Binomial Theorem

Statement I : $25^{13}+20^{13}+8^{13}+3^{13}$ is divisible by 7.

Key concept : If $n$ is odd, $x^n+y^n=(x+y)\left(x^{n-1}-x^{n-2} y+x^{n-3} y^2-\cdots-x y^{n-2}+\right. y^{n-1}$ ) So, $x^n+y^n$ is divisible by $x+y$.

In the given statement :

• $25^{13}+3^{13}$ is divisible by $25+3=28$.

• $25^{13}+3^{13}$ is also divisible by 7 because 28 is divisible by 7 .

• So, $25^{13}+3^{13}=7 \cdot I_1$.........(1) ( $I_1$ is a positive integer).

Similarly :

• $20^{13}+8^{13}$ is divisible by $20+8=28$.

• $20^{13}+8^{13}$ is divisible by 7.

• So, $20^{13}+8^{13}=7 \cdot I_2$ ..........(2) ( $I_2$ is a positive integer).

Add (1) & (2) :

• $25^{13}+3^{13}+20^{13}+8^{13}=7\left(I_1+I_2\right)$

So, $25^{13}+20^{13}+8^{13}+3^{13}$ is divisible by 7 . Statement I is true.

Statement : The integral part of $(7+4 \sqrt{3})^{25}$ is an odd number.

Let $X=(7+4 \sqrt{3})^{25}$. Let $I$ be the integral part and $f$ be the fractional part ( $0

Adding the two expansions:

$ (7+4 \sqrt{3})^{25}+(7-4 \sqrt{3})^{25}=I+f+f^{\prime} $

Using the binomial expansion $(a+b)^n+(a-b)^n=2\left[a^n+\binom{n}{2} a^{n-2} b^2+\ldots\right]$, we see that the sum is always an even integer because of the factor of 2 .

$ I+f+f^{\prime}=\text { Even Integer } $

Since $I$ is an integer, $f+f^{\prime}$ must also be an integer. Given $0 < f < 1$ and $0 < f^{\prime} < 1$, the only possible integer value for their sum is $f+f^{\prime}=1$.

$ I+1=\text { Even Integer } $

$ I=\text { Even Integer }-1=\text { Odd Integer } $

The integral part I is indeed an odd number. Statement II is True.

Let

$ S=(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\cdots+x^{1000} $

Notice that this is a Geometric Progression (GP) with:

• First term (a): $(1+x)^{1000}$

• Common ratio $(r): \frac{x}{1+x}$

• Number of terms ( $n$ ): 1001

The sum of a GP is given by $S=a \frac{\left(1-r^n\right)}{(1-r)}$. Substituting our values:

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

Simplify the denominator: $1-\frac{x}{1+x}=\frac{1+x-x}{1+x}=\frac{1}{1+x}$.

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

$ S=(1+x)^{1001}-x^{1001} $

Now we need the coefficients of $x^{499}$ and $x^{500}$ in the simplified expression $(1+x)^{1001}- x^{1001}$.

Since $x^{1001}$ only affects the term with power 1001, we only look at $(1+x)^{1001}$.

In the binomial expansion $(1+x)^n$

Coefficient of $x^r$ is ${ }^n C_r$

Coefficient of $x^{499}$ in $S$ :

$ { }^{1001} C_{499} $

Coefficient of $x^{500}$ in $S$ :

${ }^{1001} C_{500}$ (The term $x^{1001}$ does not contribute.)

$ \text { Sum of coefficients } $

$ \begin{array}{ll} { }^{1001} C_{499}+{ }^{1001} C_{500}={ }^{1002} C_{500} \end{array} $

[$\text { Using pascal's identity } $

${ }^n C_r+{ }^n C_{r+1}={ }^{n+1} C_{r+1}$]

$ S=\frac{1}{1!25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots 13 \text { terms } $

multiply in numerator and denominator by 26 !

$ \begin{aligned} & S=\frac{1}{26!}\left[\frac{26!}{1!25!}+\frac{26!}{3!23!}+\frac{26!}{5!2!}+\ldots 13 \text { term }\right] \\ & S=\frac{1}{26!}\left[{ }^{26} C_1+{ }^{26} C_3+{ }^{26} C_5+\ldots 13 \text { term }\right],\left\{{ }^n C_r=\frac{n!}{(n-r)!r!}\right\} \end{aligned} $

Lets find 13th term lower subscript are in A.P with, common difference $=2$

$ \begin{aligned} & T_{13}=1+(13-1) 2=25 \\ & \text { so, } S=\frac{1}{26!}\left[{ }^{26} C_1+{ }^{26} C_3+{ }^{26} C_5+\ldots{ }^{26} C_{25}\right] \\ & S=\frac{1}{26!}\left[2^{26-1}\right],\left\{\text { because }{ }^n C_1+{ }^n C_3+{ }^n C_5 \cdots=2^{n-1}\right\} \end{aligned} $

$ S=\frac{2^{25}}{26!} $

It is given that $13 S=\frac{2^K}{n!}$

substitute value of $S$

$ \begin{aligned} & 13 \times \frac{2^{25}}{26!}=\frac{2^K}{n!} \\ & \frac{13 \times 2^{25}}{26 \times 25!}=\frac{2^K}{n!} \\ & \frac{2^{24}}{25!}=\frac{2^K}{n!} \end{aligned} $

so $n=25, K=24$

$n+K=25+24=49$

$ \begin{aligned} & \left(1+x^2\right)^2\left(1+x^n\right) \\ & \left(1+x^4+2 x^2\right)(1+x)^n \\ & \text { Coeff of } x={ }^n C_1 \\ & \text { Coeff of } x^2={ }^n C_2+2 \cdot{ }^n C_0 \\ & \text { Coeff of } x^3={ }^n C_3+2 \cdot{ }^n C_1 \\ & \because \text { Coeff of } x, x^2, x^3 \rightarrow \mathrm{AP} \\ & \Rightarrow 2\left({ }^n C_2+2\right)={ }^n C_1+{ }^n C_3+2 n \\ & 2^n C_2+4={ }^n C_3+3 n \\ & \Rightarrow n^2-9 n^2+26 n-24=0 \\ & \Rightarrow n=2,3,4 \\ & \Rightarrow \text { Sum of all value of } n \text { is } 9 \end{aligned} $

$ \begin{aligned} & (1+x)^{100}={ }^{100} C_0+{ }^{100} C_{1 x}+{ }^{100} C_{2 x^2}+\ldots \ldots . .+{ }^{100} C_{100 x^{100}} \\ & \sum_{r=50}^{100} \frac{{ }^{100} C_r}{r+1}=\frac{1}{101} \sum_{r=50}^{100}{ }^{101} C_{r+1} \\ & \Rightarrow \frac{1}{101} \sum_{r=51}^{100}{ }^{101} C_r \\ & \Rightarrow \frac{2^{100}}{101} \end{aligned} $

$ \begin{aligned} P_n & =\sum_{r=0}^n \frac{{ }^n C_r(-2)^r}{r+1}=\sum_{r=0}^n \frac{1}{(n+1)}{ }^{n+1} C_{r+1}(-2)^r \\ & =\frac{-1}{2(n+1)} \sum_{r=0}^n{ }^{n+1} C_{r+1}(-2)^{r+1}=\frac{-1}{2(n+1)}\left[(1-2)^{n+1}-1\right] \\ P_n & =\frac{1}{2(n+1)}\left[1-(-1)^{n+1}\right] \end{aligned} $

$ \begin{aligned} & P_{2 n}=\frac{1}{2(2 n+1)}\left[1-(-1)^{2 n+1}\right] \\ & P_{2 n}=\frac{1}{2 n+1} \\ & \sum_{n=0} \frac{1}{P_{2 n}}=\sum_{n=1}^{25}(2 n+1)=3+5+\ldots . .+51=\frac{25}{2}[51+3]=25 \times 27=675 \end{aligned} $

$ \begin{aligned} & S=(1+x)+2(1+x)^2+3(1+x)^3+\ldots+100(1+x)^{100} \\ & (1+x) S=(1+x)^2+\ldots+99(1+x)^{100}+100(1+x)^{101} \\ & -x S=(1+x)\left(\frac{(1+x)^{100}-1}{x}\right)-100(1+x)^{101} \\ & S=\left(\frac{(1+x)-(1+x)^{101}}{x^2}\right)+\frac{100(1+x)^{101}}{x} \\ & \text { Coeff of } x^{48}=-{ }^{101} C_{50}+100 \times{ }^{(101)} C_{49} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \left(a x^2+b x+c\right)(1-2 x)^{26} \\ & a x^2(1-2 x)^{26}+b x(1-2 x)^{26}+c(1-2 x)^{26} \end{aligned}\\ &\text { Coefficient of }\\ &\begin{aligned} & x \Rightarrow\left[0+b \cdot{ }^{26} C_0+c \cdot{ }^{26} C_1(-2)\right] x \\ & =b-52 c \end{aligned} \end{aligned} $

Coefficient of

$ \begin{aligned} & x^2=\left(a\left({ }^{26} C_0\right)+b \cdot{ }^{26} C_1(-2)+c{ }^{26} C_2(-2)^2\right) x^2 \\ & =a-52 b+{ }^{26} C_2 \times 4=a-52 b+1300 c \end{aligned} $

Coefficient of

$ \begin{aligned} & x^3=\left[a \cdot{ }^{26} C_1(-2)+b \cdot{ }^{26} C_2(26)^2+c^{26} C_3(-2)^3\right] \\ & =-52 a+1300 b-20800 c \\ & \Rightarrow b-52 c=-56 \\ & \Rightarrow b-52 c=-56 \\ & \begin{aligned} a-52 b+1300 c=0 \quad \Rightarrow & a=1300 \\ \quad-a+25 b-400 c=0 \quad & b=100 \\ & c=3 \\ \Rightarrow & a+b+c=1403 \end{aligned} \end{aligned} $

$\begin{aligned} &\begin{aligned} & \mathrm{T}_{\mathrm{r}}={ }^{1016} \mathrm{C}_{\mathrm{r}}(5)^{\frac{1016-\mathrm{r}}{2}} 7^{\frac{\mathrm{r}}{8}} \\ & \Rightarrow \mathrm{r}=0,8,16,24, \ldots ., 1016 \\ & 1016=0+(\mathrm{n}-1) 8 \\ & \Rightarrow \mathrm{n}-1=\frac{1016}{8}=127 \end{aligned}\\ &\text { So, } \mathrm{n}=128 \text {. } \end{aligned}$

$\begin{aligned} &\begin{aligned} & 64^{64} \Rightarrow(63+1)^{64}=63 \lambda+1 \\ & 64^{664^{64}} \Rightarrow(63+1)^{64^{64}} \\ & 63 \lambda_1+1 \end{aligned}\\ &\text { Required remainder when divided } 7 \text { is } 1 \text {. } \end{aligned}$

$\sum_{r=1}^{15} r^2 \cdot{ }^{15} C_r \quad\left(r^n C_r=n^{n-1} C_{r-1}\right)$

$\begin{aligned} & =15 \sum_{r=1}^{15} r \cdot{ }^{14} C_{r-1} \\ & =15 \sum_{r=1}^{15}(r-1+1){ }^{14} C_{r-1} \\ & =15 \cdot \sum_{r=1}^{15}(r-1){ }^{14} C_{r-1}+15 \cdot \sum_{r=1}^{15}{ }^{14} C_{r-1} \\ & =15 \cdot 14 \cdot 2^{13}+15 \cdot 2^{14} \\ & =15 \cdot 2^{14}(7+1) \\ & =5 \cdot 3 \cdot 2^{17} \\ & n+m+k=17+1+1=19 \end{aligned}$

$\begin{aligned} & \text { Mean }=\frac{{ }^{2 n-3} C_0+{ }^{2 n-3} C_1+{ }^{2 n-3} C_2+\cdots{ }^{2 n-3} C_{2 n-3}}{2 n-2}=16 \\ &=2^{2 n-3}=16(2 n-2) \\ &=2^{2 n-3}=2^5(n-1) \\ & \Rightarrow n=5 \\ & \therefore \quad P(9,5) \\ & d=\left|\frac{9+5-8}{\sqrt{2}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2} \end{aligned}$

In the expansion of $(a+b)^n$

$15^{\text {th }}$ term from beginning: $T_{15}={ }^n C_{14} a^{n-14} b^{14}$

$15^{\text {th }}$ term from end: $T_{15}^{\prime}={ }^n C_{14} b^{n-14} a^{14}$

$\begin{aligned} & \therefore \quad \frac{T_{15}}{T_{15}^{\prime}}=\frac{1}{6} \\ & \left(\frac{a}{b}\right)^{n-28}=\frac{1}{6} \\ & \left(6^{\frac{1}{3}}\right)^{n-28}=6^{-1} \\ & \Rightarrow \quad \frac{n-28}{3}=-1 \\ & n=25 \\ & \therefore \quad{ }^{25} C_3=2300 \end{aligned}$

To find the sum of all rational terms in the expansion of $(2+\sqrt{3})^8$, we consider the binomial expansion:

$ S = { }^8 C_0 (2)^8 + { }^8 C_1 (2)^7 (\sqrt{3}) + \ldots + { }^8 C_8 (\sqrt{3})^8 $

We need to identify and sum only the rational terms. In the binomial expansion, a term is rational if the exponent of $\sqrt{3}$ is even.

Thus, the rational terms are:

$ \begin{aligned} &{ }^8 C_0 (2)^8 + { }^8 C_2 (2)^6 (\sqrt{3})^2 + { }^8 C_4 (2)^4 (\sqrt{3})^4 + \\ &{ }^8 C_6 (2)^2 (\sqrt{3})^6 + { }^8 C_8 (\sqrt{3})^8 \end{aligned} $

Evaluating these terms:

${ }^8 C_0 (2)^8 = 256$

${ }^8 C_2 (2)^6 (3) = { }^8 C_2 \times 64 \times 3 = 1344$

${ }^8 C_4 (2)^4 (3)^2 = { }^8 C_4 \times 16 \times 9 = 3024$

${ }^8 C_6 (2)^2 (3)^3 = { }^8 C_6 \times 4 \times 27 = 4032$

${ }^8 C_8 (3)^4 = 1 \times 81 = 81$

Summing these rational terms gives:

$ 256 + 1344 + 3024 + 4032 + 81 = 18817 $

Therefore, the sum of all rational terms in the expansion is $18,817$.

$\begin{aligned} & \sum_{r=1}^9\left(\frac{r+3}{2^r}\right) \cdot{ }^9 C_r=\sum_{r=1}^9 \frac{r}{2^r} \cdot \frac{9}{r} \cdot{ }^8 C_{r-1}+\sum_{r=1}^9 3 \cdot{ }^9 C_r\left(\frac{1}{2}\right)^r \\ & =\frac{9}{2} \sum_{r=1}^9{ }^8 C_{r-1}\left(\frac{1}{2}\right)^{r-1}+3 \sum_{r=1}^9 C_r\left(\frac{1}{2}\right)^r \\ & =\frac{9}{2} \sum_{r=0}^8{ }^8 C_{r-1}\left(\frac{1}{2}\right)^r+3 \sum_{r=1}^9{ }^9 C_r\left(\frac{1}{2}\right)^r \\ & =\frac{9}{2}\left(1+\frac{1}{2}\right)^8+3\left[\left(1+\frac{1}{2}\right)^9-{ }^9 C_0\left(\frac{1}{2}\right)^0\right] \\ & =\frac{9}{2} \cdot \frac{3^8}{2^8}+3\left[\frac{3^9}{2^9}-1\right] \\ & =\frac{3^{10}}{2^9}+\frac{3^{10}}{2^9}-3=4 \cdot \frac{3^{10}}{2^{10}}-3 \\ & =4\left(\frac{3}{2}\right)^{10}-3 \\ & =6\left(\frac{3}{2}\right)^9-3 \\ & \alpha=6, \beta=3 \Rightarrow(\alpha+\beta)^2=81 \end{aligned}$

$\begin{aligned} & V_3(50!)=\left[\frac{50}{3}\right]+\left[\frac{50}{9}\right]+\left[\frac{50}{27}\right]+\left[\frac{50}{81}\right]+\ldots+\left[\frac{50}{3^n}\right] \\ & =16+5+1+0+0+0 \ldots=22 \end{aligned}$

$\begin{aligned} & 7^{103}=7\left(7^{102}\right)=7(343)^{34}=7(345-2)^{34} \\ & 7^{103}=23 \mathrm{~K}_1+7.2^{34} \\ & \text { Now } 7.2^{34}=7 \cdot 2^2 \cdot 2^{32} \\ & =28 \cdot(256)^4 \\ & =28(253+3)^4 \\ & \therefore 28 \times 81 \Rightarrow(23+5)(69+12) \\ & 23 \mathrm{~K}_2+60 \\ & \therefore \text { Remainder }=14 \end{aligned}$

$\begin{aligned} & \text { General term }={ }^n C_r\left(7^{1 / 3}\right)^{n-r}\left(11^{1 / 12}\right)^r \\ & ={ }^n C_r(7)^{\frac{n-r}{3}}(11)^{r / 12} \end{aligned}$

For integral terms, $r$ must be multiple of 12

$\therefore \mathrm{r}=12 \mathrm{k}, \mathrm{k} \in \mathrm{~W}$

Total values of $\mathrm{r}=183$

Hence $\max r=12(182)$

$=2184$

Min value of $n=2184$

$\begin{aligned} & (\mathrm{a}+\mathrm{b})^{\frac{1}{2}} \\ & \mathrm{~T}_{\mathrm{r}}, \mathrm{~T}_{\mathrm{r}+1}, \mathrm{~T}_{\mathrm{r}+2} \rightarrow \mathrm{GP} \\ & \text { So, } \frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\frac{\mathrm{T}_{\mathrm{r}+2}}{\mathrm{~T}_{\mathrm{r}+1}} \\ & \frac{{ }^{12} \mathrm{C}_{\mathrm{r}}}{{ }^{12} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}+1}}{{ }^{12} \mathrm{C}_{\mathrm{r}}} \\ & \frac{12-\mathrm{r}+1}{\mathrm{r}}=\frac{12-(\mathrm{r}+1)+1}{\mathrm{r}+1} \\ & (13-\mathrm{r})(\mathrm{r}+1)=(12-\mathrm{r})(\mathrm{r}) \\ & -\mathrm{r}+12 \mathrm{r}+13=12 \mathrm{r}-\mathrm{r}^2 \\ & 13=0 \end{aligned}$

No value of r possible

So $\mathrm{P}=0$

$\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)^{12}=\sum{ }^{12} C_r\left(3^{\frac{1}{4}}\right)^{12-\mathrm{r}}\left(4^{\frac{1}{3}}\right)^{\mathrm{r}}$

Exponent of $\left(3^{\frac{1}{4}}\right)$ exponent of $\left(4^{\frac{1}{3}}\right)$ term

$\begin{array}{ccc} 12 & 0 & 27 \\ 0 & 12 & 256 \\ \mathrm{q}=27+256=283 & & \\ \mathrm{p}+\mathrm{q}=0+283=283 & & \end{array}$

$\begin{aligned} & A={ }^{2 n-1} C_{29} \quad B={ }^{2 n-1} C_{11} \\ & 2{ }^{2 n-1} C_{29}=5{ }^{2 n-1} C_{11} \\ & 2 \frac{(2 n-1)!}{29!(2 n-30)!}=5 \frac{(2 n-1)!}{(2 n-12)!11!} \\ & \frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 2} \\ & \frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 12} \\ & 2 n-12=30 \\ & n=21 \end{aligned}$

$\begin{aligned} & (1+\mathrm{x})^{\mathrm{n+4}} \\ & { }^{\mathrm{n}+4} \mathrm{C}_4{ }^{\mathrm{n+4}} \mathrm{C}_5{ }^{\mathrm{n}+4} \mathrm{C}_6, \rightarrow \text { A.P. } \\ & \Rightarrow 2 \times{ }^{\mathrm{n}+4} \mathrm{C}_5={ }^{\mathrm{n+4}} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_6 \\ & \Rightarrow 4 \times{ }^{\mathrm{n+4}} \mathrm{C}_5=\left({ }^{n+4} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_5\right)+\left({ }^{\mathrm{n+4}} \mathrm{C}_5+{ }^{\mathrm{n}+4} \mathrm{C}_6\right) \\ & \Rightarrow 4 \times{ }^{n+4} \mathrm{C}_5={ }^{\mathrm{n}+5} \mathrm{C}_5+{ }^{\mathrm{n}+5} \mathrm{C}_6 \\ & \Rightarrow 4 \times \frac{(\mathrm{n}+4)!}{5!.(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\mathrm{n}!} \\ & \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6 \mathrm{n}} \\ & \Rightarrow \mathrm{n}^2+11 \mathrm{n}+30=24 \mathrm{n} \\ & \Rightarrow \mathrm{n}^2-13 \mathrm{n}+30=0 \\ & \Rightarrow \mathrm{n}=3,10(\text { rejected }) \\ & \because \mathrm{n} \neq 10 \end{aligned}$

$\therefore$ Largest binomial coefficient in expansion of

$\begin{aligned} & (1+x)^7 \\ & (\because \mathrm{n}+4=7) \end{aligned}$

is coeff. of middle term

$\Rightarrow{ }^7 \mathrm{C}_4={ }^7 \mathrm{C}_3=35$

N.T.A. Ans Option (2)

$\begin{aligned} & (1+\mathrm{x})^{\mathrm{p}}(1-\mathrm{x})^{\mathrm{q}}=\left({ }^{\mathrm{p}} \mathrm{C}_0+{ }^{\mathrm{p}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{p}} \mathrm{C}_2 \mathrm{x}^2+\ldots\right)\left({ }^q \mathrm{C}_0-{ }^q \mathrm{C}_1 \mathrm{x}+{ }^q \mathrm{C}_2 \mathrm{x}^2+\ldots\right) \\ & \text { coff of } \mathrm{x} \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^{\mathrm{q}} \mathrm{C}_1-{ }^{\mathrm{p}} \mathrm{C}_1{ }^q \mathrm{C}_0=1 \\ & \mathrm{p}-\mathrm{q}=1 \\ & \text { coff of } \mathrm{x}^2 \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^q \mathrm{C}_2-{ }^{\mathrm{p}} \mathrm{C}_1{ }^q \mathrm{C}_1+{ }^{\mathrm{p}} \mathrm{C}_2{ }^{\mathrm{q}} \mathrm{C}_0=-2 \\ & \frac{\mathrm{q}(\mathrm{q}-1)}{2}-\mathrm{pq}+\frac{\mathrm{p}(\mathrm{p}-1)}{2}=-2 \\ & \mathrm{q}^2-\mathrm{q}-2 \mathrm{pq}+\mathrm{p}^2-\mathrm{p}=-4 \\ & (\mathrm{p}-\mathrm{q})^2-(\mathrm{p}+\mathrm{q})=-4 \\ & \mathrm{p}+\mathrm{q}=5 \\ & \mathrm{p}=3 \\ & \mathrm{q}=2 \\ & \text { so } \mathrm{p}^2+\mathrm{q}^2=13 \end{aligned}$

To find the sum of $ u $ and $ v $, we first need to expand the expression:

$ \left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5 $

Using the Binomial Theorem, the expansion yields:

$ = 2\left({}^5C_0 \cdot x^5 + {}^5C_2 \cdot x^3(x^3-1) + {}^5C_4 \cdot x(x^3-1)^2\right) $

Simplifying this, we obtain:

$ = 2\left(5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x\right) $

From this expansion, we can identify the coefficients:

The coefficient of $ x^7 $ is $ \alpha = 10 $

The coefficient of $ x^5 $ is $ \beta = 2 $

The coefficient of $ x^3 $ is $ \gamma = -20 $

The coefficient of $ x $ is $ \delta = 10 $

Given the equations:

$ \alpha u + \beta v = 18 $

$ \gamma u + \delta v = 20 $

Substituting in the coefficients:

$ 10u + 2v = 18 $

$ -20u + 10v = 20 $

By solving these equations, we find:

From $ 10u + 2v = 18 $, simplify to $ 5u + v = 9 $.

From $ -20u + 10v = 20 $, simplify to $ -2u + v = 2 $.

Solving these linear equations simultaneously, we find:

Subtracting equation 2 from equation 1:

$ 5u + v = 9 $

$ - (-2u + v = 2) $

This yields:

$ 7u = 7 \quad \Rightarrow \quad u = 1 $

Substitute $ u = 1 $ back into $ 5u + v = 9 $:

$ 5(1) + v = 9 \quad \Rightarrow \quad v = 4 $

Thus, the sum $ u + v $ is:

$ u + v = 1 + 4 = 5 $

$ \begin{gathered} Let\,\,\,\,m=\frac{\left|\frac{3 x}{-4 y}\right|(n+1)}{\left|\frac{3 x}{-4 y}\right|+1} \\ m=\frac{1 \times(23+1)}{2} \\ m=12 \end{gathered} $

$\Rightarrow T_{12}$ and $T_{13}$ are the greatest terms.

$ \begin{aligned} T_{12} & ={ }^{23} C_{11}(3 x)^{12} \times\left|(-4 y)^{11}\right| \\ & ={ }^{23} C_{11}\left(\frac{1}{2}\right)^{12} \times\left(\frac{1}{2}\right)^{11} \\ & ={ }^{23} C_{11}\left(\frac{1}{2}\right)^{23} \end{aligned} $

$ \text { } \begin{array}{r} T_{r+1}={ }^{6144} C_2(\sqrt{2})^{6144-r} \times(\sqrt[3]{3})^r \\ ={ }^{6144} C_r(2)^{\frac{6144-r}{2}} \times(3)^{\frac{r}{3}} \end{array} $

For $T_{r+1}$ to be integer

$ \begin{aligned} 6144-r & =2 k_1 \\ r & =3 k_2 \end{aligned} $

$\Rightarrow r$ should be multiple of 6 .

$ \begin{aligned} & \Rightarrow r=0,6,12,18, \ldots, 6144 \\ & \Rightarrow \text { Number of terms }=\frac{6144}{6}+1=1025 \end{aligned} $

To find $\alpha_{1025}-\alpha_{1026}-\alpha_{1024}$

Let $f(x)=(1+x)^{-1}\left(1+x^2\right)^{-1}\left(1+x^4\right)^{-1}$

$ \left(1+x^8\right)^{-1} \times\left(1+x^{16}\right)^{-1} $

$ \begin{aligned} & =\frac{1}{(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)} \\ & =\frac{1-x}{1-x^{32}} \\ & \Rightarrow(1-x)\left(1-x^{32}\right)^{-1} \\ & \Rightarrow\left(1-x^{32}\right)^{-1}-x \times\left(1-x^{32}\right)^{-1} \end{aligned} $

$\alpha_{1025}=$ Coefficient of $x^{1025}$ in

$\left(1-x^{32}\right)^{-1}$ - Coefficient of $x^{1024}$ in

$ \left(1-x^{32}\right)^{-1} $

$\alpha_{1025}=-$ Coefficient of $x^{1024}$ in $\left(1-x^{32}\right)^{-1}$

$\alpha_{1026}=$ Coefficient of $x^{1026}$ in $\left(1-x^{32}\right)^{-1}$

$\alpha_{1024}=$ Coefficient of $x^{1024}$ in $\left(1-x^{32}\right)^{-1}$

So, $\alpha_{1025}-\alpha_{1026}-\alpha_{1024}$

$=-2 \times$ Coefficient of $x^{1024}$ in $\left(1-x^{32}\right)^{-1}$

• Coefficient of $x^{1026}$ in $\left(1-x^{32}\right)^{-1}$

$ =-2 \times 1-0=-2 $

$ \begin{gathered} (1+x)^{10}=C_0+C_1 x+C_2 x^2+\ldots+C_{10} x^{10} \\ \left(1+\frac{1}{x}\right)^{10}=C_0+\frac{C_1}{x}+\frac{C_2}{x_2}+ \\ \ldots+\frac{C_6}{x^6}+\ldots+\frac{C_{10}}{x^{10}} \\ \begin{aligned} & \frac{(1+x)^{20}}{x^{10}}=x^{-6}\left(C_0 C_6+C_1 C_7+C_2 C_8\right. \\ &\left.+C_3 C_9+C_4 C_{10}\right) \\ & \Rightarrow C_0 C_6+C_1 C_7+C_2 C_8+C_3 C_9+C_4 C_{10} \\ &= \text { Coefficient of } x^4 \text { in }(1+x)^{20} \\ &={ }^{20} C_4=\frac{20 \times 19 \times 18 \times 17}{24} \\ &= 4845 \end{aligned} \end{gathered} $

Coefficient of $x^6$ in $(x-2)^2(1+2 x)^{-2}$

$ \begin{aligned} & =\left(x^2-4 x+4\right)\left(1-4 x+\frac{-2(-2-1) \times 4 x^2}{2}+\right. \\ & -\frac{2(-2-1)(-2-2) \times 8 x^3}{6}+ \\ & \frac{-2(-2-1)(-2-2)(-2-3) \times 16 x^4}{24}+ \\ & -\frac{2(-2-1)(-2-2)(-2-3)(-2-4) \times 32 x^5}{5!}+ \\ & -\frac{\left.2(-2-1)(-2-2)(-2-3)(-2-4)(-2-5) \times 64 x^6\right)}{6!} \end{aligned} $

So, coefficient of $x^6$ equals to

$ \begin{aligned} & =\frac{16}{24} \times(-2) \times(-3) \times(-4) \times(-5) \\ & \quad \frac{-4 \times 32}{5!}(-2) \times(-3) \times(-4) \times(-5) \times(-6) \\ & +\frac{4 \times 64}{6!} \times-2 \times(-3) \times-4 \times-5 \times(-6) \times-7 \\ & =2640 \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} &=\sum_{r=0}^5 r^3 \cdot{ }^5 C_r=\sum_{r=1}^5 r^3 \cdot \frac{5}{r} \cdot{ }^4 C_{r-1} \\ &=\sum_{r=1}^5 r^2 \cdot 5 \cdot{ }^4 C_{r-1}=t=r-1 \\ &=5 \sum_{t=0}^4(t+1)^2 \cdot{ }^4 C_t \\ & 5\left(\Sigma t^2 \cdot{ }^4 C_t+2 \Sigma t \cdot{ }^4 C_t+\Sigma^4 C_t\right) \\ &= 5\left(\Sigma t^2 \cdot \frac{4}{t} \cdot{ }^3 C_{t-1}+2 \Sigma t \cdot \frac{4}{t} \cdot{ }^3 C_{t-1}+2^t\right) \\ &= 5\left(4 \Sigma t{ }^3 C_{t-1}+8\left(2^3\right)+2^4\right) \\ & p=t-1 \\ &= 5\left(4 \cdot \Sigma(p+1){ }^3 C_p+64+16\right) \\ &= 5\left(4 \cdot \Sigma p \cdot{ }^3 C_p+4\left(2^3\right)+64+16\right) \\ &= 5\left(4 \cdot 3 \cdot 2^2+32+64+16\right) \\ &= 5(48+32+80)=5(160)=800 \end{aligned} \end{aligned} $

We need to find the coefficient of $x^{12}$ in the expansion of $\left(x^2+2x+2\right)^8$.

Step 1: General Term Formula

Each term in the expansion is made by choosing $P_1$ times $x^2$, $P_2$ times $2x$, and $P_3$ times $2$ from the $8$ factors, so $P_1 + P_2 + P_3 = 8$. The general term is:
$\dfrac{8!}{P_1!P_2!P_3!}(x^2)^{P_1}(2x)^{P_2}(2)^{P_3}$

Step 2: Power of $x$

The power of $x$ in each term is $2P_1$ (from $x^2$) plus $P_2$ (from $2x$), for a total of $2P_1 + P_2$.

Step 3: Find Values for $x^{12}$

We want $2P_1 + P_2 = 12$ and $P_1 + P_2 + P_3 = 8$.
Possible solutions are: $ \begin{array}{c|c|c} P_1 & P_2 & P_3 \\ \hline 6 & 0 & 2 \\ 5 & 2 & 1 \\ 4 & 4 & 0 \\ \end{array} $

Step 4: Write Each Term's Coefficient

For each solution, we substitute into the formula:

• When $P_1=6,\,P_2=0,\,P_3=2$:
  Coefficient: $\dfrac{8!}{6!0!2!} \cdot 2^0 \cdot 2^0 \cdot 2^2$
• When $P_1=5,\,P_2=2,\,P_3=1$:
  Coefficient: $\dfrac{8!}{5!2!1!} \cdot 2^2 \cdot 2^1$
• When $P_1=4,\,P_2=4,\,P_3=0$:
  Coefficient: $\dfrac{8!}{4!4!0!} \cdot 2^4$

Step 5: Calculate Coefficients

$ \begin{aligned} &\text{Total Coefficient} \\ &= \dfrac{8!}{6!2!}\cdot 2^2 + \dfrac{8!}{5!2!1!}\cdot 2^2 \cdot 2^1 + \dfrac{8!}{4!4!}\cdot 2^4 \\ &= 28 \times 4 + 168 \times 4 \times 2 + 70 \times 16 \\ &= 112 + 1344 + 1120 \\ &= 2576 \end{aligned} $

The coefficient of $x^{12}$ is $\boxed{2576}$.

For numerically greatest term in the expansion of $(2 x-3 y)^n$

$ \begin{aligned} r & =\frac{n+1}{1+\left|\frac{2 x}{3 y}\right|}=\frac{13+1}{1+\left|\frac{2 \times \frac{7}{2}}{3 \times \frac{3}{7}}\right|}=\frac{14}{1+\frac{49}{9}} \\ & =\frac{14 \times 9}{9+49}=\frac{126}{58} \\ \therefore \quad[r] & =2 \end{aligned} $

$\therefore T_3$ will be numerically greatest term

$ \begin{aligned} \therefore \quad T_3 & ={ }^{13} C_2(2 x)^{11}(-3 y)^2 \\ & =\frac{13 \times 12}{2} \times\left(2 \times \frac{7}{2}\right)^{11}\left(-3 \times \frac{3}{7}\right)^2 \\ & =13 \times 6 \times 7^{11} \times \frac{3^4}{7^2} \\ & =263^5 7^9 \end{aligned} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} & \sum_{r=1}^8 r^3 \cdot \frac{C_r}{C_{r-1}}=\sum_{r=1}^8 r^3 \cdot \frac{8-r+1}{r} \\ = & \sum_{r=1}^8\left(9 r^2-r^3\right)\left[\because \frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}\right] \\ = & 9\left(\frac{8}{6}(8+1)(17)\right)-\left(\frac{8}{2} \times 9\right)^2 \\ = & 3 \times 4 \times 9 \times 17-4 \times 4 \times 9 \times 9 \\ = & 4 \times 9(51-36)=36 \times 15=540 \end{aligned} \end{aligned} $

Let the given expression,

$ E=\left(1+\frac{1}{x}\right)^{20}\left[30 x(1+x)^{29}+(1+x)^{30}\right] $

Here, $30 x(1+x)^{29}+(1+x)^{30}$

$ \begin{aligned} & =(1+x)^{29}[30 x+(1+x)] \\ & =(1+x)^{29}(31 x+1) \\ \therefore \quad E & =\left(1+\frac{1}{x}\right)^{20}(1+x)^{29}(31 x+1) \\ & =\frac{(x+1)^{20}}{x^{20}}(1+x)^{29}(31 x+1) \\ & =\frac{(x+1)^{49}}{x^{20}}(31 x+1) \\ & =(x+1)^{49}\left(31 x^{-19}+x^{-20}\right) \\ & =31 x^{-19}(x+1)^{49}+x^{-20}(x+1)^{49} \end{aligned} $

For constant term $x^0$

$\therefore$ Constant term in $31 x^{-19}(x+1)^{49}$

i.e., $x^{-19} \cdot x^k=x^0 \Rightarrow k=19$

$\therefore$ Constant term $=31{ }^{49} C_{19}$

Constant term in $x^{-20}(x+1)^{49}$

i.e., $x^{-20} \cdot x^k=x^0 \Rightarrow k=20$

$\therefore$ Constant term $=1 \times{ }^{49} C_{20}$

$\therefore$ Sum of constant term

$ \begin{aligned} & =31{ }^{49} C_{19}+{ }^{49} C_{20} \\ & =30{ }^{49} C_{19}+{ }^{49} C_{19}+{ }^{49} C_{20} \\ & =30 \cdot{ }^{49} C_{19}+{ }^{50} C_{20} \end{aligned} $

We have, $x^{3 / 2}(3+x)^{1 / 2}$

$ \begin{aligned} & =x^{3 / 2} x^{1 / 2}\left(1+\frac{3}{x}\right)^{1 / 2} \\ & =x^2\left(1+\frac{3}{x}\right)^{1 / 2} \end{aligned} $

Now, coefficient of general term of $\left(1+\frac{3}{x}\right)^{1 / 2}$ is

$ \begin{aligned} & \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{1}{3}\right) \ldots\left(\frac{1}{2}-k+1\right)}{k!} \\ & \quad \frac{(-1)^{k+1} 1 \cdot 3 \cdot 5 \ldots(2 k-3)}{2 k k!} \end{aligned} $

On multiply by $x^2$, then

$ x^2\left(1+\frac{3}{x}\right)^{1 / 2}=\sum_{k=0}^{\infty}{ }^{1 / 2} C_k 3^k x^{2-k} $

For coefficient of $\frac{1}{x^n}, 2-k=-n$

$ \Rightarrow \quad k=n+2 $

Coefficient of $\frac{1}{x^n}$ is

$ \begin{aligned} & \frac{(-1)^{n+3} 1 \cdot 3 \cdot 5 \ldots(2(n+2)-3) 3^{n+2}}{2^{n+2}(n+2)!} \\ & \quad=\frac{(-1)^{n+1} 1 \cdot 3 \cdot 5 \ldots(2 n+1)}{2^{n+1}(n+2)!} 3^{n+2} \end{aligned} $

Given, coefficient of 3rd term from beginning in the expansion $\left(a x^2-\frac{8}{b x}\right)^9 =$ Coefficient of 3rd term from end in the expansion $\left(a x-\frac{2}{b x^2}\right)^9$

For $\left(a x^2-\frac{8}{b x}\right)^9$

$ \begin{aligned} T_3 & ={ }^9 C_2\left(a x^2\right)^7\left(\frac{-8}{b x}\right)^2 \\ & =\frac{9!}{2!7!} a^7 x^{14} \cdot \frac{64}{b^2 x^2} \\ & =\frac{36 \cdot a^7}{b^2} \cdot x^{12} \cdot 64 \end{aligned} $

So, its coefficient is $\frac{36 \cdot 64 a^7}{b^2}$

For $\left(a x-\frac{2}{b x^2}\right)^9$

$ \begin{aligned} T_8 & ={ }^9 C_7(a x)^2\left(\frac{-2}{b x^2}\right)^7 \\ & =\frac{9!}{2!7!} a^2 x^2 \cdot \frac{-128}{b^7 x^{14}} \end{aligned} $

So, if coefficient is

$ =-36 \times 128 \frac{a^2}{b^7} $

Now, $36 \cdot \frac{64 a^7}{b^2}=-36 \cdot 128 \frac{a^2}{b^7}$

$ \begin{aligned} \Rightarrow & & 64 \cdot a^5 b^5 & =-128 \\ \Rightarrow & & a^5 b^5 & =-2 \end{aligned} $

Given, $5^{2 n}-48 n+k$ is divisible by 24

Let $\quad E(n)=5^{2 n}-48 n+k$

and $\quad E(n) \equiv 0(\bmod 24), \forall n \in N$

$\Rightarrow \quad 5^{2 n}-48 n+k \equiv 0(\bmod 24)$

$\Rightarrow \quad k=48 n-5^{2 n}(\bmod 24)$

For $n=1$,

$ \begin{aligned} k & =48(1)-5^{2 \times 1}(\bmod 24) \\ & =48-25(\bmod 24) \\ & =23(\bmod 24) \end{aligned} $

For $n=2$

$k=48(2)-5^1(\bmod 24)$

$ \begin{aligned} & =96-625(\bmod 24) \\ & =529(\bmod 24)=23(\bmod 24) \end{aligned} $

Thus, divisibility for 24 to hold for all $n$, we must have $k \equiv 23(\bmod 24)$

Thus, the least positive integral value of $k$ is 23 .

Given $X \sim B(7, P)$ is a binomial variate and $P(X=3)=P(X=5)$

Using the binomial probability formula

$ P(X=K)=\binom{n}{K} P^K(1-P)^{n-K} $

Here $n=7$,

So, $P(X=3)=\binom{7}{3} P^3(1-P)^4$

And, $P(X=5)=\binom{7}{5} P^5(1-P)^2$

$ \begin{aligned} & \therefore\binom{7}{3} P^3(1-P)^4=\binom{7}{5} P^5(1-P)^2 \\ & \Rightarrow 35 P^3(1-P)^4=21 P^5(1-P)^2 \\ & \Rightarrow 35(1-P)^2=21 P^2, \text { where } P \neq 0, P \neq 1 \\ & \Rightarrow 5(1-P)^2=3 P^2 \\ & \Rightarrow 5\left(1-2 P+P^2\right)=3 P^2 \\ & \Rightarrow 5-10 P+5 P^2-3 P^2=0 \\ & \Rightarrow 2 P^2-10 P+5=0 \\ & \Rightarrow P=\frac{-(-10) \pm \sqrt{100-4 \times 2 \times 5}}{2 \times 2} \\ & \quad=\frac{10 \pm \sqrt{100-40}}{4} \\ & \quad=\frac{10 \pm 2 \sqrt{15}}{4}=\frac{5 \pm \sqrt{15}}{2} \end{aligned} $

But $P=\frac{5+\sqrt{15}}{2}>1$, So it can't be a valid probability.

$ \therefore P=\frac{5-\sqrt{15}}{2} $

We are given the expression $(p-q)^{14}$. We need to find the value of $\frac{p+q}{p-q}$, given that the sum of the 7th and 8th terms in its expansion is zero.

Finding the 7th and 8th Terms

In the binomial expansion of $(p-q)^{14}$, the formula for the $(r+1)$th term is: $T_{r+1} = {^{14}C_r} p^{14-r} (-q)^r$.

The 7th term, $T_7$, is $^{14}C_6 p^{8}(-q)^{6} = {^{14}C_6} p^8 q^6$.

The 8th term, $T_8$, is $^{14}C_7 p^7(-q)^7 = -{^{14}C_7} p^7 q^7$.

Setting Up the Equation

We are told that $T_7 + T_8 = 0$.

This means: ${^{14}C_6} p^8 q^6 - {^{14}C_7} p^7 q^7 = 0$.

We can factor out $p^7 q^6$ from both terms: $p^7 q^6 \left({^{14}C_6}p - {^{14}C_7}q \right) = 0$.

Because $p$ and $q$ are not zero, we must have ${^{14}C_6}p = {^{14}C_7}q$.

Solving for $\frac{p}{q}$

Divide both sides by $q$ and ${^{14}C_6}$ to get $\frac{p}{q} = \frac{^{14}C_7}{^{14}C_6}$.

Now, $^{14}C_7 = \frac{14!}{7! \cdot 7!}$ and $^{14}C_6 = \frac{14!}{6! \cdot 8!}$.

Write $\frac{^{14}C_7}{^{14}C_6}$ as $\frac{8}{7}$ (after cancelling out common terms, which are shown in the expansion above).

Express $p$ and $q$

If $\frac{p}{q} = \frac{8}{7}$, let $p = 8k$ and $q = 7k$ for some $k$.

Find $\frac{p+q}{p-q}$

Then $p+q = 8k + 7k = 15k$ and $p-q = 8k - 7k = k$.

So, $\frac{p+q}{p-q} = \frac{15k}{k} = 15$.

General term of $(x+3 y)^{13}$ is

$ \begin{aligned} & T_{r+1}={ }^{13} C_r(x)^{13-r}(3 y)^r \\ & \text { at } x=1 / 2, y=1 / 3 \end{aligned} $

So, $3 y=1$

$ \text { then, } T_{r+1}={ }^{13} C_r\left(\frac{1}{2}\right)^{13-r}(1)^r={ }^{13} C_r\left(\frac{1}{2}\right)^{13-r} $

For maximum term

$ \begin{aligned} & \begin{aligned} \left|\frac{T_{r+1}}{T_r}\right| & =\frac{{ }^{13} C_r}{{ }^{13} C_{r-1}}\left(\frac{1}{2}\right)^{13-r-(13-(r-1))} \\ & =\frac{{ }^{13} C_r}{{ }^{13} C_{r-1}}\left(\frac{1}{2}\right)^{-1}=\frac{14-r}{r} \times 2 \end{aligned} \\ & \text { put }\left|\frac{T_{r+1}}{T_r}\right|=1 \end{aligned} $

$ \Rightarrow \quad 28-2 r=r \Rightarrow r=28 / 3 \approx 9.33 $

So, maximum occurs either $r=9$ or $r=10$

for $r=9$

$ T_{10}={ }^{13} C_9\left(\frac{1}{2}\right)^4=44.6875 $

and for $r=10$

$ T_{11}={ }^{13} C_{10}\left(\frac{1}{2}\right)^3=35.75 $

Hence, the greatest term is $T_{10}$

$ T_{10}={ }^{13} C_9\left(\frac{1}{2}\right)^4 $

We have, $(2 m+1)^{2 n}=\left[(2 m+1)^2\right]^n$

$ \begin{aligned} = & \left(4 m^2+4 m+1\right)^n \\ = & (1+4 m(m+1)]^n \\ = & { }^n C_0[(4 m)(m+1)]^n \\ & \quad+{ }^n C_1[4 m(m+1)]^{n-1}+ \\ & \quad \ldots+{ }^n C_{n-1}(4 m(m+1)]^1+{ }^n C_n \\ = & { }^n C_0\left(4^n\right)(m(m+1))^n \\ & \quad+{ }^n C_1(4 m(m+1))^{n-1} \\ & \quad+\ldots+{ }^n C_{n-1}(4)(m(m+1))+1 \end{aligned} $

$\because m(m+1)$ is always even.

$ \therefore 8(k)+1 $

∴ Remainder is 1 .

$ \begin{aligned} & \text { We have, } \sum_{r=1}^{15} r^2 \frac{{ }^{15} C_r}{{ }^{15} C_{r-1}} \\ & \begin{array}{l} \because \quad \frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r} \\ \Rightarrow \quad \sum\limits_{r = 1}^{15} {{r^2}} \cdot \frac{(15-r+1)}{r} \\ \Rightarrow \quad \sum\limits_{r = 1}^{15} {{r}}(16-r) \Rightarrow \sum_{r=1}^{15}\left(16 r-r^2\right) \\ \Rightarrow \quad 16\left(\frac{15}{2} \times 16\right)-\frac{15}{6}(16)(31) \\ \quad=15(16)\left(8-\frac{31}{6}\right) \\ \quad=\frac{15 \times 16 \times 17}{6} \\ \quad=5 \times 8 \times 17=680 \end{array} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{gathered} \left(\frac{1}{81}\right)^n-{ }^{2 n} C_1 \frac{10}{81^n}+{ }^{2 n} C_2 \frac{10^2}{81^n}-\ldots+\frac{10^{2 n}}{81^n} \\ \Rightarrow \quad \frac{1}{81^n}\left({ }^{2 n} C_0-{ }^{2 n} C_1 10^1+{ }^{2 n} C_2 10^2\right. \left.+\ldots+{ }^{2 n} C_{2 n} 10^{2 n}\right) \\ \Rightarrow \quad \frac{1}{81^n}(1-10)^{2 n}=\frac{9^{2 n}}{9^{2 n}}=1 \end{gathered} \end{aligned} $

We have,

$ \begin{aligned} (1+x)^{\frac{27}{5}}= & 1+\frac{27}{5} x+\frac{\frac{27}{5}\left(\frac{27}{5}-1\right)}{2!} x^2 \\ & +\frac{\frac{27}{5}\left(\frac{27}{5}-1\right)\left(\frac{27}{5}-2\right)}{3!} x^3+ \end{aligned} $

the first negative term

$ \frac{\frac{27}{5}\left(\frac{27}{5}-1\right) \ldots\left(\frac{27}{5}-6\right)}{7!} x^7 $

that is 8 term.

Coefficient of $x^{10}$ in the expansion of $\left(x+\frac{2}{x}-5\right)^{12}$

Using multinomial theorem

$ =\frac{12!}{r_{1}!r_{2}!r_{3}!} x^n\left(\frac{2}{x}\right)^{r_2}(-5)^{r_3} $

for coefficient of $x^{10}$

$ r_1+r_2+r_3=12 \Rightarrow r_1-r_2=10 $

$ \begin{aligned} &\begin{array}{|c|c|c|} \hline \boldsymbol{r}_1 & \boldsymbol{r}_2 & \boldsymbol{r}_3 \\ \hline 10 & 0 & 2 \\ \hline 11 & 1 & 0 \\ \hline \end{array}\\ &\text { Coefficient }\\ &\begin{aligned} & =\frac{12!}{10!0!2!} 2^0 \cdot(-5)^2+\frac{12!}{11!1!!} 2^{!} .(-5)^0\\ & =\frac{12 \times 11}{2} \times 25+24 \\ & =1650+24=1674 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} S_1 & =\sum_{j=1}^{10} j(j-1) \cdot{ }^{10} C_j \\ & =\sum_{j=1}^{10} j(j-1) \frac{10}{j} \cdot \frac{9}{j-1} \cdot{ }^8 C_{j-2} \\ & =\sum_{j=2}^{10} 90 \cdot{ }^8 C_{j-2}=90 \cdot 2^8 \end{aligned}\\ &\text { } \begin{aligned}and\,\, S_2 & =\sum_{j=1}^{10} j \cdot{ }^{10} C_j \\ & =\sum_{j=1}^{10} j \cdot \frac{10}{j}{ }^9 C_{j-1}=10 \sum_{j=1}^{10}{ }^9 C_{j-1} \\ & =10 \cdot 2^9 \\ S_3 & =\sum_{j=1}^{10} j^{210} C_j=10 \sum_{j=1}^{10} j \cdot{ }^9 C_{j-1} \\ & =10 \cdot 11 \cdot 2^{9-1}=110 \cdot 2^8 \\ & \quad\left[\because \sum r \cdot{ }^n C_r=n \cdot 2^{n-1}\right] \\ & =55 \cdot 2^9 \end{aligned} \end{aligned} $

Given,

$ \begin{aligned} y & =\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots+\infty \\ & =\frac{3}{1!} \cdot \frac{1}{4}+\frac{3 \cdot 5}{2!} \cdot \frac{1}{4^2}+\frac{3 \cdot 5 \cdot 7}{3!} \cdot \frac{1}{4^3}+\ldots+\infty \end{aligned} $

Since, using binomial expansion

$ \begin{aligned} (1-x)^{-n}=1+n x+ & \frac{n(n+1)}{2!} x^2 +\frac{n(n+1)(n+2)}{3!} x^3+\ldots \end{aligned} $

Here, $n x=\frac{3}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} &\begin{aligned} & \frac{n(n+1)}{2} x^2=\frac{3 \cdot 5}{2!} \cdot \frac{1}{4^2} \\ \Rightarrow \quad & n(n+1) x^2=\frac{15}{16} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned}\\ &\text { From Eq. (i) and (ii), we get }\\ &\begin{aligned} \frac{n(n+1) x^2}{n^2 x^2} & =\frac{15 / 16}{(3 / 4)^2} \\ & =\frac{15 / 16}{9 / 16}=\frac{5}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{array}{lc} \Rightarrow & \frac{n+1}{n}=\frac{5}{3} \\ \Rightarrow & 3 n+3=5 n \\ \Rightarrow & n=\frac{3}{2} \end{array}\\ &\text { So, } x=\frac{3}{4} \cdot \frac{1}{n}=\frac{3}{4} \times \frac{2}{3}=\frac{1}{2}\\ &\begin{aligned} \therefore \quad y & =(1-x)^{-n}-1=\left(1-\frac{1}{2}\right)^{-3 / 2}-1 \\ & =\left(\frac{1}{2}\right)^{-3 / 2}-1=2^{3 / 2}-1=2 \sqrt{2}-1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { So, } \quad y+1=2 \sqrt{2}-1+1=2 \sqrt{2} \\ & \quad(y+1)^2=(2 \sqrt{2})^2=8 \\ & \Rightarrow y^2+2 y+1-8=0 \\ & \Rightarrow \quad y^2+2 y-7=0 \end{aligned}\\ &\text { Thus, the quadratic equation is }\\ &y^2+2 y-7=0 \end{aligned} $

We will expand $\left(1+x-x^2\right)^6$ and find the coefficients of $x^4$ and $x^6$.

Multimonial expansion states that

$ \begin{aligned} \left(a_1+a_2+a_3\right)^n & =\sum \frac{n!}{k_{1}!k_{2}!k_{3}!} \\ a_1^k a_2^{k_2} a_3^{k_3} & \end{aligned} $

where $k_1+k_2+k_3=n$

In our case, $a_1=1, a_2=x, a_3=-x^2$ and $n=6$

$ \begin{aligned} & \text { From } k_1+k_2+k_3=6 \\ & \Rightarrow \quad k_1=6-k_2-k_3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

To get $x^4$, we can have different combinations of $x$ and $x^2$

$k_1$ : number of times 1 is chosen

$k_2$ : number of times $x$ is chosen

$k_3$ : number of times $x^2$ is chosen

The equation is: $k_2+2 k_3=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, if $k_3=0$, then $k_2=4$ and $k_1=2$

If $k_3=1$, then $k_2=2$ and $k_1=3$

If $k_3=2$, then, $k_2=0, k_1=4$

Now, find the coefficients for each valid combination.

Coefficient of

$ \begin{aligned} & x^4=\frac{6!}{2!4!0!}-\frac{6!}{3!2!1!}+\frac{6!}{4!0!2!} \\ & =15-60+15=-30 \end{aligned} $

Now, for $x^6, k_2+2 k_3=6$ and

$ k_1+k_2+k_3=6 $

If $k_3=0$, then $k_2=6$, and $k_1=0$

If $k_3=1$, then $k_2=4$, and $k_1=1$

If $k_3=2$ then $k_2=2$ and $k_1=2$

If $k_3=3$, then $k_2=0$, and $k_1=3$

So, coefficient of $x^6=\frac{6!}{0!6!0!}-\frac{6!}{1!4!1!}+\frac{6!}{2!2!2!}-\frac{6!}{3!0!3!}$

$ =1-30+90-20=41 $

$\therefore$ Sum of the coefficient of $x^4$ and $x^6=41-30=11$

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} 11^{12}-11^2 & =k\left(5 \times 10^9+6 \times 10^9\right. + \left.33 \times 10^8+110 \times 10^7+\ldots+33\right) \\ 11^{12}-11^2 & =(10+1)^{12}-(10+1)^2 \\ & =10^{12}+12 \times 10^{11}+66 \times 10^{10}+220 \times 10^9 \cdots+66 \times 10^2+120+1-121 \end{aligned} \end{aligned} $

$ \begin{aligned} & =10^{12}+12 \times 10^{11}+66 \times 10^{10}+220 \times 10^9 \ldots+66 \times 10^2 \\ & =2 \times 10^2\left(5 \times 10^9+6 \times 10^9+33 \times 10^8\right. \left.+110 \times 10^7+\ldots+33\right) \\ & =200\left(5 \times 10^9+6 \times 10^9+33 \times 10^8\right. \left.+110 \times 10^7+\ldots+33\right) \\ & \therefore k=200 \end{aligned} $

$ \because(1+x)^n=\sum_{k=0}^n C_k x^k$

and

$ \left(C_0+C_1\right)-\left(C_2+C_3\right)+\left(C_4+C_5\right)-\cdots $

can be grouped as

$ \begin{gathered} \sum_{k=0}^{n / 2}(-1)^k\left(C_{2 k}+C_{2 k+1}\right) \\ \text { and }(1+i)^n=\sum_{k=0}^n C_k i^k \\ \operatorname{Re}\left[(1+i)^n\right]=C_0-C_2+C_4-C_6+\cdots \\ \operatorname{Im}\left[(1+i)^n\right]=C_1-C_3+C_5-C_7+\cdots \end{gathered} $

Adding both

$ \begin{aligned} & \operatorname{Re}\left[(1+i)^n\right]+\operatorname{Im}\left[(1+i)^n\right] \\ & =\left(C_0+C_1\right)-\left(C_2+C_3\right)+\left(C_4+C_5\right)-\ldots \end{aligned} $

Now, $(1+i)^n$

$ =(\sqrt{2})^n\left(\cos \left(\frac{n \pi}{4}\right)+i \sin \left(\frac{n \pi}{4}\right)\right) $

Hence, $\left(C_0+C_1\right)-\left(C_2+C_3\right)+\left(C_4+C_5\right)$

$ \begin{aligned} -\ldots & =\operatorname{Re}\left[(1+i)^n\right]+\operatorname{Im}\left[(1+i)^n\right] \\ & =2^{n / 2}\left(\cos \left(\frac{n \pi}{4}\right)+\sin \left(\frac{n \pi}{4}\right)\right) \end{aligned} $