Maths M.C.Q (1 Marks)

Equating it to $\frac{1}{4} - 3x + ....,$ we get $a = 2,b = 12$.

$ = {6^{ - 1/3}}\left[ {1 + \left( { - \frac{1}{3}} \right)\,\left( { - \frac{x}{2}} \right)x + \frac{{\left( { - \frac{1}{3}} \right)\,\left( { - \frac{4}{3}} \right)}}{{2.1}}{{\left( { - \frac{x}{2}} \right)}^2} + ....} \right]$

$ = {6^{ - 1/3}}\left[ {1 + \frac{x}{6} + \frac{{2{x^2}}}{{{6^{^2}}}} + ....} \right]$

$ { + \frac{{(r + 1)(r + 2)(r + 3)}}{6}{x^r} + ....} ]$

Therefore ${T_{r + 1}} = \frac{{(r + 1)(r + 2)(r + 3)}}{6}{x^r}$.

$ = 1 + ny + \frac{{n(n - 1)}}{{2!}}{y^2} + .....$

Comparing the terms, we get

$ny = \frac{1}{3}x,\frac{{n(n - 1)}}{{2!}}{y^2} = \frac{{1.4}}{{3.6}}{x^2}$

Solving, $n = - \frac{1}{3},y = - x$.

Hence given series $ = {(1 - x)^{ - 1/3}}$

$\therefore \,\,1 + y = 1 + 3x + 6{x^2} + 10{x^3} + ...$

==> $1 + y = {(1 - x)^{ - 3}}\,\,\,\, \Rightarrow 1 - x = {(1 + y)^{ - 1/3}}$

==> $x = 1 - {(1 + y)^{ - 1/3}}$

$ = \frac{{\sqrt {1 + {x^2}} + x}}{{1 + {x^2} - {x^2}}} = x + \sqrt {1 + {x^2}} = x + {(1 + {x^2})^{1/2}}$

$ = x + 1 + \frac{1}{2}{x^2} + \frac{1}{2}\left( { - \frac{1}{2}} \right)\frac{{{x^4}}}{{2!}} + ...$.

Obviously, coefficient of $x$ is $1$.

$\frac{{1\,.\,3\;.\,5...(2r - 1)}}{{{2^r}}}\,.\,\frac{{{{( - 1)}^r}{{( - 1)}^r}{2^r}}}{{r!}}$

= $\frac{{1\,.\,3\,.\,5...(2r - 1)}}{{r!}} = \frac{{2r!}}{{r!\,\,r!\,\,{2^r}}}$.

$ = 2\,[{x^5} + {\,^5}{C_2}{x^3}{\{ {({x^3} - 1)^{1/2}}\} ^2} + {\,^5}{C_4}x{\{ {({x^3} - 1)^{1/2}}\} ^4}]$

$=2[{x^5} + 10{x^3}({x^3} - 1) + 5x{({x^3} - 1)^2}]$

$=5{x^7} + 10{x^6} + {x^5} - 10{x^4} - 10{x^3} + 5x$,

which is a polynomial of degree $7$.

$^n{C_{r - 1}} = 28,{\,^n}{C_r} = 56$ and $^n{C_{r + 1}} = 70,$ so that

$\frac{{^n{C_r}}}{{^n{C_{r - 1}}}} = \frac{{n - r + 1}}{r} = \frac{{56}}{{28}} = 2$

and $\frac{{^n{C_{r + 1}}}}{{^n{C_r}}} = \frac{{n - r}}{{r + 1}} = \frac{{70}}{{56}} = \frac{5}{4}$

This gives $n + 1 = 3r$ and $4n - 5 = 9r$

 $\frac{{4n - 5}}{{n + 1}} = 3 \Rightarrow n = 8$

= $4\,{[^4}{C_0}{ + ^4}{C_1}(\sqrt 2 ){ + ^4}{C_2}{(\sqrt 2 )^2}{ + ^4}{C_3}{(\sqrt 2 )^3}{ + ^4}{C_4}{(\sqrt 2 )^4}]$

= $4\,\left[ {1 + 4\sqrt 2 + \frac{{4.3}}{2}.2 + \frac{{4.3.2}}{{1.2.3}}.2\sqrt 2 + 4} \right]$

= $4\,[1 + 4\sqrt 2 + 12 + 8\sqrt 2 + 4]$ = $4\,[17 + 12\sqrt 2 ]$

= $4\,[17 + (\tilde - 17)]$ = $4\,[\,34] = 136$.

$\therefore {3^{180}} = {({3^4})^{45}}$ terminates in $1$

Also ${3^3}$ $= 27$ terminates in $7$

 ${3^{183}} = {3^{180}}{3^3}$ terminates in $7.$

 $183! + {3^{183}}$ terminates in $7$

$i.e$. the digit in the unit place $= 7$.

= $5 \times (26) \times ({\rm{Positive terms)--5,}}$ So when it is divided by $13$ it gives the remainder $-5$ or $(13 -5)$ $i.e.,$ $8$.

==> ${({17^5})^6} = {(2)^6}$(mod $5$) 

==> ${(17)^{30}} = 4$ (mod $5$)Hence required remainder $= 4.$

$\sum \limits_{ k =0}^{10}{ }^{10} C _{ k } \cdot k ^2=\sum \limits_{ k =1}^{10} \frac{10}{ k } \cdot{ }^9 C _{ k -1} \cdot k ^2$

$10 \sum\limits_{ k =1}^{10}( k -1+1)^9 C _{ k -1}=10\left(\sum \limits_{ k =1}^{10} 9 \cdot{ }^8 C _{ k -2}+\sum \limits_{ k =1}^{10}{ }^9 C _{ k -1}\right)$

$10\left(9 \cdot 2^8+2^9\right)=10 \cdot 2^8(9+2)=110 \cdot 2^8$

$\frac{1}{2^{10}} \times 28 \times 110=\frac{2^9}{2^{10}} \cdot(55)=\frac{55}{2}$

$a^3+b^3+c^3=9 k_1 \Rightarrow a b c=3 k_2$

Possibility of $a, b, c$ can be

$3 k _1, 3 k _1+1$ or $3 k _1+2$

$C - I :\left(3 k_1\right)^3+\left(3 k_1+1\right)^3+\left(3 k_1+2\right)^3$

$=9 k ^{\prime}+9$

divisible by $9$

$a b c=\left(3 k_1\right)\left(3 k_1+1\right)\left(3 k_1+2\right)$ divisible by $3$

$C-II$ : $\left(3 k_1\right)^3+\left(3 k_2\right)^3+\left(3 k_3\right)^3=9 k^{\prime}$ divisible $9$

$a b c=\left(3 k_1\right)\left(3 k_2\right)\left(3 k_3\right)$ divisible by 3

$C-III$ :

$\left(3 k _1+1\right)^3+\left(3 k _2+1\right)^3+\left(3 k _3+1\right)^3+9 k ^{\prime}+3$

Not divisible by $9$

$C-IV$: $\left(3 k _1+2\right)^3+\left(3 k _2+2\right)^3+\left(3 k _3+2\right)^3$

$=9 k ^{\prime}+24$ not divisible by 9

$C - V :\left(3 k_1+2\right)^3+\left(3 k_2+2\right)^3+\left(3 k_3\right)^3$

$=9 k ^1+16$ not divisible by $9$

$C - V I :\left(3 k _1+1\right)^3+\left(3 k _2+1\right)^3+\left(3 k _3\right)^3$

$=9 k ^{\prime}+2$ not divisible by $9$

other then $C-I$ and $C-II$ all cases are not divisible by $9$

for $C-I$ and $C-II$ statement is true.

Hence statement $1$ is true.

Statement $2$

$a^3+b^3+c^3+d^3=1^3+2^3+1^3+2^3$ divisible

by 9 but $1.2$.$1.2$ not divisible by $3$

Hence statement $2$ is false

Alternate:

$a^3+b^3+c^3=9 k$

$=3 a b c+(a+b+c) \cdot\left((a+b+c)^2-3(a b+b c+c a)\right)$

$9 k =3 a b c+(a+b+c)^3-3(a+b+c)(a b+b c+c a)$

$\text { divide both side by } 3$

$3 k =a b c+\frac{(a+b+c)^3}{3}-(a+b+c)(a b+b c+c a)$

$\therefore a+b+c=3\,m \quad \because 3 \text { is prime }$

$\Rightarrow \therefore a b c=3\,t$

$\text { for second take } a=1, b=2, c=1, d=2$

$x ^2- x -1=0$

$x =\frac{1 \pm \sqrt{5}}{2}$

$\lambda=\frac{1+\sqrt{5}}{2}, 1-\lambda=\frac{1-\sqrt{5}}{2}$

$a _{ n }=\frac{1}{\sqrt{5}}\left(\lambda^{ n }-(1-\lambda)^{ n }\right) \forall n \in N$

$a _{ n }=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{ n }-\left(\frac{1-\sqrt{5}}{2}\right)^{ n }\right)$

$a _{ n }=\frac{1}{\sqrt{5} \cdot 2^{ n }}\left[(1+\sqrt{5})^{ n }-(1-\sqrt{5})^{ n }\right]$

$(1+\sqrt{5})^{ n }=1+{ }^{ n } C_1 \sqrt{5}+{ }^{ n } C_2(\sqrt{5})^2+\ldots$

$(1-\sqrt{5})^{ n }=1-{ }^{ n } C_1 \sqrt{5}+{ }^{ n } C_2(\sqrt{5})^2-\ldots$

$(1+\sqrt{5})^{ n }-(1-\sqrt{5})^{ n }$

$=2\left[{ }^{ n } C _1 \sqrt{5}+{ }^{ n } C_3 5 \sqrt{5}+{ }^{ n } C_5 5^2 \sqrt{5}+\ldots\right]$

So $a _{ n }=\frac{\left[{ }^{ n } C_1+{ }^{ n } C_3 5+{ }^{ n } C_5 5^2+\ldots\right]}{2^{ n -1}}$

$a _{ n }=\frac{2 \Sigma^{ n } C_{ r } 5^{ r }}{2^{ n }}$ where $r$ is odd

$a _{ n }=\frac{(1+5)^{ n }-(1-5)^{ n }}{2^{ n }}=\frac{6^{ n }-(-4)^{ n }}{2^{ n }}$

$a _{ n }=3^{ n }-(-2)^{ n } \in I \forall n \in N$

$\Rightarrow A \in \phi ; B \in \phi$

We have,

$(2021)^{2020}=(1+2020)^{2020}$

$={ }^{2020} C_0+{ }^{2021} C_1 \cdot 2020+{ }^{2020} C_2 \cdot(2020)^2+\ldots$ $=1+(2020)^2+(2020)^2 \times \lambda$, where $\lambda \in Z^{+}$ when divided by (2020) remainder will be $1$

Since, Statement $I$ does not hold for $n=4$

$\Rightarrow$ Statement $I$ is false.

For Statement $II$,

$n(n+1)^2 \mid n !$

$(n+1)^2 \mid(n-1) !$

Let $n=3 k-1, k > 3, k \in N$

$n+1 =3 k, n-1=3 k-2$

$(n-1) ! =(3 k-2) !$

Since, RHS contains $3^2, k^2$

Hence, it is divisible by $(3 k)^2$.

$\Rightarrow(n-1)$ ! is divisible by $(n+1)^2$

$\Rightarrow$ Statement $II$ is true.

As $N$ be the least positive integer and when a non-zero $\operatorname{digit} C$ is written af ter the last digit of $N$, the resulting number is divisible by $C$.

So, $10 N+C$ is divisible by $C$

$\therefore 10 N$ must be divisible by $C$.

Now, the least integer $(N)$ which is divisible by digit ' $C$ ' i.e. ( 1 to 9$)$ must be L.C.M of $\{1,3,4,6,7,9\}$.

$= L.C.M$ $of$ $\{4,7,9\}$

$=252=N$

and sum of digits of number ' $N$ ' is

$2+5+2=9$

It is given that, $N_2=165$ $=3 \times 5 \times 11$ and $N_1=2^{55}+1$

As we know that, if $n$ is odd integer then $x^n+y^n$ is divisible by $x+y$.

So, $N_1=2^{55}+1^{55}$ is divisible by $2+1=3$ and $N_1=2^{55}+1^{55}$

$=\left(2^5\right)^{11}+\left(1^5\right)^{11}=(32)^{11}+(1)^{11}$

is divisible by $32+1=33$

$\therefore$ the HCF of $N_1$ and $N_2$ is $33$

We have,

$F_1$ and $F_2$ be the fractional part of

$(44-\sqrt{2017})^{2017}$ and $(44+\sqrt{2017})^{2017}$

$\Rightarrow \quad(44+\sqrt{2017})^{2017}=I+F_2$

$\Rightarrow \quad(44-\sqrt{2017})^{2017}=F_1$

$[\because 0<44-\sqrt{2017}<1]$

$\therefore I+F_1+F_2$

$=2\left({ }^{2017} C_{ \theta }(44)^{2017}+{ }^{2017} C_2(44)^{2015}\right.$

$(2017)+\ldots]$

$\Rightarrow I+F_1+F_2=2$ (integer)

$\therefore I+F_1+F_2$ is an even integer.

$\therefore F_1+F_2$ is also integer.

But, $F_1+F_2$ is an integer.

$\therefore \quad F_1+F_2=1$

Hence, $F_1^{\prime}+F_2$ is lie between $0.9$ and $1.35$

Let

$\quad p(x) =1+x^2+x^4+x^6+\ldots+x^{34}$

$\Rightarrow \quad p(x) =\frac{x^{36}-1}{x^2-1}$

$\Rightarrow \quad q(x) =1+x+x^2+x^3+\ldots+x^{17}$

$\Rightarrow \quad q(x) =\frac{x^{18}-1}{x-1}$

$\therefore \quad \frac{p(x)}{q(x)} =\left(\frac{x^{36}-1}{x^2-1}\right)\left(\frac{x-1}{x^{18}-1}\right)$

$\Rightarrow \quad \frac{p(x)}{q(x)} =\frac{\left(x^{18}+1\right)\left(x^{18}-1\right)(x-1)}{(x+1)(x-1)\left(x^{18}-1\right)}$

$\Rightarrow \quad \frac{p(x)}{q(x)} =\frac{x^{18}+1}{x+1}$

$\Rightarrow \quad \frac{p(x)}{q(x)}= \frac{(x+1)\left(x^{17}-x^{16}+x^{15}-x^{14} \ldots-1\right)}{x+1}$

$\therefore$ Quotient is $x^{17}-x^{16}+x^{15}-x^{14} \ldots-1$

Let $p(x)=x^{135}+x^{125}-x^{115}+x^5+1$,

$q(x)=x^3-x$ and $p(x)=q(x) k+r(x)$

$x^{135}+x^{125}-x^{115}+x^5+1 =\left(x^3-x\right) k+a x^2+b x+c$

Put $x=0$

$\therefore \quad c=1$

Put $x=1, \quad 3=a+b+c \Rightarrow 3=a+b+1$

$\Rightarrow \quad a+b=2$

Put $x=-1,-1=a-b+c \Rightarrow-1=a-b+1$

$\Rightarrow \quad a-b=-2$ From Eqs.$(i)$ and $(ii)$, we get

$a =0, b=2$
$\therefore r(x) =2 x+1$

$\therefore \text { Degree of } r(x) =1$

We have,

$16^5 \cdot 5^{16}=16 \cdot 16^4 \cdot 5^{16}$

$=16 \times 2^{16} \cdot 5^{16}$

$=16 \times(10)^{16}$

$\therefore$ Total number of digits in $16^5 \cdot 5^{16}=18$

$12 !(1+13+13 \times 14)$

$12 !(1+13(1+14))$

$12 ! \times 196$

The number of distinct prime of $12 ! \times 196$ is $2,3,5,7,11$.

Given,

$x=(\sqrt{50}+7)^{1 / 3}-(\sqrt{50}-7)^{1 / 3}$

On cubing both sides, we get

$x^3=(\sqrt{50}+7)-(\sqrt{50}-7)-3$

$(\sqrt{50}+7)^{1 / 3}(\sqrt{50}-7)^{1 / 3}$

${\left[(\sqrt{50}+7)^{1 / 3}-(\sqrt{50}-7)^{1 / 3}\right] }$

$\Rightarrow \quad x^3=14-3(50-49)^{1 / 3}(x)$

$\Rightarrow \quad x^3=14-3 x$

$\Rightarrow \quad x^3+3 x-14=0$

$\Rightarrow(x-2)\left(x^2+2 x+7\right)=0$

$\Rightarrow \quad x=2$

Let $S=(2014)^3-(2013)^3+(2012)^3$

$-(2011)^3+\ldots+2^3-1^3$

$\Rightarrow S=2\left(2014^3+2012^3+2010^3+\ldots+2^3\right)$

$-\left(2014^3+2013^3+\ldots+2^3+1^3\right)$

$\Rightarrow S=2 \times 2^3\left(1007^3+1006^3+\ldots+1^3\right)$

$-\left(2014^3+2013^3+\ldots+2^3+1^3\right)$

$\Rightarrow S=\frac{2 \times 8(1007)^2(1008)^2 \quad 2014^2 \times 2015^2}{4}$

$\Rightarrow S=(1007)^2\left[\frac{2 \times 8 \times(1008)^2-4 \times(2015)^2}{4}\right]$

$\Rightarrow S=(1007)^2\left[(2016)^2-(2015)^2\right]$

$\Rightarrow S=(1007)^2(2016+2015)(2016-2015)$

$\Rightarrow \quad S=(1007)^2(4031)$

Thus, $S$ is divisible by $(1007)^2$.

Let $n=3 q+r$

$0 \leq r < 3$

$n=3 q, 3 q+1,3 q+2$

If $n$ is multiple of 3

i.e. $n=3 q$

Then, $n^{19}$ is also multiple of 3 .

When $n=3 q+1$ and $3 q+2$

If $n$ is multiple of $3$

i.e. $n=3 q$

Then, $n^{19}$ is also multiple of $3$ .

When $n=3 q+1$ and $3 q+2$

$n^{38}=(3 q+1)^{38}$

$=(3 q+1)^{36}(3 q+1)^2$

$=(36 k+1)\left(9 q^2+6 q+1\right)$

$=36 k\left(9 q^2+6 q+1\right)+9 q^2+6 q+1$

$=3 k+3 \lambda+1$

$\therefore n^{38}-1=3 k+3 \lambda+1-1=3\,m$

$\therefore n^{38}-1$ is multiple of 3

Similarly, when $n=3 q+2$

$n^{38}-1$ is also multiple of $3.$

We have, $\frac{1+x}{\left(1+x^2\right)} \frac{(1-x)}{}$

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

$+\frac{1}{2(1-x)}$

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

Coefficient of $x^{2012}$ in

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

$=0+\frac{1}{2}+\frac{1}{2}=1$

We have,

$6^m+9^n=(5+1)^m+(10-1)^n$

$=5 k+(1)^m+10 \lambda-(1)^n$

$=5 k+10 \lambda_\omega+(1)^m-(1)^n$

lt is multiple of 5 if $m \in\{1,2,3, \ldots, 50\}$

and $n \in\{1,3,5, \ldots, 49\} \quad[\because n$ must be odd $]$

Total order pairs of $(m, n)=50 \times 25$

$=1250$

We have,

Now, $\quad 512-253-259=0$

We know that, $a+b+c=0$ then

$a^3+b^3+c^3=3 a b c$

$\therefore \quad(512)^3-(253)^3-(259)^3$

$=3(512)(-253)(-259)$

$=3 \cdot 512 \cdot 253 \cdot 259$

$=3 \cdot 2^9 \cdot 11 \times 23 \times 7 \times 37$

$\therefore$ There are $6$ distinct prime divisors.

We have, $\left(x^{1 / 2}+\frac{1}{2 x^{1 / 4}}\right)^n$

$T_{r+1}={ }^n C_r\left(x^{1 / 2}\right)^{n-r} \frac{1}{\left(2 x^{\frac{1}{4}}\right)^r}$

$T_{r+1}={ }^n C_r x{ }^{n-r}-{ }^r 4 \cdot 2^{-r}$

$T_{r+1}={ }^n C_r \frac{2 n-3 r}{4} \cdot 2^{-r}=\frac{{ }^n C_r}{2^r} x^{\frac{2 n-3 r}{4}}$

Given $T_1, T_2, T_3$ are in AP.

$\therefore 2 T_2=T_1+T_3$

$\frac{2^n C_1}{2}={ }^n C_0+\frac{{ }^n C_2}{2^2}$

$\Rightarrow { }^n C_1={ }^n C_0+\frac{n(n-1)}{2 \cdot 2^2}$

$\Rightarrow \quad n=1+\frac{n(n-1)}{8}$

$\Rightarrow \quad n-1=\frac{n(n-1)}{8}$

$\Rightarrow n-1=0 \text { or } n=8$

When $n=8$,

$\frac{2 n-3 r}{4}=\frac{16-3 r}{4}$ is integer

If $\quad r=0,4,8$

Given,

$a_0=0$

$a_n=3 a_{n-1}+1, \forall n \geq 1$

$a_1=3 a_0+1=1$

$a_2=3 a_1+1=3+1$

$a_3=3 a_2+1=3(3+1)+1$

$=3^2+3+1$

and $\quad a_4=3 a_3+1=3\left(3^2+3+1\right)+1=3^3$

$+3^2+3+1$

Similarly,

$a_n=1+3+3^2+\ldots+3^{n-1}=\frac{3^n-1}{3-1}=\frac{3^n-1}{2}$

$a_{2010}=\frac{3^{2010}-1}{2}$

$a_{2010}$ is divided by $11$

$\therefore \quad \frac{3^{2010}-1}{2} \div 11=\frac{3^{2010}-1}{22}$

$3^{2010}=\left(3^5\right)^{402}$

$=(243)^{402}=(242+1)^{402}$

$=402(242) k+1$

$\quad[\because k$ is positive integer $]$

$=402 \times 22 \times 11 \times k+1$

$=22 m+1$

$\therefore \frac{3^{2010}-1 \quad 22\,m+1-1}{22}=\frac{22\,m}{22}$

$\therefore$ Remainder $=0$

We have,

$(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$

Put $x=1$,

$3^{20}=a_0+a_1+a_2+\ldots+a_{20}$

Put $x=-1$,

$1=a_0-a_1+a_2-a_3+\ldots+a_{20} \ldots \text { (ii) }$

On adding Eqs.$(i)$ and $(ii)$, we get $\frac{3^{20}+1}{2}=a_0+a_2+a_4+\ldots+a_{20}$

On subtracting Eq.$(ii)$ from Eq.$(i)$,

we get $\frac{3^{20}-1}{2}=a_1+a_3+a_5+\ldots+a_{19}$

Now, we have

$3 a_0+2 a_1+3 a_2+2 a_3+\ldots+2 a_{19}+3 a_{20}$

$=3\left(a_0+a_2+a_4+\ldots+a_{20}\right)$

$+3\left(\frac{3^{20}+2\left(a_1+a_3+\ldots+a_{19}\right)}{2}\right)+2\left(\frac{3^{20}-1}{2}\right)$

$=\frac{5 \cdot 3^{20}+1}{2}$

$ \text { General term }={ }^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{ax}})^{10-\mathrm{r}}\left(\frac{1}{2 \mathrm{x}^3}\right)^{\mathrm{r}} $

$ 20-2 \mathrm{r}-3 \mathrm{r}=0 $

$ \mathrm{r}=4 $

$ { }^{10} \mathrm{C}_4 \mathrm{a}^3 \cdot \frac{1}{16}=105 $

$ \mathrm{a}^3=8 $

$ \mathrm{a}^2=4$

then $A=(1-3+10)^n=8^n$ (put $\left.x=1\right)$

and sum of coefficients in the expansion of $\left(1+x^2\right)^n=B$

then $B=(1+1)^n=2^n$

$A=B^3$

$ (1-x)\left(1-x^3\right)^{2007} $

$ (1-x)\left({ }^{2007} C_0-{ }^{2007} C_1\left(x^3\right)+\ldots \ldots .\right)$

General term

$ (1-x)\left((-1)^r{ }^{2007} C_r x^{3 r}\right) $

$ (-1)^{r 2007} C_r x^{3 r}-(-1)^{r 2007} C_r x^{3 r+1} $

$ 3 r=2012 $

$ r \neq \frac{2012}{3} $

$ 3 r+1=2012 $

$ 3 r=2011 $

$ r \neq \frac{2011}{3}$

Hence there is no term containing $\mathrm{x}^{2012}$.

So coefficient of $x^{2012}=0$

For integral term, $r$ must be multiple of $6$ .

Hence $r=0,6,12, \ldots \ldots \ldots........ .822$

Binomial coefficients are

$\begin{array}{ll} & { }^n C_r,{ }^n C_{r+1},{ }^n C_{r+2},{ }^n C_{r+3} \text { respectively } \\ \Rightarrow & { }^n C_r+{ }^n C_{r+1}=2 \\ \Rightarrow & { }^{n+1} C_{r+1}=2 \quad \ldots \ldots .(1) \\ & \quad \text { Also, } \quad{ }^n C_{r+2}+{ }^n C_{r+3}=2 \\ \Rightarrow & { }^{n+1} C_{r+3}=2 \quad \ldots \ldots .(2) \\ & \quad \text { From }(1) \text { and (2) } \\ & { }^{n+1} C_{r+1}={ }^{n+1} C_{r+3}\end{array}$

$\begin{aligned} \Rightarrow \quad & 2 \mathrm{r}+4=\mathrm{n}+1 \\ & \mathrm{n}=2 \mathrm{r}+3 \\ & { }^{2 \mathrm{r}+4} \mathrm{C}_{\mathrm{r}+1}=2\end{aligned}$
Data Inconsistent

$(1+x)^{17}-x(1+x)^{17}$

$=0-1$

$=-1$

$\operatorname{coeff}\left(\mathrm{x}^{-13}\right)$ in the expansion $\approx \operatorname{coeff}\left(\mathrm{x}^2\right)$ in

$(1+x)^{17}-x(1+x)^{17}$

$=\left(\begin{array}{c}17 \\ 2\end{array}\right)-\left(\begin{array}{c}17 \\ 1\end{array}\right)$

$=17 \times 8-17$

$=17 \times 7$

$=119$

Hence Answer $=119-1=118$

$ t_7={ }^{18} c_6\left(\frac{x^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{x^{\frac{-2}{3}}}{2}\right)^6={ }^{18} c_6 \frac{1}{(3)^{12}} \cdot \frac{1}{2^6} $

$ \mathrm{t}_{13}={ }^{18} c_{12}\left(\frac{x^{\frac{1}{3}}}{3}\right)^6\left(\frac{x^{\frac{-2}{3}}}{2}\right)^{12}={ }^{18} c_{12} \frac{1}{(3)^6} \cdot \frac{1}{2^{12}} \cdot x^{-6} $

$ \mathrm{~m}={ }^{18} \mathrm{c}_6 \cdot 3^{-12} \cdot 2^{-6}: \mathrm{n}={ }^{18} \mathrm{c}_{12} \cdot 2^{-12} \cdot 3^{-6} $

$ \left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}=\left(\frac{2^{-12} \cdot 3^{-6}}{3^{-12} \cdot 2^{-6}}\right)^{\frac{1}{3}}=\left(\frac{3}{2}\right)^2=\frac{9}{4}$

$ ={ }^{15} \mathrm{C}_{\mathrm{r}} 5^{\frac{\mathrm{r}}{3}} \cdot 2^{\frac{15-\mathrm{r}}{5}} $

$ \mathrm{R}=3 \lambda, 15 \mu $

$ \Rightarrow \mathrm{r}=0,15 $

$ 2 \text { rational terms } $

$ \Rightarrow{ }^{15} \mathrm{C}_0 2^3+{ }^{15} \mathrm{C}_{15}(5)^5 $

$ =8+3125=3133$

$ \text { Coeff. of } x^5={ }^n C_5 $

$ \text { Coeff. of } x^6={ }^n C_6 $

$ { }^n C_4,{ }^n C_5,{ }^n C_6 \ldots . A P $

$ 2 . C_5={ }^n C_4+{ }^n C_6 $

$ 2=\frac{{ }^n C_4}{{ }^n C_5}+\frac{{ }^n C_6}{{ }^n C_5} \quad\left\{\frac{{ }^n C_r}{{ }^n C_r}=\frac{n-r+1}{r}\right\} $

$ 2=\frac{5}{n-4}+\frac{n-5}{6} $

$ 12(n-4)=30+n^2-9 n+20 $

$ n^2-21 n+98=0 $

$ (n-14)(n-7)=0 $

$ n_{\max }=14 \quad n_{\min }=7$

General term $\mathrm{m}\left(\frac{3}{2} \mathrm{x}^2-\frac{1}{3 \mathrm{x}}\right)^9$

$={ }^9 \mathrm{C}_{\mathrm{r}} \cdot \frac{3^{9-2 \mathrm{r}}}{2^{9-\mathrm{r}}}(-1)^{\mathrm{r}} \cdot \mathrm{x}^{18-3 \mathrm{r}}$

Put $r=6$ to get coeff. of $x^0={ }^9 \mathrm{C}_6 \cdot \frac{1}{6^3} \cdot x^0=\frac{7}{18} x^0$

Put $r=7$ to get coeff. of $x^{-3}={ }^9 C_r \cdot \frac{3^{-5}}{2^2}(-1)^7 \cdot x^{-3}$

$ =-{ }^9 C_7 \cdot \frac{1}{3^5 \cdot 2^2} \cdot x^{-3}=\frac{-1}{27} x^{-3} $

$ \left(1+2 x-3 x^3\right)\left(\frac{7}{18} x^0-\frac{1}{27} x^{-3}\right) $

$ \frac{7}{18}+\frac{3}{27}=\frac{7}{18}+\frac{1}{9}=\frac{7+2}{18}=\frac{9}{18}=\frac{1}{2} $

$ \therefore 108 \cdot \frac{1}{2}=54$

$ \mathrm{T}_{\mathrm{r}+1}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}(3)^{\frac{12-\mathrm{r}}{5}}(2)^{\mathrm{r}}(\mathrm{x})^{2 \mathrm{r}-12}}{(5)^{\mathrm{r} / 3}} $

$ \mathrm{r}=6 $

$ \mathrm{~T}_7=\frac{{ }^{12} \mathrm{C}_6(3)^{6 / 5}(2)^6}{5^2}=\left(\frac{9 \times 11 \times 7}{25}\right) 2^8 \cdot 3^{1 / 5} $

$ 25 \alpha=693$

$ { }^n C_2 x^{n-2} y^2=30 $ ............($ii$)

$ { }^n C_3 x^{n-3} y^3=\frac{10}{3} $ ............($iii$)

$ \text { By } \frac{(i)}{(i i)} $

$ \frac{{ }^n C_1}{{ }^n C_2} \frac{x}{y}=\frac{9}{2} $ ............($iv$)

$ \text { By } \frac{(\text { ii) }}{(\text { iii) }} $

$ \frac{{ }^n C_2}{{ }^n C_3} \frac{x}{y}=9 $ ...............($v$)

$ \text { By } \frac{(\text { iv) }}{(v)} $

$ \frac{{ }^n C_1 C_3}{{ }^n C_2{ }^n C_2}=\frac{1}{2}$

$ \frac{2 n^2(n-1)(n-2)}{6}=\frac{n(n-1)}{2} \frac{n(n-1)}{2} $

$ 4 n-8=3 n-3 $

$ \Rightarrow n=5$

put in ($v$)

$ \frac{x}{y}=9 $

$ x=9 y $

$ \text { put in (i) } $

$ { }^5 C_1 x^4\left(\frac{x}{9}\right)=135 $

$ x^5=27 \times 9 $

$ \Rightarrow x=3, \quad y=\frac{1}{3} $

$ 6\left(n^3+x^2+y\right) $

$ =6\left(125+9+\frac{1}{3}\right) $

$ =806$

$ \frac{(n+1) !}{(r+1) !(n-r)} ! \frac{r !(n-r) !}{n !}=\frac{11}{7} $

$ \frac{(n+1)}{r+1}=\frac{11}{7}$

$ 7 \mathrm{n}=4+11 \mathrm{r} $

$ \frac{{ }^n C_r}{{ }^{n-1} C_{r-1}}=\frac{35}{21} $

$ \frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}=\frac{(\mathrm{r}-1) !(\mathrm{n}-\mathrm{r}) !}{(\mathrm{n}-1) !}=\frac{5}{3} $

$ \frac{\mathrm{n}}{\mathrm{r}}=\frac{5}{3} $

$ 3 \mathrm{n}=5 \mathrm{r} $

By solving $ \mathrm{r}=6 \quad \mathrm{n}=10 $

$ 2 \mathrm{n}+5 \mathrm{r}=50$