Maths M.C.Q (1 Marks)

$ ={ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}(\mathrm{r})^{\left(6-\frac{2 \mathrm{r}-2 \mathrm{r}}{3}\right)}$

for coefficient of $x^{2 / 3}$, put $6-\frac{2 r}{3}-\frac{2 r}{5}=\frac{2}{3}$

$\Rightarrow \mathrm{r}=5$

$\therefore$ Coefficient of $\mathrm{x}^{2 / 3}$ is $={ }^9 \mathrm{C}_3\left(\frac{1}{5}\right)^5$

For coefficient of $x^{-2 / 5}$, put $6-\frac{2 r}{3}-\frac{2 r}{5}=-\frac{2}{5}$

$\Rightarrow \mathrm{r}=6$

Coefficient of $\mathrm{x}^{-2 / 5}$ is ${ }^9 \mathrm{C}_6\left(\frac{1}{2}\right)^6$

$\text { Sum }={ }^9 \mathrm{C}_5\left(\frac{1}{2}\right)^5+{ }^9 \mathrm{C}_6\left(\frac{1}{2}\right)^6=\frac{21}{4}$

$ =\frac{1}{12} \sum_{\mathrm{r}=1}^9{ }^{12} \mathrm{C}_{\mathrm{r}+1}$

$ =\frac{1}{12}\left[2^{12}-26\right]=\frac{2035}{6} $

$\therefore \mathrm{m}+\mathrm{n}=2041$

$ (x+2)^2+\ldots \ldots . .+(x+2)^{n-1} $

$\sum \alpha_r=4^{n-1}+4^{n-2} \times 3+4^{n-3} \times 3^2 \ldots \ldots+3^{n-1} $

$=4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}\right)^2 \ldots . .+\left(\frac{3}{4}\right)^{n-1}\right]$

$=4^{\mathrm{n}-1} \times \frac{1-\left(\frac{3}{4}\right)^{\mathrm{n}}}{1-\frac{3}{4}} $

$=4^{\mathrm{n}}-3^{\mathrm{n}}=\beta^{\mathrm{n}}-\gamma^{\mathrm{n}} $

$ \beta=4, \gamma=3 $

$\beta^2+\gamma^2=16+9=25$

coeff. of  $x^{36}$ in $(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8$

General term

${ }^6 \mathrm{C}_{\mathrm{r}_1}{ }^7 \mathrm{C}_{\mathrm{r}_2}{ }^8 \mathrm{C}_{\mathrm{r}_3}(-1)^{\mathrm{I}_3} \mathrm{x}^{\mathrm{I}_1+2 \mathrm{r}_2+3 \mathrm{r}_3}$

$\mathrm{r}_1+2 \mathrm{r}_2+3 \mathrm{r}_3=36$

Case-$I$ :    $r_1+2 r_2=12\left(\right.$ Taking $\left.r_3=8\right)$

case-$II$       $r_1+2 r_2=15\left(\right.$ Taking $\left.r_3=7\right)$

case-$III$      $r_1+2 r_2=18\left(\right.$ Taking $\left.r_3=6\right)$

$\begin{aligned} & \text { Coeff. }=7+(15 \times 21)+(15 \times 35)+(35) \\ & -(6 \times 8)-(20 \times 7 \times 8)-(6 \times 21 \times 8)+(15 \times 28) \\ & +(7 \times 28)=-678=\alpha \\ & |\alpha|=678\end{aligned}$

$ \frac{\mathrm{e}^{(1+\mathrm{x})}}{1+\mathrm{x}}=\frac{1}{1+\mathrm{x}}+1+\frac{(1+\mathrm{x})}{2 !}+\frac{(1+\mathrm{x})^2}{3 !}+\frac{(1+\mathrm{x})^2}{4 !} $

$ \text { coef } \mathrm{x}^2 \text { in RHS : } 1+\frac{{ }^2 \mathrm{C}_2}{3}+\frac{{ }^3 \mathrm{C}_2}{4}+\ldots=\mathrm{a} $

$ \text { coeff. } \mathrm{x}^2 \text { in L.H.S. } $

$ \mathrm{e}\left(1+\mathrm{x}+\frac{\mathrm{x}^2}{2 !}\right) \ldots .\left(1-\mathrm{x}+\frac{\mathrm{x}^2}{2 !} \ldots \ldots .\right) $

$ \text { is } \mathrm{e}-\mathrm{e}+\frac{\mathrm{e}}{2 !}=\mathrm{a} $

$ \mathrm{b}=1+\frac{2}{1 !}+\frac{2^2}{2 !}+\frac{2^3}{3 !}+\ldots \ldots=\mathrm{e}^2 $

$ \frac{2 \mathrm{~b}}{\mathrm{a}^2}=8$

$ \alpha=4 \sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r}^2 \cdot \frac{\mathrm{n}}{\mathrm{r}} \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}+2 \sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r} \cdot \frac{\mathrm{n}}{\mathrm{r}} \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-}+\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} $

$ +4 n \sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}+2 \mathrm{n} \sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}+\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} $

$ \alpha=4 \mathrm{n}(\mathrm{n}-1) \cdot 2^{\mathrm{n}-2}+4 \mathrm{n} \cdot 2^{\mathrm{n}-1}+2 \mathrm{n} \cdot 2^{\mathrm{n}-1}+2^{\mathrm{n}} $

$ \alpha=2^{\mathrm{n}-2}[4 \mathrm{n}(\mathrm{n}-1)+8 \mathrm{n}+4 \mathrm{n}+4] $

$ \alpha=2^{\mathrm{n}-2}\left[4 \mathrm{n}^2+8 \mathrm{n}+4\right] $

$ \alpha=2 \mathrm{n}(\mathrm{n}+1)^2 $

$ \beta=\sum_{\mathrm{r}=0}^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1}+\frac{1}{\mathrm{n}+1} $

$ =\sum_{\mathrm{r}=0}^{\mathrm{n}} \frac{{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+1} $

$ =\frac{1}{\mathrm{n}+1}\left(1+{ }^{\mathrm{n}+1} \mathrm{C}_1+\ldots .+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1}\right) $

$ =\frac{2^{\mathrm{n}+1}}{\mathrm{n}+1} $

$ \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n+1)^2}{2^{n+1}} \cdot(n+1)=(n+1)^3 $

$ 140<(\mathrm{n}+1)^3<281 $

$ \mathrm{n}=4 \Rightarrow(\mathrm{n}+1)^3=125 $

$ \mathrm{n}=5 \Rightarrow(\mathrm{n}+1)^3=216 $

$ \mathrm{n}=6 \Rightarrow(\mathrm{n}+1)^3=343 $

$ \therefore \mathrm{n}=5 $

$ x^{54}(1+x)^{46}$

Coeff. of $\mathrm{x}^{70}:{ }^{98} \mathrm{C}_{68}+{ }^{97} \mathrm{C}_{67}+{ }^{96} \mathrm{C}_{66}+\ldots \ldots \ldots$.

$ { }^{47} \mathrm{C}_{17}+{ }^{46} \mathrm{C}_{16} $

$ ={ }^{46} \mathrm{C}_{30}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30} $

$ =\left({ }^{46} \mathrm{C}_{31}+{ }^{46} \mathrm{C}_{30}\right)+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31} $

$ ={ }^{47} \mathrm{C}_{31}+{ }^{47} \mathrm{C}_{30}+\ldots \ldots \ldots . .{ }^{98} \mathrm{C}_{30}-{ }^{46} \mathrm{C}_{31} $

$ \ldots \ldots $

$ ={ }^{99} \mathrm{C}_{31}-{ }^{46} \mathrm{C}_{31}={ }^{99} \mathrm{C}_{\mathrm{p}}-{ }^{46} \mathrm{C}_{\mathrm{q}}$

Possible values of $(p+q)$ are $62,83,99,46$

$\Rightarrow p+q=83$

$ x^4 \cdot\left(\frac{1-x^6}{1-x}\right)^4 $

$ x^4 \cdot\left(1-x^6\right)^4 \cdot(1-x)^{-4} $

$ x^4\left[1-4 x^6+6 x^{12} \ldots .\right]\left[(1-x)^{-4}\right] $

$ \left(x^4-4 x^{10}+6 x^{16} \ldots\right)(1-x)^{-4} $

$ \left(x^4-4 x^{10}+6 x^{16}\right)\left(1+{ }^{15} C_{12} x^{12}+{ }^9 C_6 x^6 \ldots\right) $

$ \left({ }^{15} C_{12}-4 \cdot{ }^9 C_6+6\right) x^{16} $

$ \left({ }^{15} C_3-4 \cdot{ }^9 C_6+6\right) $

$ =35 \times 13-6 \times 8 \times 7+6 $

$ =455-336+6 $

$ =125$

$ 64^{32^{32}}=64^t=8^{2 t}=(9-1)^{2 \mathrm{t}} $

$ =9 \mathrm{k}+1$

Hence remainder $=1$

$ =(21 \times 20+8)^{2024} $

$ =21 \mathrm{~m}+8^{2024} $

$ \text { Now } 8^{2024}=\left(8^2\right)^{1012} $

$ =(64)^{1012} $

$ =(63+1)^{1012} $

$ =(21 \times 3+1)^{1012} $

$ =21 \mathrm{n}+1 $

$ \Rightarrow \text { Remainder is } 1 .$

$ =\frac{1}{n+1} \sum_{k=0}^n{ }^{n+1} C_{k+1} \cdot{ }^n C_{n-k} $

$ =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1} $

$ =\sum_{k=0}^{n-1}{ }^n C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} $

$ \frac{1}{n+1} \sum_{k=0}^{n-1} C_{n-k} \cdot{ }^{n+1} C_{k+2} $

$ =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2}$

$\alpha=\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+1}$

$ \beta=\sum_{k=0}^{n-1} C_k \cdot \frac{{ }^n C_{k+1}}{k+2} \frac{n+1}{n+1} $

$ \frac{1}{n+1} \sum_{k=0}^{n-1} C_{n-k} \cdot{ }^{n+1} C_{k+2}$

$ =\frac{1}{n+1} \cdot{ }^{2 n+1} C_{n+2} $

$ \frac{\beta}{\alpha}=\frac{{ }^{2 n+1} C_{n+2}}{{ }^{2 n+1} C_{n+1}}=\frac{2 n+1-(n+2)+1}{n+2} $

$ \frac{\beta}{\alpha}=\frac{n}{n+2}=\frac{5}{6} $

$n=10$

$T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r}$

$\therefore 15-5 r =5$

$\therefore r =2$

$T_3=10\left(\frac{8}{9}\right) x^5$

So, coefficient is $\frac{80}{9}$

$22+7+2=31$

Sum of coefficients of first three terms

${ }^n C_0-{ }^n C_1 \cdot 3+{ }^n C_2 3^2=376$

$\Rightarrow 3 n^2-5 n-250=0$

$\Rightarrow(n-10)(3 n+25)=0$

$\Rightarrow n =10$

Now general term ${ }^{10} C _{ r } x ^{10- r }\left(\frac{-3}{ x ^2}\right)^{ r }$

$={ }^{10} C _{ r } x ^{10- r }(-3)^{ r } \cdot x ^{-2 r }$

$={ }^{10} C _{ r }(-3)^{ r } \cdot x ^{10-3 r }$

Coefficient of $x^4 \Rightarrow 10-3 r=4$

$\Rightarrow r =2$

${ }^{10} C _2(-3)^2=405$

For constant term,

$n _1+2 n _3=7 n _2$

$n _1+ n _2+ n _3=5$

Only possibility $n _1=1, n _2=1, n _3=3$

$\Rightarrow$ constant term $=1080$

$\because$ Both are equal

$\therefore \frac{11}{C_6} \cdot \frac{\alpha^5}{\beta^6}=-\frac{11}{C_5} \cdot \frac{\alpha^6}{\beta^5}$

$\Rightarrow \frac{1}{\beta}=-\alpha$

$\Rightarrow \alpha \beta=-1$

$\Rightarrow(\alpha \beta)^2=1$

$\Rightarrow \frac{\frac{n !}{(r-1) !(n-r+1) !}}{\frac{n !(2)}{r !(n-r) !}}=\frac{2}{5}$

$\Rightarrow \frac{r}{n-r+1}=\frac{4}{5} \Rightarrow 5 r=4 n-4 r+4$

$\Rightarrow \frac{{ }^n C_r(2)^r}{{ }^n C_{r+1}(2)^{r+1}}=\frac{5}{8}$

$\Rightarrow \frac{\frac{n !}{r !(n-r) !}}{\frac{n !}{(r+1) !(n-r-1) !}}=\frac{5}{4} \Rightarrow \frac{r+1}{n-r}=\frac{5}{4}$

$\Rightarrow 4 r+4=5 n-5 r \Rightarrow 5 n-4=9 r \ldots$

From (1) and (2)

$\Rightarrow 4 n +4=5 n -4 \Rightarrow n =8$

$(1)$ $\Rightarrow r=4$

so, coefficient of middle term is

${ }^8 C_4 2^4=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120$

$T_{r+1}={ }^{15} C_r\left(a x^3\right)^{15-r}\left(\frac{1}{b x^{1 / 3}}\right)^r$

$45-3 r-\frac{r}{3}=15$

$30=\frac{10 r}{3}$

$r=9$

$\text { Coefficient of } x^{15}={ }^{15} C_9 a^6 b^{-9}$

$\text { Coefficient of } x^{-15} \text { in }\left(a x^{1 / 3}-\frac{1}{b x^3}\right)^{15}$

$T_{r+1}={ }^{15} C_r\left(a x^{1 / 3}\right)^{15-r}\left(-\frac{1}{b x^3}\right)^r$

$5-\frac{r}{3}-3 r=-15$

$\frac{10 r}{3}=20$

$r =6$

Coefficient $={ }^{15} C _6 a ^9 \times b ^{-6}$

$\frac{ a ^9}{ b ^6}=\frac{ a ^6}{ b ^9}$

$a ^3 b ^3=1 \Rightarrow ab =1$

$={ }^{30} C _{ r } \cdot 2^{ r } \cdot x ^{\frac{60-11 r }{3}}$

$\frac{60-11 r }{3} < 0 \Rightarrow 11 r > 60 \Rightarrow r >\frac{60}{11} \Rightarrow r =6$$T _7={ }^{30} C _6 \cdot 2^6 x ^{-2}$

We have also observed $\beta={ }^{30} C _6(2)^6$ is a natural number.

$\therefore \alpha=2$

Now, $T _{ r +1}={ }^9 C _{ r } \cdot\left(\frac{4 x }{5}\right)^{9- r }\left(\frac{5}{2 x ^2}\right)^{ r }$ $={ }^9 C _{ r } \cdot\left(\frac{4}{5}\right)^{9- r }\left(\frac{5}{2}\right)^{ r } \cdot x ^{9-3 r }$

Coefficient of $x^{-6}$ i.e. $9-3 r=-6 \Rightarrow r=5$

So, Coefficient of $x^{-6}={ }^9 C _5\left(\frac{4}{5}\right)^4 \cdot\left(\frac{5}{2}\right)^0=5040$

$T_{ r +1}={ }^9 C_{ r } \frac{\left(x^{5 / 2}\right)^{9- r }}{2^{9- r }}\left(\frac{-4}{x^{\ell}}\right)^{ r }$

$=(-1)^{ r } \frac{{ }^9 C_{ r }}{2^{9- I }} 4^{ r } x ^{\frac{45}{2}-\frac{ r }{2} Ir }$

$=45-5 r -21 r =0$

$r =\frac{45}{5+21}..........(1)$

Now, according to the question, $(-1)^{ T } \frac{{ }^9 C _{ r }}{2^{9- I }} 4^{ T }=-84$

$=(-1)^{ I }{ }^9 C _{ r } 2^{3 T -9}=21 \times 4$

Only natural value of $r$ possible if $3 r-9=0$ $r =3$ and ${ }^9 C _3=84$

$\therefore 1=5$ from equation $(1)$

Now, coefficient of $x^{-31}=x^{\frac{45}{2}-\frac{5 r }{2}}$-ir at $l=5$, gives

$r=5$

$\therefore{ }^9 c_5(-1) \frac{4^5}{2^4}=2^\alpha \times \beta$

$=-63 \times 2^7$

$\Rightarrow \alpha=7, \beta=-63$

$\therefore \quad \text { value of }|\alpha \ell-\beta|=98$

${ }^{ m } C _1,{ }^{ m } C _2,{ }^{ m } C _3$ are in $A.P.$

$2.$ ${ }^{ m } C _2={ }^{ m } C _1+{ }^{ m } C _3$

Solving for $m$, we get

$m =$ $2$(rejected), $7$

Put in equation $(1)$

$21 .\left(10-3^x\right) \frac{3^x}{9}=21$

$3^x=3^0, 3^2$

$x =0,2$

$={ }^{22} C _{ r } \cdot x ^{\frac{44}{3}-\frac{2 r }{3}-3 r }(\alpha)^{ r }$

$\frac{44}{3}=\frac{11 r }{3}$

$r =4$

${ }^{22} C _4 \cdot \alpha^4=7315$

$\frac{22 \times 21 \times 20 \times 19}{24} \cdot \alpha^4=7315$

$\alpha=1$

$\Rightarrow n =10$

So $T_3={ }^{10} C _2 2^{\frac{1}{4} \cdot 8} \cdot 3^{-\frac{1}{4}-2}=\frac{45.4}{\sqrt{3}}=60 \sqrt{3}$

$T_{r+1}={ }^{15} C_r\left(x^4\right)^{15-r}\left(\frac{-1}{x^3}\right)^r$

$60-7 r=18$

$r=6$

Hence coeff. of $x^{18}={ }^{15} C _6=5005$

$T _{ r +1}={ }^{11} C _{ r }\left( ax ^2\right)^{11- r } \cdot\left(\frac{1}{2 bx }\right)^{ r }$

$={ }^{11} C _{ r } a ^{11- r } \cdot\left(\frac{1}{2 b }\right)^{ I } \cdot x ^{22-2 r - r }={ }^{11} C _{ r } a^{11- r } \cdot\left(\frac{1}{2 b }\right)^{ I } \cdot x ^{22-3 r }$

$\therefore 22-3 r=7$

$3 r =15$

$r=5$

Again $\left(a x-\frac{1}{3 b x^2}\right)^{11}$

$T _{ r +1}={ }^{11} C _{ r }( ax )^{11- r }\left(-\frac{1}{3 b x^2}\right)^{ r }$

$={ }^{11} C _{ r } a ^{11- r } \cdot\left(\frac{-1}{3 b }\right)^{ r } \cdot x ^{11- r -2 r }$ $\therefore 11-3 r=-7$

$3 r =18$

$r=6$

$\text { Now, } \frac{{ }^{11} C _5 a ^6}{32 b^5}=\frac{{ }^{11} C _6 \cdot a ^5}{3^6 \cdot b ^6}$

$729 ab =32$

$\frac{ n - r +1}{ r }=5 \quad n =5 r +4 \ldots(2)$

$n=6 r-1 \ldots(1)$

$\therefore n=29, r=5$

$\text { Coeff of } 4^{\text {th }} \text { term }={ }^{29} C _3$

$=3654$

$T _{ r +1}={ }^7 C _{ r }\left(3 x ^2\right)^{7- r }\left(-\frac{1}{2 x ^5}\right)^{ r }$

$14-2 r -5 r =14-7 r =0$

$\therefore r =2$

$\therefore T _3={ }^7 C _2 \cdot 3^5\left(-\frac{1}{2}\right)^2=\frac{21 \times 243}{4}=1275.75$

$\therefore[\alpha]=1275$

$={ }^{11} C _{ r } 2^{11-2 r } x ^{22-3 r }$

$22-3 r =10 \quad \text { and } \quad 22-3 r =7$

$r =4 \quad \text { and } \quad r =5$

$\text { Coefficient of } x ^{10}={ }^{11} C _4 \cdot 2^3$

$\text { Coefficient of } x ^7={ }^{11} C _5 \cdot 2^1$

$\text { difference }={ }^{11} C _4 \cdot 2^3-{ }^{11} C _5 \cdot 2$

$=\frac{11 \times 10 \times 9 \times 8}{24} \times 8-\frac{11 \times 10 \times 9 \times 8 \times 7}{120} \times 2$

$=11 \times 10 \times 3 \times 8-11 \times 3 \times 4 \times 7$

$=11 \times 3 \times 4 \times(20-7)$

$=11 \times 12 \times 13$

$=12(12-1)(12+1)$

$=12\left(12^2-1\right)$

$=12^3-12$

$={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r}$

$1 3 - 3 r = 7 \Rightarrow r=2$

Coefficient of $x^7={ }^{13} C_2(a)^{11} \cdot \frac{1}{b^2}$

In the other expansion $T_{r+1}={ }^{13} C_r(a x)^{13-r}\left(\frac{1}{b x^2}\right)^r$

$13-3 r=-5 \Rightarrow r=6$

Coefficient of $x^{-5}={ }^{13} C_6(a)^7 \cdot \frac{1}{b^6}$

${ }^{13} C_2 \frac{a^{11}}{b^2}={ }^{13} C_6 \frac{a^7}{b^6}$

$a^4 b^4=\frac{{ }^{13} C_6}{{ }^{13} C_2}=22$

where $r_1+r_2+r_3=10$ and $r_2+3 r_3=7$

$\begin{array}{lll}r_1 & r_2 & r_3 \\ 3 & 7 & 0 \\ 5 & 4 & 1 \\ 7 & 1 & 2\end{array}$

Required coefficient

$=\frac{10 !}{3 ! .7 !}(-1)^7+\frac{10 !}{5 ! .4 !}(-1)^4(2)+\frac{10 !}{7 ! \cdot 2 !}(-1)^1(2)^2$

$=-120+2520-1440=960$

General term $={ }^{680} C _{ r }\left(3^{\frac{1}{2}}\right)^{680- r }\left(5^{\frac{1}{4}}\right)^{ r }$

$={ }^{680} C _{ r } 3^{\frac{680- r }{2}} 5^{\frac{ r }{4}}$

Value's of $r$, where $\frac{r}{4}$ goes to integer

$r =0,4,8,12, \ldots \ldots \ldots .680$

All value of $r$ are accepted for $\frac{680-r}{2}$ as well so No of integral terms $=171$.

$={ }^{2022} C_{1010}\left(\frac{4 x}{5}\right)^{1012}\left(\frac{-5}{2 x }\right)^{1010}$

$T _{1011}$ from end

$={ }^{2022} C_{1010}\left(\frac{-5}{2 x }\right)^{1012}\left(\frac{4 x }{5}\right)^{1010}$

Given : ${ }^{2022} C _{\text {1010 }}\left(\frac{-5}{2 x }\right)^{1012}\left(\frac{4 x }{5}\right)^{1010}$

$=2^{10} \cdot{ }^{2022} C _{1000}\left(\frac{-5}{2 x }\right)^{1010}\left(\frac{4 x }{5}\right)^{1012}$

$\left(\frac{-5}{2 x}\right)^2=2^{10}\left(\frac{4 x}{5}\right)^2$

$x^4=\frac{5^4}{2^{16}}$

$|x|=\frac{5}{16}$

$\frac{ n - k }{2}-\frac{3}{2} k =0$

$n -4 k =0$

$(-5)^{ n }-\left({ }^{ n } C _{\frac{ n }{}}(-6)^{\frac{ n }{4}}\right)=649$

By observation $(625+24=649)$, we get $n=4$

$\because n=4 \quad k=1$

Required is coefficient of $x^{-4}$ is $\left(\sqrt{4}-\frac{6}{x^{\frac{3}{2}}}\right)^4$ ${ }^4(-6)^3$

By calculating we will get $\lambda=36$

$={ }^{45} C _{23}$

$\sum \limits_{r=0}^n r^{2 n} C_r=n(n+1) \cdot 2^{n-2}$

$=2023 \times \alpha \times 2^{2022} \text { So, }$

$\Rightarrow \alpha=1012$

$ S =30 \cdot(^{30} C _0)^2+29 \cdot{ }^{30} C _1)^2+28 \cdot{ }^{30} C _2)^2$

$+\ldots \ldots+0 \cdot{ }^{30} C _0)^2$

$\left.2 S =30 \cdot{ }^{30} C _0^2++^{30} C _1^2+\ldots \ldots \cdot+\cdot{ }^{30} C _{30}{ }^2\right)$

$S =15 \cdot{ }^{60} C _{30}=15 \cdot \frac{60 !}{(30 !)^2}$

$\frac{15 \cdot 10 !}{(30 !)^2}=\frac{\alpha \cdot 60 !}{(30 !)^2}$

$\Rightarrow \alpha=15$

$\Rightarrow \sum \limits_{ r =1}^{10} r ^3\left(\frac{{ }^{10} C _{ r }}{{ }^{10} C _{ r -1}}\right)^2=\sum \limits_{ r =1}^{10} r ^3\left(\frac{11- r }{ r }\right)^2=\sum \limits_{ r =1}^{10} r (11- r )^2$

$=\sum \limits_{ r =1}^{10}\left(121 r + r ^3-22 r ^2\right)=1210$

$(1+ x )^{99}= C _0+ C _1 x + C _2 x ^2+\ldots+ C _{99} x ^{99}$

$K = C _1+ C _3+\ldots . .+ C _{99}=2^{98}$

$a$ Middle in the expansion of $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$

$\frac{T_{200}}{2}+1 ={ }^{200} C _{100}(2)^{100}\left(\frac{1}{\sqrt{2}}\right)^{100}$

$={ }^{200} C _{100} \cdot 2^{50}$

So, $\frac{{ }^{200} C _{99} \times 2^{98}}{{ }^{900} C _{100} \times 2^{50}}=\frac{100}{101} \times 2^{48}$

So, $\frac{25}{101} \times 2^{50}=\frac{ m }{ n } 2^{\prime}$

$\therefore m , n$ are odd so

$(\ell, n )$ become $(50,101)$ Ans.

$=(1+x)^{500} \cdot\left\{\frac{1-\left(\frac{x}{1+x}\right)^{501}}{1-\frac{x}{1+x}}\right\}$

$=(1+x)^{500} \frac{\left((1+x)^{501}-x^{501}\right)}{(1+x)^{501}} \cdot(1+x)$

$=(1+x)^{501}-x^{501}$

Coefficient of $x^{301}$ in $(1+x)^{501}-x^{501}$ is given by ${ }^{501} C _{301}={ }^{501} C _{200}$

$=\frac{1}{51 !}\left\{{ }^{51} C _1+{ }^{51} C _3+\ldots .+{ }^{51} C _{51}\right\}$

$=\frac{1}{51 !}\left(2^{50}\right)$

$\frac{{ }^{n+2} C_{ r -1}}{{ }^{ n +2} C _{ r }}=\frac{1}{3}$

$n =4 r -3 \ldots \ldots \text { (i) }$

$\frac{{ }^{n+2} C _{ r }}{{ }^{ n +2} C _{ r +1}}=\frac{3}{5}$

$8 r -1=3 n \text {...... (ii) }$

$\text { From, (i) and (ii) }$

$r =2 \text { and } n =5$

$\text { Required sum }=63$

$C _3 x ^3+\ldots C _{99 x }{ }^{99}+ C _{100 x } x ^{10}$

$\Rightarrow Co ^{- C _1}+ C _2- C _3+\ldots \ldots- C _{99}+ C _{100}=0$

$2\left( Co - C _1+ C _2+\ldots \ldots- C _9\right)+ C _{50}=0$

$C _0- C _1+ C _2+\ldots . C _{99}=-\frac{1}{2}{ }^{100} C _{50}$

$-\frac{1}{2} \frac{100 !}{50 ! 50 !}=-\frac{1}{2} \times \frac{100 \times 99 !}{50 ! 50 !}=-{ }^{99} C _{49}$

$=\frac{1}{ n +1}\left(2^{ n +1}-1\right)=\frac{1023}{10}$

$n +1=10 \Rightarrow n =9$

Coefficient of $x^1=20$

$20=\frac{10 !}{9 ! 1 !} \times a^9 \times b^1$

$a^9 . b =2$

$a=1, b=2$

Coefficient of $x ^2=210$

$210=\frac{10 !}{9 ! 1 !} \times a^9 \times c^1+\frac{10 !}{8 ! 2 !} \times a^8 b^2$

$210=10 . c+45 \times 4$

$10 c=30$

$c=3$

$2(a+b=c)=12$

$\left(1+p x+\frac{p(p-1)}{2 !} x^2+\ldots\right)$

$\left(1-q x+\frac{q(q-1)}{2 !} x^2-\ldots\right)$

$p-q=4$

$\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-p q=-5$

$p^2+q^2-p-q-2 p q=-10$

$(q+4)^2+q^2-(q+4)-q-2(4+q) q=-10$

$q^2+8 q+16-q^2-q-4-q-8 q-2 q^2=-10$

$-2 q=-22$

$q=11$

$p=15$

$2(15)+3(11)$

$30+33=63$

Of $x ^2={ }^9 C _2 2^7$

Of $x ^7={ }^9 C _7 \cdot 2^2$

Mean $=\frac{{ }^9 C _1 \cdot 2^8+{ }^9 C _2 \cdot 2^7 \ldots . .+{ }^9 C _7 \cdot 2^2}{7}$

$=\frac{(1+2)^9-{ }^9 C _0 \cdot 2^9-{ }^9 C _8 \cdot 2^1-{ }^9 C _9}{7}$

$=\frac{3^9-2^9-18-1}{7}$

$=\frac{19152}{7}=2736$

$x ^{\prime}=(8 \sqrt{3}-13)^{13}={ }^{13} C _0(8 \sqrt{3})^{13}-{ }^{13} C _1(8 \sqrt{3})^{12}(13)^1+\ldots$

$x - x ^{\prime}=2\left[{ }^{13} C _1 \cdot(8 \sqrt{3})^{12}(13)^1+{ }^{13} C _3(8 \sqrt{3})^{10} \cdot(13)^3 \ldots\right]$

therefore, $x-x^{\prime}$ is even integer, hence $[x]$ is even

$\text { Now, } y =(7 \sqrt{2}+9)^9={ }^9 C _0(7 \sqrt{2})^9+{ }^9 C _1(7 \sqrt{2})^8(9)^1$

$+{ }^9 C _2(7 \sqrt{2})^7(9)^2 \ldots \ldots$

$y ^{\prime}=(7 \sqrt{2}-9)^9={ }^9 C _0(7 \sqrt{2})^9-{ }^9 C _1(7 \sqrt{2})^8(9)^1$

$+{ }^9 C _2(7 \sqrt{2})^7(9)^2 \ldots \ldots$

$y - y ^{\prime}=2\left[{ }^9 C _1(7 \sqrt{2})^8(9)^1+{ }^9 C _3(7 \sqrt{2})^6(9)^3+\ldots\right.$

$y - y ^{\prime}=$ Even integer, hence $[ y ]$ is even

$=(2030-7)^{2023}$

$=(35 K -7)^{2023}$

$={ }^{2023} C _0(35 K )^{2023}(-7)^0+{ }^{2023} C _1(35 K )^{2022}(-7)+\ldots .+$

$\ldots \ldots+{ }^{2023} C _{2023}(-7)^{2023}$

$=35 N -7^{2023}$

$\text { Now, }-7^{2023}=-7 \times 7^{2022}=-7\left(7^2\right)^{1011}$

$=-7(50-1)^{1011}$

$=-7\left({ }^{1011} C _0 50^{1011}-{ }^{1011} C _1(50)^{1010}+\ldots \ldots .{ }^{1011} C _{1011}\right)$

$=-7(5 \lambda-1)$

$=-35 \lambda+7$

$\therefore$ when $(2023)^{2023}$ is divided by $35$ remainder is $7$

$=25 K -\left(6^{25}+1\right)=25 K -\left((5+1)^{25}+1\right)$

$=25 K _1-2 \quad \text { Remainder }=23$