightarrow 17 \mathrm{r}=3 \mathrm{n}-14$

Solving $(1)$ and $(2)$ $n=16, r=2$

Average of coefficient $=\frac{^{16} \mathrm{C}_{1}+^{16} \mathrm{C}_{2}+^{16} \mathrm{C}_{3}}{3}$

$=\frac{16+120+560}{3}$

$=232$

$\Rightarrow^{15} \mathrm{C}_{2} \times 9-45 \mathrm{a}+\mathrm{b}=0.........(1)$

Coefficient of $x^{3}=-27 \times^{15} \mathrm{C}_{3}+9 \mathrm{a} \times^{15} \mathrm{C}_{2}-3 \mathrm{b} \times^{15} \mathrm{C}_{1}=0$

$\Rightarrow-273+21 a-b=0.........(2)$

$(1)+(2)$ give

$-24 a+672=0$

$\Rightarrow \quad a=28, b=315$

$2n - 5r = 1\Rightarrow 2n = 5r + 1$ for $r = 15, n = 38$ smallest value of $n$ is $38$.

$\left(1-x^{2}\right)\left(1-x^{3}\right)^{9}$

$^{9} \mathrm{C}_{6}=84$

Term independent of $x$ will be $\frac{1}{60} \times$

independent of $x$ in $\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}-\frac{1}{8} \times$

Termof $x^{-8}$ in $\left(2 x^{2} \frac{3}{x^{3}}\right)^{6}$

$\mathrm{T}_{r+1}$ in $\left(2 \mathrm{x}^{2}-\frac{3}{\mathrm{x}^{2}}\right)^{6}$ will be

$\mathrm{T}_{r+1}=^{6} \mathrm{C}_{r}\left(2 \mathrm{x}^{2}\right)^{6-\mathrm{f}}\left(-\frac{3}{\mathrm{x}^{2}}\right)^{r}$

$=^{6} \mathrm{C}_{r} 2^{6-5}(-1)^{r} \times 3^{r} \times \mathrm{x}^{12-2 r-2 r}$

Case $I:$ For term independent of $x, 12-4 r=0$ $\Rightarrow r=3$

$T_{4}=^{6} C_{3} \times 2^{3} \times 3^{3} x^{6}=-20 \times 2^{3} \times 3^{3}$

case $II :$ For term of $x^{-8}$

${12-4 r=-8}$

${4 r=20 \Rightarrow r=5}$

$T_{6}=^{6} C_{5} \cdot 2^{1}(-1) \cdot 3^{5} \cdot x^{-8}$

Required Answer $=\frac{1}{60} \times(-20) 2^{3} \times 3^{3}-\frac{1}{81} \times 6 \times 2 \times(-1) \times 3^{5}$

$=-72+36=-36$

Hence the correct answer is option $(B).$

Now $\frac{{^{20}{C_{I - 1}}}}{{^{20}{C_1} + {\,^{20}}{C_{I - 1}}}} = \frac{{^{20}{C_{I - 1}}}}{{^{20}{C_1}}} = \frac{1}{{21}}$

Let given sum be $S$, so

$S = \sum\limits_{I = 1}^{20} {\frac{{{{\left( i \right)}^3}}}{{{{21}^3}}}}  = \frac{1}{{{{(21)}^3}}}{\left( {\frac{{20.21}}{2}} \right)^2} = \frac{{100}}{{21}}$

Given $S = \frac{k}{{21}} \Rightarrow k = 100$

$ = \sum\limits_{r = 1}^{25} {\frac{{\left| {50} \right.}}{{\left| {r\left| {25 - r\left| {25} \right.} \right.} \right.}}} $

$ = \frac{{\left| {50} \right.}}{{\left| {25} \right.}}\sum\limits_{r = 1}^{25} {\frac{1}{{\left| {r\left| {25 - r} \right.} \right.}}} $

$ = \frac{{\left| {50} \right.}}{{\left| {25\left| {25} \right.} \right.}}\sum\limits_{r = 1}^{25} {{\,^{25}}{C_r} = {\,^{50}}{C_{25}}} \left( {{2^{25}} - 1} \right)$

$ = \sum\limits_{r = 0}^{20} {(3r + 2){\,^{20}}} {C_r}$

$ = 3\sum\limits_{r = 0}^{20} r { \cdot ^{20}}{C_r} + 2\sum\limits_{r = 0}^{20} {{\,^{20}}} {C_r}$

$ = 3\sum\limits_{r = 0}^{20} {r\left( {\frac{{20}}{r}} \right)} {\,^{19}}{C_{r - 1}} + {2.2^{20}}$

$=60.2^{19}+2.2^{20}=2^{25}$

Differential equation w.r.t. $x$

$20(1+x)^{19}=$

$^{20} C_{1} \cdot 1+2.^{20} C_{2} x+\ldots \ldots+20^{20} C_{20} x^{19}.........(ii)$

Multiply equation $(2)$ by $x$

$20x \times(1+x)^{19}=$

$^{20} \mathrm{C}_{1} \mathrm{x}+2 \cdot^{20} \mathrm{C}_{2} \mathrm{x}^{2}+\ldots \ldots+20^{20} \mathrm{C}_{20} \mathrm{x}^{20}.........(iii)$

Differential equation $(3)$ w.r.t. $x$

$20\left[(1+x)^{19}+19 x(1+x)^{18}\right]=$

$20 \times ({2^{19}} + {19.2^{18}}) = {1^2}{\,^{20}}{C_1} + {2^2}{\,^{20}}{C_2}+ ...... + {({20^2})^{20}}{C_{20}}$

Put $x=1$ in equation $(iv)$

$20\left(2^{10}+19.2^{18}\right)=1^{200} \mathrm{C}_{1}+2^{2} \mathrm{C}_{2}+\ldots \ldots+\left(20^{2}\right)^{20} \mathrm{C}_{20}$

$=20 \times 2^{18}(2+19)=20 \times 21 \times 2^{18}$

$=420 \times 2^{18}$

$A=420, \beta=18$

Hence correct Option is $(A)$.