Maths M.C.Q (1 Marks)

$=625\left[5^5\right]^{19}$

$=625[3125]^{19}$

$=625[3124+1]^{19}$

$=625[11 k \times 19+1]$

$=625 \times 11\,k \times 19+625$

$=11\,k _1+616+9$

$=11\left( k _2\right)+9$

Remainder $=9$

$\Rightarrow 2\left[{ }^{100} C _0 21^{200}+200 C _2 21^{198} \cdot 2^2+\ldots . .+{ }^{200} C _{198} 21^2\right.$

$\left.2^{198}+2^{200}\right]$

$\Rightarrow 2\left[49 I _1+2^{200}\right]=49 I _1+2^{201}$

Now, $2^{201}=(8)^{67}=(1+7)^{67}=49 I _2+{ }^{67} C _0{ }^{67} C _1 \cdot 7=$ $49 I _2+470=49 I _2+49 \times 9+29$

$\therefore$ Remainder is $29$

$\Rightarrow{ }^{2022} C _1(1999)^{2021}(24)+{ }^{2022} C _2(1999)^{2020}(24)^2+\ldots \text { so on }$

$S _1$ is divisible by $8$

$S _2: 13\left(13^{ n }\right)-11 n -13$

$13^{ n }=(1+12)^{ n }=1+12 n +{ }^{ n } C _2 12^2+{ }^{ n } C _3 12^3 \ldots \ldots$

$13\left(13^{ n }\right)-11 n -13=145 n +{ }^{ n } C _2 12^2+{ }^{ n } C _3 12^3 \ldots \ldots .$

If $( n =144 m , m \in N )$, then it is divisible by $144$ For infinite value of $n$.

$19^{190}-2^{190}$ is divisible by $19-2=17$

$25^{190}-19^{190}$ is divisible by $25-19=6$

$8^{190}-2^{190}$ is divisible by $8-2=6$

$L.C.M.$ of $1746=34$

$\therefore$ divisible by $34$ but not by $14$

$=\left\{\frac{(1+15)^{1011}}{15}\right\}$

$=\frac{1}{15}$

$=7 \times(49)^{51}$

$=7 \times(51-2)^{51}$

Remainder :- $7 \times(-2)^{51}$

$\Rightarrow-7\left(2^3 \cdot(16)^{12}\right)$

$\Rightarrow-56(17-1)^{12}$

$\text { Remainder }=-56 \times(-1)^{12}=-56+68=12$

divided by $3$

$(21+1)^{2022}+(2022)^{22}$

$=3 k+1 \quad(\alpha=1)$

Divided by $7$

$(21+1)^{2022}+(2023-1)^{22}$

$7 k +1+1 \quad(\beta=2)$

$7 k +2$

So $\alpha^2+\beta^2 \Rightarrow 5$

$\beta={ }^{n} C_{2}+{ }^{n} C_{3} 5+{ }^{n} C_{4} 5^{2}+\ldots$

option $(3)$ will be the answer.

$t _{ r +1}={ }^{12} C _{ r }\left(2 x ^{3}\right)^{ r }\left(\frac{3}{ x ^{ k }}\right)^{12- r }$

$x ^{3 r -(12- r ) k } \rightarrow constant$

$\therefore 3 r -12 k + rk =0$

$\Rightarrow k =\frac{3 r }{12- r }$

$\therefore$ possible values of $r$ are $3,6,8,9,10$ and corresponding values of $k$ are $1,3,6,9,15$

Now ${ }^{12} C _{ r }=220,924,495,220,66$

$\therefore$ possible values of $k$ for which we will get $2^{8}$ are $3. 6$

$m ={ }^{15} C _{10} 2^{5}$

$n =-1$

$\text { so } mn ^{2}={ }^{15} C _{5} 2^{5}$

General term of $\left(\frac{5}{2} x ^{3}-\frac{1}{5 x ^{2}}\right)^{11}$ is

${ }^{11} C_{r}\left(\frac{5}{2} x^{3}\right)^{11-r}\left(-\frac{1}{5 x^{2}}\right)^{ r }$

General term is ${ }^{11} C _{ r }\left(\frac{5}{2}\right)^{11- r }\left(-\frac{1}{5}\right)^{ r } x ^{33-5 r }$

Now, term independent of $x$

$1 \times$ coefficient of $x^{0}$ in $\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}$

$-1 \times$ coefficient of $x^{-2}$ in $\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}+$

$3 \times$ coefficient of $x^{-3}$ in $\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}$

$\begin{array}{ll}\text { for coefficient of } x ^{0} & 33-5 r =0 \text { not possible } \\ \text { for coefficient of } x ^{-2} & 33-5 r =-2 \\ & 35=5 r \Rightarrow r =7 \\ \text { for coefficient of } x ^{-3} & 33-5 r =-3 \\ & 36=5 r \text { not possible }\end{array}$

So term independent of $x$ is

$(-1)^{11} C_{7}\left(\frac{5}{2}\right)^{4}\left(-\frac{1}{5}\right)^{7}=\frac{33}{200}$

$T_{r+1}={ }^{15} C_{r}\left(t^{2} x^{\frac{1}{5}}\right)^{15-r} \cdot \frac{(1-x)^{\frac{r}{10}}}{t^{r}}$

For independent of $t$,

$30-2 r-r=0$

$r =10$

So, Maximum value of ${ }^{15} C _{10} x (1- x )$ will be at

$x=\frac{1}{2}$

i.e. $6006$

${ }^{p} C _{1}-{ }^{9} C _{1}=-3$

$p - q =-3$

$\text { Comparing coefficients of } x ^{2}$

${ }^{9} C _{1}{ }^{9} C _{1}+{ }^{ p } C _{2}+{ }^{9} C _{2}=-5$

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

Solving $(1)$ and $(2)$

$p=8, q=11$

Coefficient of $x ^{3}$ is

$-{ }^{4} C_{3}+{ }^{P} C_{3}+{ }^{P} C_{1}^{9} C_{2}-{ }^{P} C_{2}^{9} C_{1}$

$=-{ }^{11} C_{3}+{ }^{8} C_{3}+{ }^{8} C_{1}^{11} C_{2}-{ }^{8} C_{2}^{11} C_{1}$

$=23$

$\Rightarrow 2^{\frac{n-8}{4}} 3^{\frac{n-8}{4}}=6^{1 / 4}$

$\Rightarrow 6^{n-3}=6$

$\Rightarrow n-8=1 \Rightarrow n=9$

$T_{6}={ }^{9} C_{5}\left(2^{1 / 4}\right)^{4}\left(3^{-1 / 4}\right)^{5}=\frac{84}{\sqrt[4]{3}}$

$\therefore \alpha=84$

$=\frac{(5+x)^{501}-x^{501}}{(5+x)-x}=\frac{(5+x)^{501}-x^{501}}{5}$

$\Rightarrow$ coefficient $x ^{101}$ in given expression

$=\frac{{ }^{501} C _{101} 5^{400}}{5}={ }^{501} C _{101} 5^{399}$

$={ }^{10} C_{r} 2^{10-r} 3^{r} x^{30-4 r}$

Put $r=0,1,2, \ldots 7$ and we get $\beta=83$

$3 n-20 \geq 0 \cap 2 n-25<0 \cap n \in I ~\\ \therefore \quad 7 \leq n \leq 12 ~\\ \text { Sum }=7+8+9+10+11+12=57$

$={ }^{31} C _{1} \cdot{ }^{31} C _{0}+{ }^{31} C _{2} \cdot{ }^{31} C _{1}+\ldots .+{ }^{31} C _{31} \cdot{ }^{31} C _{30}$

$={ }^{31} C _{0} \cdot{ }^{31} C _{30}+{ }^{31} C _{1} \cdot{ }^{31} C _{29}+\ldots .+{ }^{31} C _{30} \cdot{ }^{31} C _{0}$

$={ }^{62} C _{30} \cdot$

Similarly

$\sum\limits_{ R =1}^{30}\left({ }^{30} C _{ R } \cdot{ }^{30} C _{ R -1}\right)={ }^{60} C _{29}$

${ }^{62} C _{30}-{ }^{60} C _{29}=\frac{62 !}{30 ! 32 !}-\frac{60 !}{29 ! 31 !}$

$=\frac{60 !}{29 ! 31 !}\left\{\frac{62 \cdot 61}{30 \cdot 32}-1\right\}$

$=\frac{60 !}{30 ! 31 !}\left(\frac{2822}{32}\right)$

$\therefore 16 \alpha=16 \times \frac{2822}{32}=1411$

$m=1, n=98$

$m + n =99$

$\sum_{ K =1}^{10}\left( K ^{10} C _{ K }\right)^{2}=\sum_{ K =1}^{10}\left(10 \cdot{ }^{9} C _{ K -1}\right)^{2}$

$=100 \sum_{ K =1}^{9} C _{ K -1} \cdot{ }^{9} C _{10- K }$

$=100\left({ }^{18} C _{9}\right)=100\left(\frac{18 !}{9 ! 9 !}\right)$

$\Rightarrow 4862000=22000 L$

Hence $L =221$

$\left(\frac{1}{41}+1\right){ }^{41} C _{1}+{ }^{42} C _{2}+\ldots \ldots$

$\left[\frac{42}{41}\left(\frac{2}{42}\right)+1\right]{ }^{42} C _{2}+{ }^{43} C _{3}+\ldots .$

$\left(\frac{2}{41}+1\right)^{42} C _{2}+{ }^{43} C _{3}+\ldots . .$

$\left(\frac{43}{41} \times \frac{3}{43}+1\right){ }^{43} C _{3}+{ }^{44} C _{4}+\ldots \ldots .$

$\frac{3+41}{41}{ }^{43} C _{3}+\ldots \ldots .$

Similarly :

$\frac{20+41}{41}$

$\Rightarrow m =61 ; n =41$

$m + n =102$

$T _{ r +1}=\frac{! 10}{! r _{1}! r _{2}! r _{3}}(3)^{ r _{1}}(-2)^{ r _{2}}(5)^{ r _{3}}( x )^{3 r _{1}+2 r _{2}-5 r _{3}}$

$3 r_{1}+2 r_{2}-5 r_{3}=0$      $\dots(1)$

$r_{1}+r_{2}+r_{3}=10$      $\dots(2)$

from equation $(1)$ and $(2)$

$r_{1}+2\left(10-r_{3}\right)-5 r_{3}=0$

$r _{1}+20=7 r _{3}$

$\left( r _{1}, r _{2}, r _{3}\right)=(1,6,3)$

constant term $=\frac{!10}{!16!{6}!3}(3)^{1}(-2)^{6}(5)^{3}$

$=2^{9} \cdot 3^{2} \cdot 5^{4} \cdot 7^{1}$

$l=9$

$=\frac{(10-1)^{1011}-1}{2}$

$=\frac{100 \lambda+10110-1-1}{2}$

$=50 \lambda+\frac{10108}{2}$

$=50 \lambda+5054$

$=50 \lambda+50 \times 101+4$

$\operatorname{Rem}(50)=4 .$

$=5(2 m-1)+4( m \text { is integer })$

Remainder $=4$

$={ }^{2023} C_{0}(7 A )^{2023}-\ldots{ }^{2023} C _{2023} 2^{2023}$

$=7 t -2^{2023}$

$\therefore-2^{2023}=-2 \times 2^{2022}$

$=-2 \times\left(2^{3}\right)^{674}$

$=-2(1+7 \mu)^{674}$

$=-(7 \alpha+2)$

$\Rightarrow \text { remainder }=-2 \text { or }+5$

$=(2023-2)^{\text {an2 }}+(2023-1)^{3031}$

$=7 n_{1}+2^{m 2 n 2}+7 n _{2}-1$

$=7\left( n _{1}+ n _{2}\right)+8^{674}-1$

$=7\left(n_{1}+n_{2}\right)+(7-1)^{674}-1$

$=7\left(n_{1}+n_{2}\right)+7 n_{3}+1-1$

$=7\left(n_{1}+n_{2}+n_{3}\right)$

$\therefore$ Given number is divisible by $7$ hence remainder is zero

$=(49)^{1011}+(9)^{1011}$

$=(50-1)^{1011}+(10-1)^{1011}$

$=5 \lambda-1+5 K -1$

$=5\,m -2$

Remainder $=5-2=3$

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

$={ }^{60} C_{r} 5 \frac{3 r-60}{4} x \frac{180-5 r}{6}$

$\frac{180-5 r}{6}=10 \Rightarrow r=24$

Coeff. of $x^{10}={ }^{60} C_{24} 5^{3}=\frac{\mid 60}{|24| 36} 5^{3}$

Powers of $5$ in $={ }^{60} C_{24} \cdot 5^{3}=\frac{5^{14}}{5^{4} \times 5^{8}} \times 5^{3}=5^{5}$

$T _{ r +1}{ }^{ n } C _{ r } 3^{ n - r } \cdot(6 x ) r $

$={ }^{ n } C _{ r } 3^{ n - r } \cdot 6^{ I } \cdot x ^{ r } 3^{ n - r } \cdot 3^{ r } \cdot 2^{ r } \cdot\left(\frac{3}{2}\right)^{ r }$

$={ }^{ n } C _{ r } 3^{ n } \cdot 3^{ r } \quad \text { [for } x =\frac{3}{2}]$

$T_{9}$ is greatest of $x =\frac{3}{2}$

So, $T _{9}> T _{10}$ and $T _{9}> T _{8}$

(concept of numerically greatest term)

Here, $\frac{T_{9}}{T_{10}}>1$ and $\frac{T_{9}}{T_{8}}>1$

$\frac{{ }^{n} C_{8} 3^{n} \cdot 3^{8}}{{ }^{n} C_{9} 3^{n} \cdot 3^{9}}>1$ and $\frac{{ }^{n} C_{8} 3^{n} \cdot 3^{8}}{{ }^{n} C_{7} 3^{n} \cdot 3^{7}}>1$

and $\frac{{ }^{ n } C _{8}}{{ }^{ n } C _{7}}>\frac{1}{3}$

and $\frac{ n -7}{8}>\frac{1}{3}$

$\frac{29}{3}< n <11 \Rightarrow n =10= n _{0}$

So, in $(3+6 x )^{ n }$ for $n = n _{0}=10$

i.e., in $(3+6 x )^{10}$, here $T _{ r +1}={ }^{10} C _{ r } 3^{10- r } 6^{ T } x ^{ r }$

$T _{7}={ }^{10} C _{6} 3^{4} \cdot 6^{6} \cdot x ^{6}=210 \cdot 3^{10} \cdot 2^{6} x ^{6}$

$T _{4}={ }^{10} C _{3} 3^{7} 6^{3} x ^{3}=120.3^{10} \cdot 2^{3} x ^{3}$

Ratio of coefficient of $x ^{6}$ and coefficient of $x ^{3}= k$ $\therefore k =\frac{210 \cdot 3^{10} 2^{6}}{120 \cdot 3^{10} \cdot 2^{3}}=\frac{7}{4} \times 2^{3}=14$

So, $k + n _{0}=14+10=24$

$=\sum_{ i =0}^{ n }{ }^{ n } C _{ i } \cdot \sum_{ j =0}^{ n }{ }^{ n } C _{ j }-\sum_{ i = j =0}^{ n }\left({ }^{ n } C _{ i }\right)^{2}$

$=\left(2^{ n }\right)\left(2^{ n }\right)-{ }^{2 n } C _{ n }$

$=2^{2 n }-{ }^{2 n } C _{ n }$

For $\operatorname{Re}\left(\frac{2^{1011}}{9}\right)$

$2^{1011}=(9-1)^{337}={ }^{337} C_{0} 9^{337}(-1)^{0}$

$+{ }^{337} C_{1} 9^{336}(-1)^{1}$

$+{ }^{337} C_{2} 9^{335}(-1)^{2}+\ldots \ldots$

$+{ }^{337} C_{337} 9^{0}(-1)^{337}$

so, remainder is $8$ and $\operatorname{Re}\left(\frac{3^{11}}{9}\right)=0$

So, remainder is $8$

$={ }^{n} C _{ r } a ^{ r } x ^{ n -3 r }$

${ }^{ n } C _{2} a ^{2}:{ }^{ n } C _{3} a ^{3}:{ }^{ n } C _{4} a ^{4}=12: 8: 3$

After solving

$n =6, a =\frac{1}{2}$

For term independent of $x ^{\prime} \Rightarrow n =3 r$

$r =2$

$\therefore$ Coefficient is ${ }^{6} C _{2}\left(\frac{1}{2}\right)^{2}=\frac{15}{4}$

Nearest integer is $4 .$

$T_{r+1}=(-1)^{r} \cdot{ }^{12} C_{r}\left(\frac{x}{4}\right)^{12-\mathrm{r}}\left(\frac{12}{x^{2}}\right)^{r}$

$T_{r+1}=(-1)^{r} \cdot{ }^{12} C_{r}\left(\frac{1}{4}\right)^{12-r}(12)^{r} \cdot(x)^{12-3 r}$

Term independent of $x \Rightarrow 12-3 r=0 \Rightarrow r=4$

$\mathrm{T}_{5}=(-1)^{4} \cdot{ }^{12} \mathrm{C}_{4}\left(\frac{1}{4}\right)^{8}(12)^{4}=\frac{3^{6}}{4^{4}} \cdot \mathrm{k}$

$\Rightarrow \mathrm{k}=55$

$T_{r+1}={ }^{120} C_{r}\left(2^{1 / 2}\right)^{120-r}(5)^{r / 6}$

for rational terms $\mathrm{r}=6 \lambda\,\,\,\, 0 \leq \mathrm{r} \leq 120$

so total no of forms are $21$

 Sum of Coeff. of two middle terms in

$(1+x)^{19}={ }^{19} C_{9}+{ }^{19} C_{10}$

So required ratio $=\frac{{ }^{20} \mathrm{C}_{10}}{{ }^{19} \mathrm{C}_{9}+{ }^{19} \mathrm{C}_{10}} \frac{{ }^{20} \mathrm{C}_{10}}{{ }^{19} \mathrm{C}_{9}+{ }^{19} \mathrm{C}_{10}}=\frac{{ }^{20} \mathrm{C}_{10}}{{ }^{20} \mathrm{C}_{10}}=1$

$\mathrm{T}_{\mathrm{r}+1}$ will be rational number

$\text { Where } r=0,3,6,9,12$

$\&\, r=0,4,8,12$

$\Rightarrow r=0,12$

$T_{1}+T_{13}=1 \times 3^{3}+1 \times 2^{4} \times 1$

$\Rightarrow 27+16=43$

$\Rightarrow \frac{n !}{(n-7) ! 7 !} 2^{n-7} \frac{1}{3^{7}}=\frac{n !}{(n-8) ! 8 !} 2^{n-8} \frac{1}{3^{8}} \Rightarrow \frac{1}{(n-7)}=\frac{1}{8} \cdot \frac{1}{2} \cdot \frac{1}{3}$

$\Rightarrow n-7=48 \Rightarrow n=55$

${ }^{25} C _{ k }+{ }^{25} C _{ k +1}$

${ }^{26} C _{ k +1}^{  }$

as ${ }^{ n } C _{ r }$ is defined for all values of $n$ as will as r so ${ }^{26} C _{ k +1}$ always exists

Now $k$ is unbounded so maximum value is not defined.

$\Rightarrow 2^{\mathrm{n}}=2^{12} \quad\quad\quad\quad\quad\quad 2^{11}=2048$

$\mathrm{n}=12 \quad\quad\quad\quad\quad\quad\quad\quad 2^{12}=\underline{4096}$

${ }^{12} \mathrm{C}_{6}=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$= 11 \times 3 \times 4 \times 7$

$= 924$

$n$ can be any odd number

$\therefore$ Number of odd two digit numbers $=45$

$\text { General term }={ }^{10} C_{R}\left(2 x^{r}\right)^{10-R} x^{-2 R}$

$\Rightarrow 2^{10-R 10} C_{R}=180 \ldots \ldots . .(1)$

$\,(10-R) r-2 R=0$

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

$r=\frac{2(R-10)}{10-R}+\frac{20}{10-R}$

$\Rightarrow r=-2+\frac{20}{10-R} \ldots . . . \text { (2) }$

$\mathrm{R}=8$ or $5$ reject equation $(1)$ not satisfied At $R=8$

$2^{10-R 10} \mathrm{C}_{\mathrm{R}}=180 \Rightarrow \mathrm{r}=8$

$\Rightarrow x ^{\left(4+3 \log _{2}^{x}\right)}=2^{7}$

$\Rightarrow \quad(4+3 t ) t =7 ; t =\log _{2} x$

$\Rightarrow t =1, \frac{-7}{3} \Rightarrow x =2$

$\left( x ^{1 / 3}- x ^{-1 / 2}\right)^{10}$

$T _{ r +1}={ }^{10} C _{ r }\left( x ^{1 / 3}\right)^{10- r }\left(- x ^{-1 / 2}\right)^{ r }$

$\frac{10- r }{3}-\frac{ r }{2}=0 \Rightarrow 20-2 r -3 r =0$

$\Rightarrow r =4$

$T _{5}={ }^{10} C _{4}=\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210$

Coefficient of $y=1 .{ }^{n} C_{1}+{ }^{m} C_{1}(-1)$

$=n-m=10$ $\ldots(1)$

Coefficient of $\mathrm{y}^{2}\left(\mathrm{a}_{2}\right)$

$=1 .{ }^{n} \mathrm{C}_{2}-{ }^{n} \mathrm{C}_{1} \cdot{ }^{n} C_{1 .}+1 \cdot{ }^{m} C_{2}=10$

$=\frac{n(n-1)}{2}-m \cdot n+\frac{m(m-1)}{2}=10$

$m^{2}+n^{2}-2 m n-(n+m)=20$

$n+m=80$

$(n-m)^{2}-(n+m)=20$

By equation $(1)\,  \,(2)$

$\mathrm{m}=35, n=45$

$=\left(x^{1 / 3} \frac{1}{x^{1 / 2}}\right)^{10}$

Now General Term

$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{r}\left(\mathrm{x}^{1 / 3}\right)^{10-\mathrm{r}} \cdot\left(-\frac{1}{\mathrm{x}^{1 / 2}}\right)^{\mathrm{r}}$

For independent term

$\frac{10-r}{3}-\frac{r}{2}=0 \Rightarrow r=4$

$\Rightarrow T_{5}={ }^{10} C_{4}=210$

$r=0,1,2, \ldots, 10$

$T_{r+1}$ will be independent of $x$

When $10-2 r=0 \Rightarrow r=5$

$T_{6}={ }^{10} C_{5}(x \sin \alpha)^{5} \times\left(\frac{a \cos \alpha}{x}\right)^{5}$

$=^{10} C_{5} \times a^{5} \times \frac{1}{2^{5}}(\sin 2 \alpha)^{5}$

will be greatest when $\sin 2 \alpha=1$

$\Rightarrow^{10} C_{5} \frac{a^{5}}{2^{5}}={ }^{10} C_{5} \Rightarrow a=2$

${ }^{11} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{11-\mathrm{r}} \cdot\left(\frac{1}{\mathrm{bx}}\right)^{\mathrm{r}}$

${ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{22-3 \mathrm{r}} \cdot \frac{1}{\mathrm{~b}^{r}}$

$22-3 \mathrm{r}=7$

$r=5$

$\therefore{ }^{11} \mathrm{C}_{5} \cdot \frac{1}{\mathrm{~b}^{5}} \cdot \mathrm{x}^{7}$

Coefficient of $x^{-7}$ in $\left(x-\frac{b}{b x^{2}}\right)^{11}$

${ }^{11} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{11-\mathrm{r}} \cdot\left(-\frac{1}{\mathrm{bx}^{2}}\right)^{\mathrm{r}}$

${ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{11-3 r} \cdot \frac{(-1)^{r}}{\mathrm{~b}^{r}}$

$11-3 \mathrm{r}=-7 \therefore \mathrm{r}=6$

${ }^{11} \mathrm{C}_{6} \cdot \frac{1}{b^{6}} \mathrm{x}^{-7}$

${ }^{11} \mathrm{C}_{5} \cdot \frac{1}{\mathrm{~b}^{5}}={ }^{11} \mathrm{C}_{6} \cdot \frac{1}{\mathrm{~b}^{6}}$

Since $b \neq 0 \therefore b=1$

$\Rightarrow a_{0}+a_{1}+a_{2}+\ldots .+a_{40}=2^{20}$

$a_{0}-a_{1}+a_{2}+\ldots+a_{40}=2^{20}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{39}=\frac{4^{20}-2^{20}}{2}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{39}-2^{19}-a_{39}$

here $a_{39}=\frac{20 !(2)^{19} \times 1}{19 !}=20 \times 2^{19}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{19}\left(2^{20}-1-20\right)$

$=2^{19}\left(2^{20}-21\right)$

$={ }^{6} C _{0} \cdot{ }^{6} C _{6}+{ }^{6} C _{1} \cdot{ }^{6} C _{5}+\ldots \ldots+{ }^{6} C _{6} \cdot{ }^{6} C _{0}$

Now,

$(1+x)^{6}(1+x)^{6}$

$=\left(\begin{array}{l}\left.{ }^{6} C_{0}+{ }^{6} C_{1} x+{ }^{6} C_{2} x^{2}+\ldots . . .+{ }^{6} C_{6} x^{6}\right) \\ \left({ }^{6} C_{0}+{ }^{6} C_{1} x+{ }^{6} C_{2} x^{2}+\ldots \ldots+{ }^{6} C_{6} x^{6}\right)\end{array}\right.$ 

Comparing coefficeint of $x^{6}$ both sides

$\begin{array}{l} { }^{6} C _{0} \cdot{ }^{6} C _{6}+{ }^{6} C _{1}+{ }^{6} C _{5}+\ldots \ldots .+{ }^{6} C _{6} \cdot{ }^{6} C _{0}={ }^{12} C _{6} \\ =924 \end{array}$

$=\sum_{ r =1}^{10}[\{( r +3) !-( r +1) !\}-8\{( r +1) !- r !\}]$

$=(13 !+12 !-2 !-3 !)-8(11 !-1)$

$=(12.13+12-8) \cdot 11 !-8+8$

$=(160)(11) !$

Hence $\alpha=160$