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{aligned} & T_{r+1}={ }^9 C_r\left(\frac{x^{-2 / 5}}{2}\right)^r\left(x^{2 / 3}\right)^{9-r} \\ & ={ }^9 C_r \frac{1}{2^r} x^{\frac{2}{3}(9-r)+\left(\frac{-2 r}{5}\right)} \\ & ={ }^9 C_r \cdot \frac{1}{2^r} \cdot x^{6-\frac{16 r}{15}} \end{aligned} $

For coefficient of $x^{2 / 3}$

$\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{2}{3} \\ & \Rightarrow 90-16 r=10 \\ & \Rightarrow r=5 \end{aligned}$

For coefficient of $x^{-2 / 5}$

$\begin{aligned} & \Rightarrow 6-\frac{16 r}{15}=\frac{-2}{5}\\ & \Rightarrow 90-16 r=-6 \\ & \Rightarrow r=6 \end{aligned}$

Sum of coefficient of $x^{2 / 3}$ & $x^{-2 / 5}$

$\begin{aligned} & ={ }^9 C_5 \cdot \frac{1}{2^5}+{ }^9 C_6 \cdot \frac{1}{2^6} \\ & =\frac{9!}{5!4!}\left(\frac{1}{2^5}\right)+\frac{9!}{6!3!}\left(\frac{1}{2^6}\right)=\frac{21}{4} \end{aligned}$

$x^2(1+x)^{98}+x^3(1+x)^{97}+\ldots+x^{54}(1+x)^{46}$

It is a G.P. with first term $=x^2(1+x)^{98}$

and common ratio $=\frac{x}{1+x}$

sum of these term $=x^2(1+x)^{98}\left(\frac{\left(\frac{x}{1+x}\right)^{53}-1}{\frac{x}{1+x}-1}\right)$

$=x^2(1+x)^{98}\left((1+x)-x^{53}(1+x)^{-52}\right)$

$\begin{aligned} & ={ }^{99} \mathrm{C}_{68}-{ }^{46} \mathrm{C}_{15} \\ & \Rightarrow p=68, q=15 \\ & \Rightarrow p+q=83 \end{aligned}$

$\begin{aligned} & \left(\sqrt{a} x^2+\frac{1}{2 x^3}\right)^{10} \\ & T_{r+1}={ }^{10} C_r\left(\sqrt{a} x^2\right)^{10-r}\left(\frac{1}{2 x^3}\right)^r \end{aligned}$

Independent of $x \Rightarrow 20-2 r-3 r=0$

$r=4$

Independent of $x$ is ${ }^{10} C_4(\sqrt{a})^6\left(\frac{1}{2}\right)^4=105$

$\begin{gathered} \frac{210}{2 \times 8} a^3=105 \\ \Rightarrow \quad a=2 \\ a^2=4 \end{gathered}$

$\begin{aligned} & \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[5]{3}}\right)^{12} \\ & T_{r+1}={ }^{12} C r\left(\frac{\sqrt[5]{3}}{x}\right)^{12-r}\left(\frac{2 x}{\sqrt[3]{5}}\right)^r \end{aligned}$

For constant term $-12+r+r=0$

$\begin{aligned} & \Rightarrow \quad r=6 \\ & \therefore \quad \text { Constant term }={ }^{12} C_6 \frac{(3)^{\frac{6}{5}}}{(5)^{\frac{6}{3}}}(2)^6 \end{aligned}$

$\begin{aligned} & ={ }^{12} C_6 \times \frac{2^6}{25} \times 3.3^{\frac{1}{5}} \\ & =\frac{231}{25} \times 2^8 \cdot 3^{\frac{1}{5}} \cdot 3 \\ & =\frac{693}{25} \cdot 2^8 \sqrt[5]{3} \\ & \therefore \quad \alpha=\frac{693}{25} \\ & 25 \alpha=693 \end{aligned}$

$\begin{aligned} & (1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots{ }^n C_n x^n \\ & { }^n C_4,{ }^n C_5 \&{ }^n C_6 \text { are in A.P. } \\ & { }^n C_5-{ }^n C_4={ }^n C_6-{ }^n C_5 \\ & \Rightarrow \frac{n!}{5!(n-5)!}-\frac{n!}{4!(n-4)!}=\frac{n!}{6!(n-6)!}-\frac{n!}{5!(n-5)!} \\ & \Rightarrow 30(n-9)(n-6)=5(n-4)(n-11) \\ & \Rightarrow 30 n^2-450 n+1620=5 n^2 \\ & \Rightarrow \frac{1}{n-5}\left[\frac{n-4-5}{5(n-4)}\right]=\frac{1}{5}\left[\frac{n-5-6}{6(n-5)}\right] \\ & \Rightarrow \frac{n-9}{5(n-4)}=\frac{1}{5}\left[\frac{n-11}{6}\right] \\ & \Rightarrow n^2-21 n+98=0 \\ & n_{\max }=14 \end{aligned}$

$\begin{aligned} & T_{r+1}={ }^{15} \mathrm{C}_r\left(2^{1 / 5}\right)^{15-r}\left(5^{1 / 3}\right)^r \\ & ={ }^{15} C_r 5^{r / 3} 2^{\left(3-\frac{r}{5}\right)} \end{aligned}$

For rational terms,

$\frac{r}{3}$ and $\frac{r}{5}$ must be integer

3 and 5 divide $r \Rightarrow 15$ divides $r \Rightarrow r=0$ and $r=15$

${ }^{15} C_0 5^0 2^3+{ }^{15} C_{15} 5^5 2^{(0)}$

$\begin{aligned} & =8+3125 \\ & =3133 \end{aligned}$

Put $x=1$

$\therefore \mathrm{a}=1$

$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\int_\limits0^{\mathrm{x}} \frac{\ln (1+\mathrm{t})}{1+\mathrm{t}^{2024}} \mathrm{dt}}{\mathrm{x}^2}$

Using L' HOPITAL Rule

$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\ln (1+\mathrm{x})}{\left(1+\mathrm{x}^{2024}\right)} \times \frac{1}{2 \mathrm{x}}=\frac{1}{2}$

Now, $\mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0, \mathrm{x}^2+\mathrm{x}+4=0$

$(\mathrm{D}<0)$

$\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$

$2-p, p, 2-\alpha, \alpha$

Binomial coefficients are

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

Also, ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+2}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+3}=2$

$\Rightarrow \quad{ }^{n+1} C_{r+3}=2$ $\quad\text{..... (2)}$

From (1) and (2)

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

Data Inconsistent

${ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}}=(\mathrm{k}^2-8){ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}$

$\underbrace{\mathrm{r}+1 \geq 0, \quad \mathrm{r} \geq 0}_{\mathrm{r} \geq 0}$

$\begin{aligned} & \frac{{ }^{n-1} C_r}{{ }^n C_{r+1}}=k^2-8 \\ & \frac{r+1}{n}=k^2-8 \\ & \Rightarrow k^2-8>0 \\ & (k-2 \sqrt{2})(k+2 \sqrt{2})>0 \end{aligned}$

$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)\quad \text{.... (I)}$

$\begin{aligned} \therefore & \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1 \\ \Rightarrow & \mathrm{k}^2-8 \leq 1 \\ & \mathrm{k}^2-9 \leq 0 \\ & -3 \leq \mathrm{k} \leq 3 \quad \text{.... (II)} \end{aligned}$

From equation (I) and (II) we get

$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$

Sum of coefficients in the expansion of $\left(1-3 \mathrm{x}+10 \mathrm{x}^2\right)^{\mathrm{n}}=\mathrm{A}$

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

and sum of coefficients in the expansion of

$\begin{aligned} & \left(1+x^2\right)^n=B \\ & \text { then } B=(1+1)^n=2^n \\ & A=B^3 \end{aligned}$

We are given that $\left(a+bx+cx^2\right)^{10} = \sum_{i=0}^{20} p_i x^i$, and we are given that $p_1 = 20$ and $p_2 = 210$.

We need to find the value of $2(a+b+c)$.

Now we are given $p_1 = 20$ and $p_2 = 210$. So, $ 10a^9 b = 20 \implies a^9 b = 2 $

and $ 45a^8 b^2 + 10a^9 c = 210 $

Now, divide the second equation by $a^8$: $ 45b^2 + 10ac = 210 $

We know that $a^9 b = 2$. Taking the $9^{th}$ root of both sides: $ ab = \sqrt[9]{2} $

Now, let $k = ab = \sqrt[9]{2}$. We can rewrite the equation for $x^2$ term as: $ 45k^2 + 10k^9 = 210 $

From the equation $ab = k = \sqrt[9]{2}$, we know that $a$ and $b$ are positive integers. Thus, $k = 2$ (as both $a$ and $b$ must be factors of 2). Now we have:

$ a+b = 2 $

and from the equation $a^9 b = 2$, we get $a = 1, b = 2$ or vice versa.

Now we need to find the value of $c$. We can use the equation for the $x^2$ term again:

$ 45a^8 b^2 + 10a^9 c = 210 $

Using $a=1$ and $b=2$, we get:

$ 45(1)^8 (2)^2 + 10(1)^9 c = 210 \implies 180 + 10c = 210 \implies c = 3 $

So, $a=1$, $b=2$, and $c=3$. Now, we need to find the value of $2(a+b+c)$:

$ 2(a+b+c) = 2(1+2+3) = 2(6) = 12 $

Therefore, the answer is $\boxed{12}$.