Maths M.C.Q (1 Marks)

$ = \frac{{8\left( {^{100}{{\text{C}}_0}{ + ^{100}}{{\text{C}}_1}(15){ + ^{100}}{{\text{C}}_2}{{(15)}^2} +  \ldots  \ldots } \right)}}{{15}}$

$\frac{8}{15}+8\left(^{100} \mathrm{C}_{1}(15)+^{100} \mathrm{C}_{2}(15)^{2}+\ldots \ldots .\right)$

Remainder is $8 .$

After rationalise the polynomial we get

=$\left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}} \times \frac{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}\right]^{8}$

$+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}} \times \frac{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}\right]^{8}$

=$\left[\frac{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}{\left(5 x^{3}+1\right)-\left(5 x^{3}-1\right)}\right]^{8}+\left[\frac{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}{\left(5 x^{3}+1\right)-\left(5 x^{3}-1\right)}\right]^{8}$

$=\frac{1}{2^{8}}\left[(\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1})^{8}+(\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1})^{8}\right]$

=$^{8} C_{0}(\sqrt{5 x^{3}+1})^{8}+^{8} C_{2}(\sqrt{5 x^{3}+1})^{6}(\sqrt{5 x^{3}-1})^{2}$$\frac{1}{2^{8}} \cdot C_{4}(\sqrt{5 x^{3}+1})^{4}(\sqrt{5 x^{3}-1})^{4}+$

$^{\prime} C_{6}\left(\sqrt{5 x^{3}+1}^{2}(\sqrt{5 x^{3}})^{6}+^{6} C_{8}\left(y^{\sqrt{5 x^{3}-1}}\right.\right.$

$\frac{1}{2^{8}}\left[\begin{array}{l}{^{8} C_{0}\left(5 x^{3}+1\right)^{4}+^{8} C_{2}\left(5 x^{3}+1\right)^{3}\left(5 x^{3}-1\right)+8_{C_{4}}} \\ {\left(5 x^{3}+1\right)^{2}\left(5 x^{3}-1\right)^{2}+} \\ {^{8} C_{6}\left(5 x^{3}+1\right)\left(5 x^{3}-1\right)^{3}+8_{C_{4}}\left(5 x^{3}-1\right)^{4}}\end{array}\right]$

So, the degree of polynomial is $12$ , Now, coefficient of $x^{12}=\left[^{8} \mathrm{C}_{0} 5^{4}+^{8} \mathrm{C}_{2} 5^{4}+^{8} \mathrm{C}_{4} 5^{4}+^{8} \mathrm{C}_{6} 5^{4}+^{8} \mathrm{C}_{8} \mathrm{5}^{4}\right]$

$=5^{4} \times \frac{2^{8}}{2}=5^{4} \times 2^{4} \times \frac{2^{2}}{2}$

$=10^{4} \times 2^{3}=8(10)^{4}$

$\left(1+x^{2}\right)^{3}=1+3 x^{2}+3 x^{4}+x^{6},$

and $\left(1+x^{3}\right)^{4}=1+4 x^{3}+6 x^{6}+4 x^{9}+x^{12}$

So, the possible combinations for $x^{10}$ are:

$x \cdot x^{9}, x \cdot x^{6} \cdot x^{3}, x^{2} \cdot x^{2} \cdot x^{6}, x^{4} \cdot x^{6}$

Corresponding coefficientsare $2 \times 4,2 \times 1 \times 4,1 \times 3 \times 6,3 \times 6$

or $8,8,18,18$

$\therefore$ Sum of the coefficient is

$8+8+18+18=52$

Therefore, the coefficient of $x^{10}$ in the expansion of $(1+x)^{2}\left(1+x^{2}\right)^{3}\left(1+x^{3}\right)^{4}$ is $52$

$\therefore $ Coefficient of $x^{2}$ in the expansion of the product

$\left(2-x^{2}\right)\left(\left(1+2 x+3 x^{2}\right)^{6}+\left(1-4 x^{2}\right)^{6}\right)$

$=2\left(\text { Coefficient of } x^{2} \text { in a }\right)-1$ (Constant of expansion)

In the expansion of

$\left(\left(1+2 x+3 x^{2}\right)^{6}+\left(1-4 x^{2}\right)^{6}\right)$

Constant $=1+1=2$

Coefficient of $x^{2}=$

[Coefficient of $x^2$ in ${(^6}{C_0}{(1 + 2x)^6}{(3{x^2})^0}$ ] $+$ [Coefficient of $x^2$ in ${(^6}{C_1}{(1 + 2x)^5}{(3{x^2})^1}$ ] $-$ ${[^6}{C_1}(4{x^2})]$ 

$=60+6 \times 3-24=54$

$\therefore \quad$ The coefficient of $x^{2}$ in $\left(2-x^{2}\right)$

${\left(\left(1+2 x+3 x^{2}\right)^{6}+\left(1-4 x^{2}\right)^{6}\right)}$

${=2 \times 54-1(2)=108-2=106}$

${(x+a)^{5}+(x-a)^{5}}$

${=2\left[^{5} C_{0} x^{5}+^{5} C_{2} x^{3} \cdot a^{2}+^{5} C_{4} x \cdot a^{4}\right]}$

$\therefore \quad(x+\sqrt{x^{3}-1})^{5}+(x-\sqrt{x^{3}-1})^{5}$

$=2\left[^{5} \mathrm{C}_{0} \mathrm{x}^{5}+^{5} \mathrm{C}_{2} \mathrm{x}^{3}\left(\mathrm{x}^{3}-1\right)+^{5} \mathrm{C}_{4} \mathrm{x}\left(\mathrm{x}^{3}-1\right)^{2}\right]$

$\Rightarrow 2\left[x^{5}+10 x^{6}-10 x^{3}+5 x^{7}-10 x^{4}+5 x\right]$

$\therefore $ Sum of coefficients of odd degree terms $=2$

$ = {({x^{1/3}} + 1 - 1 - 1/{x^{1/2}})^{10}}$

$ = {({x^{1/3}} - 1/{x^{1/2}})^{10}}$

${T_{r + 1}}{ = ^{10}}{C_r}{x^{\frac{{20 - 5r}}{6}}}$

for $r = 10$

${T_{11}}{ = ^{10}}{C_{10}}{x^{ - 5}}$

${\text { Coefficient of } x^{-5}=^{10} C_{10}(1)(-1)^{10}=1}$

$-\left(^{10} \mathrm{C}_{1}+^{10} \mathrm{C}_{2} \ldots . .^{10} \mathrm{C}_{10}\right)$

$=\frac{1}{2}\left[\left(^{21} \mathrm{C}_{1}+\ldots+^{21} \mathrm{C}_{10}\right)+\left(^{21} \mathrm{C}_{11}+\ldots^{21} \mathrm{C}_{20}\right)\right]-\left(2^{10}-1\right)$

$\left(\because 10 \mathrm{C}_{1}+^{10} \mathrm{C}_{2}+\ldots .+^{10} \mathrm{C}_{10}=2^{10}-1\right)$

$=\frac{1}{2}\left[2^{21}-2\right]-\left(2^{10}-1\right)$

$=\left(2^{20}-1\right)-\left(2^{10}-1\right)=2^{20}-2^{10}$

$ \Rightarrow \quad \frac{{28\lambda  - 7 + 7 - 1}}{7} = \frac{{7(4\lambda  - 1) + 6}}{7}$

$\therefore \quad \operatorname{Rem}=6$

$=^{18} C_{r} x^{6-\frac{2 r}{3}} \frac{1}{2^{r}}$

$\left\{ \begin{gathered}
  6 - \frac{{2r}}{3} =  - 2 \Rightarrow r = 12 \hfill \\
  \& \,6 - \frac{{2r}}{3} =  - 4 \Rightarrow r = 15 \hfill \\ 
\end{gathered}  \right\}$

$\Rightarrow \quad \frac{\text { coefficient of } x^{-2}}{\text { coefficient of } x^{-4}}=\frac{^{18} C_{12} \frac{1}{2^{12}}}{^{18} C_{15} \frac{1}{2^{15}}}=182$

$\left(1-\frac{2}{x}+\frac{4}{x^{2}}\right)^{n}$ is $\frac{(n+2)(n+1)}{2}$ or $^{n+2} C_{2}$

[assuming $\left.\frac{1}{x} \text { and } \frac{1}{x^{2}} \text { distinct }\right]$

$\therefore \frac{(n+2)(n+1)}{2}=28$

$ \Rightarrow $ $(n+2)(n+1)=56=(6+1)(6+2) \Rightarrow n=6$

Hence, sum of coefficients $=(1-2+4)^{6}=3^{6}$

$=729$

$+\ldots+x^{2015}(1+x)+x^{2016}........(i)$

$\left(\frac{x}{1+x}\right) S=x(1+x)^{2015}+x^{2}(1+x)^{2014}$

$+\ldots +x^{2016}+\frac{x^{2017}}{1+x}........(ii)$

Subtracting $(i)$ from $(ii)$

$\frac{S}{1+x}=(1+x)^{2016}-\frac{x^{2017}}{1+x}$

$\therefore \quad S=(1+x)^{2017}-x^{2017}$

$a_{17}=$ coefficient of $x^{17}=^{2017} C_{17}$

$=\frac{2017 !}{17 ! 2000 !}$

By solving we get $r=6$

so, it is $7^{\text {th }}$ term

$^{8} \mathrm{C}_{\mathrm{r}}\left(2 \mathrm{x}^{2}\right)^{8-\mathrm{r}}\left(\frac{-1}{\mathrm{x}}\right)^{\mathrm{r}}$

$\therefore $ Given expression is equal to

$\left(1-\frac{1}{x}+3 x^{5}\right)^{8} C_{r}\left(2 x^{2}\right)^{8-\tau}\left(-\frac{1}{x}\right)^{r}$

${ = ^8}{{\text{C}}_{\text{r}}}{\left( {2{{\text{x}}^2}} \right)^{8 - {\text{r}}}}{\left( { - \frac{1}{{\text{x}}}} \right)^{\text{r}}} - {\frac{1}{{\text{x}}}^8}{{\text{C}}_{\text{r}}}{\left( {2{{\text{x}}^2}} \right)^{8 - {\text{r}}}}{\left( { - \frac{1}{{\text{x}}}} \right)^r}$

$ + 3{{\text{x}}^5}{ \cdot ^8}{{\text{C}}_{\text{r}}}{\left( {2{{\text{x}}^2}} \right)^{8 - {\text{r}}}}{\left( { - \frac{1}{{\text{x}}}} \right)^{\text{r}}}$

${ = ^8}{{\text{C}}_{\text{T}}}{2^{8 - {\text{r}}}}{( - 1)^{\text{r}}}{{\text{x}}^{16 - 3{\text{r}}}}{ - ^8}{{\text{C}}_{\text{r}}}{2^{8 - {\text{r}}}}{( - 1)^{\text{r}}}{{\text{x}}^{15 - 3{\text{r}}}}$

$ + 3{ \cdot ^8}{{\text{C}}_{\text{r}}}{2^{(8 - {\text{r}})}}{\left( { - \frac{1}{{\text{x}}}} \right)^{\text{r}}}{( - 1)^{\text{r}}}{{\text{x}}^{21 - 3{\text{r}}}}$

For the term independent of $x,$ we should have

$16-3 r=0,15-3 r=0,21-3 r=0$

From the simplification we get $r=5$ and $r=7$

$\therefore-^{8} C_{5}\left(2^{3}\right)(-1)^{5}-3 \cdot^{8} C_{7} \cdot 2$

$+\left[\frac{8 !}{5 ! 3 !} \times 8\right]-3 \times\left[\frac{8 !}{7 ! 1 !} \times 2\right]$

$=(56\times 8)-48$

$=448-6\times 8=448-48=400$

$s = {^{50}}{c_0} + {^{50}}{c_2} \cdot {2^2} + {^{50}}{c_4}{\left( 2 \right)^4} + .... + {^{50}}c_{502}^{50}$

${\left( {1 + x} \right)^{50}} = 1 + {^{50}}{C_1}{x^1} + {^{50}}{c_2}{x^2} + ...$

$x = 2, - 2$

${\left( {1 + 2} \right)^{50}} = 1 + {^{50}}{c_1}\left( 2 \right) + {^{50}}{c_2}{\left( 2 \right)^2} + ..$

${\left( {1 - 2} \right)^{50}} = 1 - {^{50}}{c_1}\left( 2 \right) + $${^{50}}{c_2}{\left( 2 \right)^2} - {^{50}}{C_3}{\left( 2 \right)^3} + .....$

${3^{50}} + 1$$ = 2\left[ {{^{50}}{C_0} + {^{50}}{c_2}{{\left( 2 \right)}^2} + {^{50}}{c_4}{{\left( 2 \right)}^4} + ..} \right]$

$\frac{{{3^{50}} + 1}}{2}$$ = {^{50}}{C_0} + {^{50}}{c_2}\left( 2 \right) + {^{50}}{c_4}{\left( 2 \right)^4} + ...$

Coefficient of $x^{3}$ in $\left(1+a x+b x^{2}\right)(1-2 x)^{18}$

$=$ Coefficient of $x^{3}$ in $(1-2 x)^{18}$

${+\text { Coefficient of } x^{2} \text { in a }(1-2 x)^{\text {18 }}}$

${\text { + Coefficient of } x \text { in } b(1-2 x)^{\text {18 }}}$

$=-^{18} \mathrm{C}_{3} \cdot 2^{3}+\mathrm{a}^{18} \mathrm{C}_{2} \cdot 2^{2}-\mathrm{b}^{18} \mathrm{C}_{1} \cdot 2$

$\because-^{18} C_{3} \cdot 2^{3}+a^{18} C_{2} \cdot 2^{2}-b^{18} C_{1} \cdot 2=0$

$\Rightarrow \frac{18 \times 17 \times 16}{3 \times 2} \cdot 8+a \cdot \frac{18 \times 17}{2} 2^{2}-b \cdot 18 \cdot 2=0$

$\Rightarrow 17 a-b=\frac{34 \times 16}{3}..........(i)$

Similarly, coefficient of $x^{4}$

$^{18} \mathrm{C}_{4} \cdot 2^{4}-\mathrm{a} \cdot^{18} \mathrm{C}_{3} 2^{3}+b \cdot^{18} \mathrm{C}_{2} \cdot 2^{2}=0$

$32 a-32 b=240..........(ii)$

On solving Eqs. $(i)$ and $(ii)$, we get

$a=16 \text { and } b=\frac{272}{3}$

$\mathrm{S}=(1+x)^{1000}+x(1+x)^{999}+x^{2}$

$(1+x)^{998}+\ldots+\ldots+x^{1006}$

Put $1+x=t$

$\mathrm{S}=t^{1000}+x t^{999}+x^{2}(t)^{998}+\ldots+x^{1000}$

This is a $G.P$ with common ratio $\frac{x}{t}$

$S=\frac{t^{1000}\left[1-\left(\frac{x}{t}\right)^{1001}\right]}{1-\frac{x}{t}}$

${ = \frac{{{{(1 + x)}^{1000}}\left[ {1 - {{\left( {\frac{x}{{1 + x}}} \right)}^{1001}}} \right]}}{{1 - \frac{x}{{1 + x}}}}}$

$ = \frac{{{{(1 + x)}^{1001}}\left[ {{{(1 + x)}^{1001}} - {x^{1001}}} \right]}}{{{{(1 + x)}^{1001}}}}$

$=\left[(1+x)^{100 t}-x^{1001}\right]$

Now coeff of $x^{50}$ in above expansion is equal to coeff of $x^{50}$ in $(1+x)^{1001}$ which is 

$^{1001} C_{50}=\frac{-(1001) !}{50 !(951) !}$

$\left(2+\frac{x}{3}\right)^{55}=2^{55}\left(1+\frac{x}{6}\right)^{55}$

$(r+1)^{\text {th }}$ term $=2^{55\, 55} \mathrm{C}_{r}\left(\frac{x}{6}\right)^{r}$

Coefficient of $x^{r}$ is $2^{55\, 55} \mathrm{C}_{r} \frac{1}{6^{r}}$

$(r+2)^{t h}$ term $=2^{55\,55} C_{r+1}\left(\frac{x}{6}\right)^{r+1}$

Coefficient of $x^{r+1}$ is $2^{55\, 55} \mathrm{C}_{r+1} \cdot \frac{1}{6^{r+1}}$

Both coefficients are equal

$2^{55\,55} C_{r} \frac{1}{6^{r}}=2^{55\, 55} \mathrm{C}_{r+1} \frac{1}{6^{r+1}}$

$\frac{1}{{r!55 - r!}} = \frac{1}{{r - 1!54 - r!}}.\frac{1}{6}$

$6(r+1)=55-r$

$6 r+6=55-r$

$7 r=49$

$r=7$

$(r+1)=8$

Coefficient of $8^{th}$ and $9^{th}$ terms are equal.

$ = {a_0} + {a_1}{(1 + x)^1} + {a_2}{(1 + x)^2} + {a_3}{(1 + x)^\beta }$

$ + {a_4}{(1 + x)^4} + {a_5}{(1 + x)^5}$

$ \Rightarrow 1 + {x^4} + {x^5}$

$ = {a_0} + {a_1}(1 + x) + {a_2}\left( {1 + 2x + {x^2}} \right)$

$ + {a_3}\left( {1 + 3x + 3{x^2} + {x^3}} \right)$

$ + {a_4}\left( {1 + 4x + 6{x^2} + 4{x^3} + {x^4}} \right)$

$ + {a_5}(1 + 5x + 10{x^2} + 10{x^3} + 5{x^4} + {x^5})$

$ \Rightarrow 1 + {x^4} + {x^5}$

$ = {a_0} + {a_1} + {a_1}x + {a_2} + 2{a_2}x + {a_2}{x^2}$

$ + {a_3} + 3{a_3}x + 3{a_3}{x^2} + {a_3}{x^3}$

$ + {a_4} + 4{a_4}x + 6{a_4}{x^2} + 4{a_4}{x^3} + {a_4}{x^4}$

$ + {a_5} + 5{a_5}x + 10{a_5}{x^2} + 10{a_5}{x^3} + 5{a_5}{x^4} + {a_5}{x^5}$

$\Rightarrow 1+x^{4}+x^{5}$

$\left(a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)$

$=+x\left(a_{1}+2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}\right)$

$+x^{2}\left(a_{2}+3 a_{3}+6 a_{4}+10 a_{5}\right)+x^{3}\left(a_{3}+4 a_{4}+10 a_{5}\right)$

$+x^{4}\left(a_{4}+5 a_{5}\right)+x^{5}\left(a_{5}\right)$

On comparing the like coefficients, we get

$\boxed{{a_5} = 1}.........(1)$

$\boxed{{a_4} + 5{a_5} = 1}.........(2)$

$\boxed{{a_3} + 4{a_4} + 10{a_5} = 0}.........(3)$

and $\boxed{{a_2} + 3{a_3} + 6{a_4} + 10{a_5} = 0}.........(4)$

from$(1)$ and $(2)$, we get

$\boxed{{a_4} =  - 4}.........(5)$

from $(1),(3)$  and  $(5),$ we get

$\boxed{{a_3} =  + 6}.........(6)$

Now, from $(1)$, $(5)$ and $(6)$, we get 

$a_{2}=-4$

Let $x^{1012}=(1)^{a}\left(x^{n}\right)^{b} \cdot\left(x^{253}\right)^{c}$

Here $a, b, c, n$ are all $+ve$ integers and

$a$ $\leq 10, b \leq 10, c \leq 4, n \leq 22, a+b+c=10$

Now $b n+253 c=1012$

$\Rightarrow b n=253(4-c)$

For $c<4$ and $n \leq 22 ; b>10,$ which is not possible.

$\therefore c=4, b=0, a=6$

$\therefore x^{1012}=(1)^{6} \cdot\left(x^{12}\right)^{0} \cdot\left(x^{253}\right)^{4}$

Hence the coefficient of $x^{1012}$

$=\frac{10 !}{6 ! 0 ! 4 !}=10 \mathrm{C}_{4}$

$=(1+x)(1+x)^{100}\left(1-x+x^{2}\right)^{100}$

$=(1+x)\left[(1+x)\left(1-x+x^{2}\right)\right]^{100}$

$=(1+x)\left[\left(1-x^{3}\right)^{100}\right]$

Expansion $\left(1-x^{3}\right)^{100}$ will have

$100+1$ $=101$ terms

$\mathrm{So},(1+x)\left(1-x^{3}\right)^{100}$ will have

$2 \times 101$

$=202$ terms

$\left(x^{13}-x^{-1 / 2}\right)^{10}$

${T_{r + 1}}{ = ^{10}}{C_r}{({x^{1/3}})^{10 - r}}{( - {x^{ - 1/2}})^r}$

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

$\Rightarrow \quad 20-2 r-3 r=0$

$\Rightarrow \quad r=4$

$T_{5}=^{10} C_{4}=\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210$

$=15 \mathrm{C}_{r} \times(2)^{r} \times x^{30-3r}$

For independent term, $30-3 r=0$

$\Rightarrow r=10$

Hence the term independent of $x$.

$\mathrm{T}_{11}=^{15} \mathrm{C}_{10} \times(2)^{10}$

For term involve $x^{15}, 30-3 r=15$

$ \Rightarrow r=5$

Hence coefficient of $x^{15}=^{15} C_{5} \times(2)^{5}$

Required ratio

$ = \frac{{{\text{15}}{{\text{C}}_5} \times {{(2)}^5}}}{{15{C_{10}} \times {{(2)}^{10}}}} = \frac{{\frac{{15!}}{{10!5!}}}}{{\frac{{15!}}{{5!10!}} \times {{(2)}^5}}}$

$ = 1:32$

$+^{10} \mathrm{C}_{1}\left(2^{1 / 2}\right)^{9}\left(3^{1 / 5}\right)+\ldots \ldots+^{10} \mathrm{C}_{10}\left(3^{1 / 5}\right)^{10}$

There are onlytwo rational terms - first term and last term.

Now sum of two rational terms

$=(2)^{5}+(3)^{2}=32+9=41$

$+\ldots \ldots+^{2 n} C_{r} x^{r}+^{2 n} C_{r+1} x^{r+1}+\ldots \ldots+^{2 n} C_{2 n} x^{2 n}$

As given $^{2n}{{\text{C}}_{r + 2}}{ = ^{2n}}{{\text{C}}_{3r}}$

$ \Rightarrow \frac{{(2n)!}}{{(r + 2)!(2n - r - 2)!}} = \frac{{(2n)!}}{{(3r)!(2n - 3r)!}}$

$\Rightarrow(3 r) !(2 n-3 r) !=(r+2) !(2 n-r-2) !.........(1)$

Now, put value of $n$ from the given choices.

Choice $(a)$ put $n=2 r+1$ in $(1)$

$LHS:$ $(3 r) !(4 r+2-3 r) !=(3 r) !(r+2) !$

$\mathrm{RHS}:(r+2) !(3 r) !$

$\Rightarrow \mathrm{LHS}=\mathrm{RHS}$

$(1+\sqrt{2})^{ n }= m _1+\sqrt{2} n _1, m _1, n _1 \in Z$

$\Rightarrow Z \cup T _1 \cup T _2 \subseteq S$

but $b \sqrt{2} \in S$ for negative $b \in Z$.

So $\quad Z \cup T _1 \cup T _2 \subset S$

$(B)(\sqrt{2}-1)^{ n }=\frac{1}{(\sqrt{2}+1)^2}<\frac{1}{2024}$

$\Rightarrow 2024<(\sqrt{2}+1)^{ n }, \exists n \in N$

$\Rightarrow T _1 \cap\left(0, \frac{1}{2024}\right) \neq \phi$

$(C)(1+\sqrt{2})^2>2024, \exists n \in N$

$\Rightarrow T _2 \cap(2024, \infty) \neq \phi$

$(D)\sin (\pi(a+b \sqrt{2})=0) \Rightarrow b=0, a \in Z$.

$\Rightarrow$ Options $(A), (C), (D)$ are Correct.

$={ }^{+} \mathrm{C}_{\mathrm{r}} \cdot \mathrm{a}^{4-\tau} \cdot \frac{70^r}{(27 \mathrm{~b})^r} \mathrm{x}^{s-3 t}$

here $8-3 r=5$

$8-5=3 r \Rightarrow r=1$

$\therefore \text { coeff. }=4 a^3 \cdot \frac{70}{27 b}$

$T_{r+1}={ }^7 C_r(a x)^{7-r}\left(\frac{-1}{b x^2}\right)^r$

$={ }^7 C_r \cdot a^{7-r}\left(\frac{-1}{b}\right)^r \cdot x^{7-3 r}$

$7-3 r=-5 \Rightarrow 12=3 r \Rightarrow r=4$

coeff. : ${ }^7 C_4 \cdot a^3 \cdot\left(\frac{-1}{b}\right)^4=\frac{35 a^3}{b^4}$

now $\frac{35 a^5}{b^4}=\frac{280 a^3}{27 b}$

$b^3=\frac{35 \times 27}{280}=b=\frac{3}{2} \Rightarrow 2 b=3$

$X=\sum_{r=0}^{10}(10-r)\left({ }^{10} C_{10-r}\right)^2$

$2 X=\sum_{r=0}^{10} 10\left({ }^{10} C_r\right)^2$

$X=5 \cdot{ }^{20} C_{10} \Rightarrow \frac{X}{1430}=646.00$

$=1+{ }^3 C_2+{ }^4 C_2+.{ }^5 C_2+\ldots \ldots+{ }^{49} C_2+$

${ }^{50} C_2 \cdot m^2$

$\text { [as } .^n C_r+.{ }^n C_{r-1}=.{ }^{n+1} C_r \text { ] }$

$=\left(.{ }^3 C_5+.{ }^3 C_2\right)+.{ }^4 C_2+.{ }^5 C_2+\ldots .+.{ }^{49} C_2+$

${ }^{50} C_2 m^2$

$=\left(.{ }^4 C_3+.{ }^4 C_2\right)+\ldots . .+{ }^{50} C_2 m^2$

$=.{ }^5 C_3+.{ }^{50} C_2 m^2+.{ }^{50} C_2 m^2$

$=.{ }^{50} C_3+.{ }^{50} C_2 m^2+.{ }^{50} C_2-.{ }^{50} C_2$

$=.{ }^{51} C_3+.{ }^{50} C_2\left(m^2-1\right)$

$=(3 n+1) .{ }^{51} C_3 \text { (given) }$

$\therefore 3 n . \frac{51}{3} .{ }^{50} C_2=.{ }^{50} C_2\left(m^2-1\right)$

$\frac{m^2-1}{51}=n$

Value of $n$ is $5$ .

i.e. $x ^9, x ^{1+8}, x ^{2+7}, x ^{3+6}, x ^{4+5}, x ^{1+2+6}, x ^{1+3+5}, x ^{2+3+4}$ and coefficient in each case is $1$ $\Rightarrow$ Coefficient of $x^9=1+1+1+\ldots \ldots \ldots+1=8$

Coefficent of $x^{11} \equiv\left(1-x^8\right)^4\left(1+x^4\right)^8\left(1+x^3\right)^7\left(1-x^2\right)^{-4}$

$=\left(1-4 x^8\right)\left(1+x^4\right)^8\left(7 x^3+35 x^9\right)\left(1-x^2\right)^{-4} $

$=\left(7 x^3+35 x^9-28 x^{11}\right)\left(1+x^4\right)^8\left(1-x^2\right)^{-4}$

Coefficent of $x^8=\left(7 x+35 x^6-28 x^8\right)\left(1+8 x^4+28 x^8\right)\left(1-x^2\right)^{-4}$

$=\left(7+35 x^6-28 x^3+56 x^4+196 x^8\right)\left(1-x^2\right)^{-4}$

Coefficent of $t ^4 \equiv\left(7+56 t ^2+35 t ^3+168 t ^4\right)(1- t )^{-4}$

$=7 \cdot{ }^7 C _3+56 \cdot{ }^5 C _3+35 \cdot{ }^4 C _3+168 $

$=245+700+168=1113 .$

$Alterantive :$

$2 x+3 y+4 z=11 $

$(x, y, z)=(0,1,2){ }^4 C_0 \times{ }^7 C_1 \times{ }^{12} C_2 $

$(1,3,0)^4 C_1 \times{ }^7 C_3 $

$(2,1,1)^4 C_2 \times{ }^7 C_1 \times{ }^{12} C_1 $

$(4,1,0)^7 C_1 $

$\text { coefficient of } x^{11} =66 \times 7+35 \times 4+42 \times 12+7 $

$\quad=1113$ Ans.

$\Rightarrow \quad \frac{{ }^{n+5} C_r}{{ }^{n+5} C_{r-1}}=\frac{10}{5} \quad \& \quad \frac{{ }^{n+5} C_{r+1}}{{ }^{n+5} C_r}=\frac{14}{10} $

$\Rightarrow \quad \frac{(n+5)-r+1}{r}=2 \quad \& \quad \frac{(n+5)-(r+1)+1}{r+1}=\frac{7}{5}$

$\Rightarrow \quad \frac{n+6}{r}=3 \quad  \&  \quad \frac{n+6}{r+1}=\frac{12}{5} $

$\Rightarrow \quad 3 r=\frac{12}{5}(r+1) $

$\Rightarrow \quad r=4 $

$\therefore \quad n+6=12 \quad \Rightarrow \quad n=6 $

$ \sum_{\mathrm{r}=1}^{10} \mathrm{~A}_{\mathrm{r}} \mathrm{B}_{\mathrm{r}}=\text { coefficient of } \mathrm{x}^{20} \text { in }\left((1+\mathrm{x})^{10}(\mathrm{x}+1)^{20}\right)-1 $

$ =C_{20}-1=\mathrm{C}_{10}-1 \text { and } \sum_{\mathrm{r}=1}^{10}\left(\mathrm{~A}_{\mathrm{r}}\right)^2=\text { coefficient of } \mathrm{x}^{10} \text { in }\left((1+\mathrm{x})^{10}(\mathrm{x}+1)^{10}\right)-1=\mathrm{B}_{10}-1 $

$ \Rightarrow \mathrm{y}=\mathrm{B}_{10}\left(\mathrm{C}_{10}-1\right)-\mathrm{C}_{10}\left(\mathrm{~B}_{10}-1\right)=\mathrm{C}_{10}-\mathrm{B}_{10} .$

${(x + 1)^{30}} = {\,^{30}}{C_0}{x^{30}} + {\,^{30}}{C_1}{x^{29}} + {\,^{30}}{C_2}{x^{28}}+ ...... + {\,^{30}}{C_{10}}{x^{20}} + .... + {\,^{30}}{C_{30}}{x^0}$....$(ii)$

Multiplying (i) and (ii) and equating the coefficient of $x^{20}$ on both sides, we get required sum 

= coefficient of $x^{20}$ in $(1 -x^2)^{30}={^{30}}{C^{10}}$.

$(1 + {t^{12}} + {t^{24}} + {t^{36}})$  

$\therefore$ Coefficient of ${t^{24}}{ = ^{12}}{C_6} + 2.$

$ = 1 + (m - n)x + \left[ {\frac{{{n^2} - n}}{2} - mn + \frac{{({m^2} - m)}}{2}} \right]{x^2}$+........

Given, $m -n = 3$ or $n = m -3$ 

Hence $\frac{{{n^2} - n}}{2} - mn + \frac{{{m^2} - m}}{2} = - 6$

==> $\frac{{(m - 3)(m - 4)}}{2} - m(m - 3) + \frac{{{m^2} - m}}{2} = - 6$

==> ${m^2} - 7m + 12 - 2{m^2} + 6m + {m^2} - m + 12 = 0$

==> $ - 2m + 24 = 0\,\,\, \Rightarrow m = 12$

==> ${2.^n}{C_2} = {\,^n}{C_1} + {\,^n}{C_3}$

==> $2.\frac{{n(n - 1)}}{{1.2}} = \frac{n}{1} + \frac{{n(n - 1)(n - 2)}}{{1.2.3}}$

==> $6(n - 1) = 6 + (n - 2)(n - 1)$

==> ${n^2} - 9n + 14 = 0$

==> $n = 2$or $n = 7$.

But $n = 2$ is not acceptable because, when $n=2$, there are only three terms in the expansion of ${(1 + x)^2}$, $\therefore $ $n = 7$.

$\therefore \,\,\,\frac{{{T_{r + 1}}}}{{{T_r}}} \ge 1\,\,\, \Rightarrow 102 - 2r \ge 15r \Rightarrow r \le 6$

Therefore by hypothesis ${({\alpha ^2} - 2\alpha + 1)^{51}} = 0 \Rightarrow \alpha = 1$

==> $5\sqrt 5 - 11 = \frac{4}{{5\sqrt 5 + 11}}$,,$\,\,\,\,0 < 5\sqrt 5  - 11 < 1 \Rightarrow \,0 < {(5\sqrt 5  - 11)^{2n + 1}} < 1$ for positive integral $n$.

Again, ${(5\sqrt 5 + 11)^{2n + 1}} - {(5\sqrt 5 - 11)^{2n + 1}}$

$ = 2\left\{ {^{2n + 1}{C_1}{{(5\sqrt 5 )}^{2n}}.11 + {\,^{2n + 1}}{C_3}{{(5\sqrt 5 )}^{2n - 2}}} \right.$

$\left. { \times {{11}^3} + ....{ + ^{2n + 1}}{C_{2n + 1}}{{11}^{2n + 1}}} \right\}$

$ = 2\,\left\{ {^{2n + 1}{C_1}{{(125)}^n}.11{ + ^{2n + 1}}} \right.{C_3}{(125)^{n - 1}}{11^3} + ..$

$\left. {....{ + ^{2n + 1}}{C_{2n + 1}}{{11}^{2n + 1}}} \right\}$

$= 2k$, (for some positive integer $k$)

Let $f' = {(5\sqrt 5 - 11)^{2n + 1}}$, then $[R] + f - f' = 2k$

==> $f - f' = 2k - [R]\,\,\, \Rightarrow \,\,\,f - f'$ is an integer.

But, $0 \le f < 1;\,0 < f' < 1\,\,\, \Rightarrow - 1 < f - f' < 1$

==> $f - f' = 0$(integer) ==> $f = f'$

$\therefore \,\,Rf = Rf' = {(5\sqrt 5 + 11)^{2n + 1}}{(5\sqrt 5 - 11)^{2n + 1}}$

$={[{(5\sqrt 5 )^2} - {11^2}]^{2n + 1}} = {4^{2n + 1}}$.

$ = [C_0^2 - C_1^2 + C_2^2 - .... + {( - 1)^n}C_n^2] - [C_1^2 - 2C_2^2 + 3C_3^2.... - {( - 1)^n}nC_n^2]$

=${( - 1)^{n/2}}\frac{{n!}}{{(n/2)!(n/2)!}} - {( - 1)^{(n/2) - 1}}.\frac{1}{2}n\,{\,^n}{C_{n/2}}$

=${( - 1)^{n/2}}.\frac{{n!}}{{(n/2)!(n/2)!}}.\left( {1 + \frac{n}{2}} \right)$

$\therefore$ the value of the given expression is

$\frac{{2(n/2)\,!\,(n/2)!}}{{n!}} \times {( - 1)^{n/2}}.\frac{{(n)!}}{{(n/2)!(n/2)!}}\left( {1 + \frac{n}{2}} \right)$

$ = {( - 1)^{n/2}}(2 + n)$

==> $1 + \frac{n}{1}ax + \frac{{n(n - 1)}}{{1.2}}{a^2}{x^2} + .... = 1 + 8x + 24{x^2} + ....$

$ \Rightarrow \,\,\,\,na = 8,\frac{{n(n - 1)}}{{1.2}}{a^2} = 24 \Rightarrow na(n - 1)a = 48$

==> $8(8 - a) = 48$==> $8 - a = 6\,\, \Rightarrow a = 2\,\, \Rightarrow n = 4$.

the general term is ${T_{r + 1}} = {\,^{10}}{C_r}{\left( {\frac{x}{2}} \right)^{10 - r}}.\,\,{\left( { - \frac{3}{{{x^2}}}} \right)^r}$ 

${ = ^{10}}{C_r}{( - 1)^r}.\frac{{{3^r}}}{{{2^{10 - r}}}}{x^{10 - r - 2r}}$

Here, the exponent of $x$ is $10 - 3r = 4 \Rightarrow r = 2$

$\therefore \,\,\,\,{T_{2 + 1}}{ = ^{10}}{C_2}{\left( {\frac{x}{{\rm{2}}}} \right)^8}{\left( { - \frac{3}{{{x^2}}}} \right)^2}$

= $\frac{{10.9}}{{1.2}}.\frac{1}{{{2^8}}}{.3^2}.{x^4}$ 

= $\frac{{405}}{{256}}{x^4}$

$\therefore $ The required coefficient $ = \frac{{405}}{{256}}$.

we get sum of the coefficients as ${(1 + 1 - 3)^{2163}} = {( - 1)^{2163}} = - 1$.

and ${99^{50}} = {(100 - 1)^{50}} = {100^{50}} - {50.100^{49}} + \frac{{50.\,49}}{{2.1}}{100^{48}} - ..…..(ii)$ 

Subtracting $(ii)$ from $(i)$, we get 

${101^{50}} - {99^{50}} = {100^{50}} + 2\frac{{50.49.48}}{{1.2.3}}{100^{47}} > {100^{50}}$

Hence ${101}^{50}$ > ${100}^{50}$+${99}^{50}$.

Putting $n = 2$

${3^{2 \times 2 + 2}} - 8 \times 2 - 9 = 729 - 16 - 9 = 704$

It is divisible by $16.$

Then ${a_1} = {\,^n}{C_r},{a_2} = {\,^n}{C_{r + 1}},{a_3} = {\,^n}{C_{r + 2,}}{a_4} = {\,^n}{C_{r + 3}}$

Now $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}} = \frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}$$ + \frac{{^n{C_{r + 2}}}}{{^n{C_{r + 2}} + {\,^n}{C_{r + 3}}}}$

$ = \frac{{^n{C_r}}}{{^{n + 1}{C_{r + 1}}}} + \frac{{^n{C_{r + 2}}}}{{^{n + 1}{C_{r + 3}}}}$$ = \frac{{^n{C_r}}}{{\frac{{n + 1}}{{r + 1}}{\,^n}{C_r}}} + \frac{{^n{C_{r + 2}}}}{{\frac{{n + 1}}{{r + 3}}{\,^n}{C_{r + 2}}}}$
                                                

                                                 $({^n}{C_r} = \frac{n}{r}{\,^{n - 1}}{C_{r - 1}})$

= $\frac{{r + 1}}{{n + 1}} + \frac{{r + 3}}{{n + 1}} = \frac{{2(r + 2)}}{{n + 1}}$

$ = 2\frac{{^n{C_{r + 1}}}}{{^{n + 1}{C_{r + 2}}}} = 2\frac{{^n{C_{r + 1}}}}{{^n{C_{r + 1}} + {\,^n}{C_{r + 2}}}} = \frac{{2{a_2}}}{{{a_2} + {a_3}}}$

$ = a.0 + 0 = 0$

Only option $(a)$ gives the value.

$\Rightarrow 11 - r - 2r = - 7$

$ \Rightarrow r = 6$

Thus coefficient of $x^{-7}$ is $^{11}{C_6}{(a)^5}{\left( { - \frac{1}{b}} \right)^6} $

$= \frac{{462}}{{{b^6}}}{a^5}$

==> $\frac{{{{(1 + x)}^{15}} - 1}}{x} = {C_1} + {C_2}x + .... + {C_{15}}{x^{14}}$

Differentiating both sides with respect to $ x$, 

we get $ = \frac{{x.15{{(1 + x)}^{14}} - {{(1 + x)}^{15}} + 1}}{{{x^2}}}$

= ${C_2} + 2{C_3}x + ...... + {\,^{14}}{C_{15}}{x^{13}}$

Putting $x = 1$, 

we get ${C_2} + 2{C_3} + .... + 14{C_{15}} = {15.2^{14}} - {2^{15}} + 1 = {13.2^{14}} + 1.$