Binomial Theorem

$ \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} $

The mean of a binomial distribution is $x$, and the variance is $5$.

We know that the mean, $\mu = n p = x$. This means $n$ times $p$ is $x$.

The formula for variance is $\sigma^2 = n p(1-p) = 5$. This means $n$ times $p$ times $(1-p)$ is $5$.

From the mean formula, $p = \frac{x}{n}$.

Substitute $p = \frac{x}{n}$ into the variance formula:

$ x\left(1-\frac{x}{n}\right) = 5 $

If we open the bracket, we get: $ x - \frac{x^2}{n} = 5 $

Take $\frac{x^2}{n}$ to the other side: $ \frac{x^2}{n} = x - 5 $

Now, solve for $n$: $ n = \frac{x^2}{x - 5} $

To get a whole number for $n$, $x-5$ must divide $x^2$ exactly. Let's check different $x$ values:

If $x = 6$: $ n = \frac{36}{1} = 36 $ $36$ is an integer.

If $x = 10$: $ n = \frac{100}{5} = 20 $ $20$ is an integer.

If $x = 30$: $ n = \frac{900}{25} = 36 $ $36$ is an integer.

So, the possible values for $x$ are: $6, 10, 30$.

As we know that, coefficient of $x^{11}$ in the expansion of $(1+x)^{11}$ is ${ }^{11} C_{11}$.

Now, we have to find coefficient of $x^{11}$ in the expansion of $\left(1+\alpha x+\beta x^2\right)(1+x)^{11}$. When we choosing 1 from $\left(1+\alpha x+\beta x^2\right)$ then we need $x^{11}$ from $(1+x)^{11}$.

So, coefficient $={ }^{11} C_{11}=1$

Similarly, we choosing $\alpha x$ from $\left(1+\alpha x+\beta x^2\right)$ then we need $x^{10}$ from $(1+x)^{11}=$ Coefficient $=\alpha \times{ }^{11} C_{10}=11 \alpha$ and when we choosing $\beta x^2$ from $\left(1+\alpha x+\beta x^2\right)$ then we need $x^9$ from $(1+x)=$ Coefficient $=\beta \times{ }^{11} C_9=55 \cdot \beta$

$\therefore$ Coefficient of $x^{11}$ in the expansion of $\left(1+\alpha x+\beta x^2\right)$

$(1+x)^{11}=1+11 \alpha+55 \beta=144$ ( given )          ...(i)

Just, in similar way, we have to find coefficient of $x^{10}$ such that

$ 11+55 \alpha+165 \beta=396 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

$ \begin{gathered} \alpha=-2 \text { and } \beta=3 \\ \therefore \alpha^2+\beta^2=(-2)^2+(3)^2=13 \end{gathered} $

Given, $\frac{-2}{3} < x < \frac{2}{3}$

$ \Rightarrow-2<3 x<2 \Rightarrow-1<\frac{3 x}{2}<1 $

Now, $\frac{1}{3 \sqrt{2-3 x}}=\frac{1}{(2-3 x)^{1 / 3}}$

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

$ =\frac{1}{3 \sqrt{2}}\left[1+\left(\frac{1}{3}\right)\left(\frac{3 x}{2}\right)+\frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}+1\right)}{2 \times 1}\left(\frac{3 x}{2}\right)^2\right. +\frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{3}+2\right)}{3 \times 2 \times 1}\left(\frac{3 x}{2}\right)^3\left.+\frac{\frac{1}{3}\left(\frac{1}{3}+1\right)\left(\frac{1}{3}+2\right)\left(\frac{1}{3}+3\right)}{4 \times 3 \times 2 \times 1}\left(\frac{3 x}{2}\right)^4+\ldots\right]$

$ \begin{aligned} &\therefore 5 \text { th term when } x=\frac{1}{2} \text { is given by }\\ &\begin{array}{r} \frac{1}{\sqrt[3]{2}}\left[\frac{1}{3} \times \frac{4}{3} \times \frac{7}{3} \times \frac{10}{3} \times \frac{1}{24} \times \frac{81}{16} \times \frac{1}{16}\right] \\ =\frac{1}{3 \sqrt{2}}\left[\frac{7 \times 5}{24 \times 2 \times 6}\right]=\frac{35}{768(\sqrt[3]{2})} \end{array} \end{aligned} $

General term for $\left(x+y^2\right)^{13}$

$ ={ }^{13} C_n x^{13-n} y^{2 n} $

The general term for $\left(x^2+y\right)^{14}$

$ ={ }^{14} C_{r_2} x^{2 r_2} y^{14-r_2} $

Equate the power of $x$ and $y$

$ \begin{aligned} & 13-r_1=2 r_2 \\ & 2 r_1=14-r_2 \end{aligned} $

Solving these equation, we get

$ r_1=5, r_2=4 $

The common term is $x^{13-5} y^{2(9)}=x^8 y^{10}$ The common term is $x^8 y^{10}$

$ \Rightarrow \quad r=8, s=10 $

$\alpha=1$, since there is only one common term

$ \begin{aligned} & \Rightarrow \quad \alpha(r+s)=1(8+10)=18 \\ & \therefore \text { Sum } \alpha \sum_{r, s}(r+s)=18 \end{aligned} $

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

$ \begin{gathered} \Rightarrow(1+3 x)^{-3}=1-3(3 x)+\frac{(-3)(-4)(3 x)^2}{2!} \\ +\frac{(-3)(-4)(-5)(3 x)^3}{3!}+\ldots \\ \Rightarrow(1+3 x)^{-3}=1-9 x+54 x^2-270 x^3+\ldots \\ \begin{aligned} (1+4 x & \left.-3 x^2\right)\left(1-9 x+54 x^2-270 x^3+\ldots\right) \\ = & 1\left(-270 x^3\right)+4 x\left(5 x^2\right)-3 x^2(-9 x) \\ = & -270 x^3+216 x^3+27 x^3 \\ = & -270+216+27=-27 \end{aligned} \end{gathered} $

$\Rightarrow$ The coefficient of $x^3=-27$

$9^{11}+11^9=(10-1)^{11}+(10+1)^9$

$ \begin{aligned} ={ }^{11} C_0 10^{11} & +{ }^{11} C_1 10^9+\ldots{ }^{11} C_{11} \\ & +{ }^9 C_0 10^{10}+{ }^9 C_1 10^8+\ldots+{ }^9 C_9 \end{aligned} $

Sum of last two terms of both expansion as other term have at least $10^2$ as a factor

$ (110-1)+90+1=200 $

$\therefore \quad 9^{11}+11^9=$ Multiple of

$ 100+200=\text { Multiple of } 100 $

Since $200=2 \times 10^2$

So, $9^{11}+11^9$ is divisible by $10^2$

$ \therefore \quad k=2 $

Given expansion $\left(x^{\frac{1}{2}}+y^{\frac{1}{k}}\right)^{10}$

General term $={ }^{10} C_r x^{\frac{10-r}{2}} y^{\frac{r}{k}}$

Given expansion have only 9 irrational term

$\therefore$ Expansion has only two rational term Two rational term is possible

LCM of 2 and $k$ is 10

$ \therefore k=5 $

$\left(x^2+x-2\right)^5$

Apply multinomial theorem

$ \Sigma \frac{5!}{p!q!r!}\left(x^2\right)^p(x)^q(-2)^r $

Such that $p+q+r=5$ and $p, q, r$ are non-negative integer.

$ \begin{aligned} & 2 p+q=2 \\ & (p, q, r)=(1,0,4)(0,2,3) \\ & \frac{5!}{4!}\left(-4^4+\frac{5!}{2!3!}\left(-2^3\right.\right. \\ & 5 \times 16+10(-8)=80-80=0 \end{aligned} $

We have, $P_n={ }^n C_0 \cdot{ }^n C_1{ }^n C_2 \ldots{ }^n C_n$ and

$ \begin{aligned} & P_{n+1}={ }^{n+1} C_0 \cdot{ }^{n+1} C_1{ }^{n+1} C_2 \cdots{ }^{n+1} C_{n+1} \\ & \therefore \frac{P_{n+1}}{P_n}=\frac{{ }^{n+1} C_0 \cdot{ }^{n+1} C_1{ }^{n+1} C_2 \cdots{ }^{n+1} C_{n+1}}{{ }^n C_0 \cdot{ }^n C_1 \cdot{ }^n C_2 \cdots{ }^n C_n} \end{aligned} $

$ \begin{aligned} & =\left(\frac{{ }^{n+1} C_1}{{ }^n C_0}\right)\left(\frac{{ }^{n+1} C_2}{{ }^n C_1}\right)\left(\frac{{ }^{n+1} C_3}{{ }^n C_2}\right) \cdots\left(\frac{{ }^{n+1} C_{n+1}}{{ }^n C_n}\right) \\ & =\left(\frac{n+1}{1}\right)\left(\frac{n+1}{2}\right)\left(\frac{n+1}{3}\right) \cdots\left(\frac{n+1}{n+1}\right) \\ & =\frac{(n+1)^{n+1}}{(n+1)!} \end{aligned} $

Given, $\frac{x^4+1}{\left(x^2+1\right)(x-1)}$

$ \begin{aligned} & =\frac{x^4\left(1+\frac{1}{x^4}\right)}{x^3\left(1+\frac{1}{x^2}\right)\left(1-\frac{1}{x}\right)} \\ & =x\left(1+\frac{1}{x^4}\right)\left(1+\frac{1}{x^2}\right)^{-1}\left(1-\frac{1}{x}\right)^{-1} \\ & =x\left(1+\frac{1}{x^4}\right)\left(1-\frac{1}{x^2}\right)\left(1+\frac{1}{x}+\frac{1}{x^2}+\ldots\right) \end{aligned} $

Clearly, there is no terms containing $x^3$ coefficient of $x^3=0$

The given sum can be rewritten as

$ \sum_{k=0}^n \sum_{r=0}^k C_r $

Now, changing order of summarion

$ \sum_{r=0}^n \sum_{k=r}^n C_r \quad\left[\because \sum_{r=0}^n C_r=2^n\right] $

$ \begin{aligned} & \sum_{r=0}^n C_r \sum_{k=r}^n 1=\sum_{r=0}^n C_r(n-r+1) \\ & \Rightarrow(n+1) \sum_{r=0}^n C_r-\sum_{r=0}^n r \cdot C_r \\ & \quad=(n+1) 2^n-\sum_{r=0}^n r \cdot{ }^{n-1} C_{r-1} \times \frac{n}{r} \\ & \quad=(n+1) 2^n-n \sum_{r=0}^n n-1 C_r \\ & \quad=(n+1) 2^n-n \cdot 2^{n-1} \\ & \quad=n \cdot 2^n+2^n-n \cdot 2^{n-1} \\ & \quad=2^{n-1}(n+2) \end{aligned} $

$ \begin{aligned} &\text { We have, }\left(\frac{3 x-5}{4 x^2+3}\right)^{-\frac{4}{5}}=\left(\frac{4 x^2+3}{3 x-5}\right)^{\frac{4}{5}}\\ &\begin{aligned} & =\left\{\frac{4 x^2\left(1+\frac{3}{4 x^2}\right)}{3 x\left(1-\frac{5}{3 x}\right)}\right\}^{\frac{4}{5}}=\left(\frac{4 x}{3}\right)^{\frac{4}{5}} \frac{\left(1+\frac{3}{4 x^2}\right)^{4 / 5}}{\left(1-\frac{5}{3 x}\right)^{4 / 5}} \\ & \Rightarrow\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{3}{4 x^2}\right)^{\frac{4}{5}} \cdot\left(1-\frac{5}{3 x}\right)^{-\frac{4}{5}} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{4}{5} \cdot \frac{3}{4 x^2}+\ldots\right) \\ & \left(1+\frac{5}{3 x} \cdot \frac{4}{5}+\frac{4}{5}\left(+\frac{9}{5}\right)\left(\frac{5}{3 x}\right)\right)^2 \frac{1}{2!}+\ldots \ldots \\ & =\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{3}{5 x^2}+\ldots\right)\left(1+\frac{4}{3 x}+\frac{2}{x^2}\right) \\ & =\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{4}{3 x}+\left(\frac{3}{5}+2\right)^{\frac{1}{x^2}}\right) \\ & =\left(\frac{4 x}{3}\right)^{\frac{4}{5}}\left(1+\frac{4}{3 x}+\frac{13}{5 x^2}\right) \end{aligned} $

$ \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

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

$(x-2 y+3 z)^6$

By multinomial theorem,

$ \begin{aligned} & \Sigma \frac{6!}{p!q!r!}(x)^p(-2 y)^q(3 z)^r \\\\ \Rightarrow \quad & \Sigma \frac{6!}{p!q!r!}(-2)^q(3)^4 x^p y^f z^r \end{aligned} $

Then, coefficient of $x y^2 z^3$ is

$ \begin{aligned} & p=1, q=2 r=3 \\\\ \Rightarrow \quad & \frac{6!}{1!2!3!}(-2)^2(3)^3 \\\\ & =\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 2} \times 4 \times 27 \\\\ & =240 \times 27=6480 \end{aligned} $

$\left(125 x^2-\frac{27}{x}\right)^{-\frac{2}{3}}$

$\Rightarrow \quad 5^{-2}\left(x^2-\frac{27}{125 x}\right)^{-\frac{2}{3}}$

$\Rightarrow\left|\frac{27}{125 x^3}\right| \leq 1 \Rightarrow-1 \leq \frac{27}{125 x^3} \leq 1$

$\Rightarrow \quad-1 \leq\left(\frac{3}{5 x}\right)^3 \leq 1 \Rightarrow-1 \leq \frac{3}{5 x}<1$

$\Rightarrow \quad-\frac{5}{3}<\frac{1}{x}<\frac{5}{3} \Rightarrow \frac{1}{x}<\frac{5}{3}$

$ \begin{array}{l}\Rightarrow \quad \frac{3}{5} < x < \infty \\\\ \text { and } \quad \frac{1}{x} > -\frac{5}{3} \\\\ \Rightarrow \quad-\infty < x < -\frac{3}{5} \\\\ \therefore \quad x \in\left(-\infty,-\frac{3}{5}\right) \cup\left(\frac{3}{5}, \infty\right)\end{array} $

Given the expression $3^{2n+2} - 8n - 9$, we want to determine the maximum power of $2$ that divides this expression for all $n \in \mathbb{N}$.

First, we can rewrite $3^{2n+2}$ as $9^{n+1}$:

$ 9^{n+1} - 8n - 9 $

Considering $9 = 1 + 8$, we can apply the binomial expansion to $ (1 + 8)^{n+1} $:

$ (1 + 8)^{n+1} = {^{n+1}C_0} + {^{n+1}C_1} \cdot 8 + {^{n+1}C_2} \cdot 8^2 + {^{n+1}C_3} \cdot 8^3 + \ldots $

Simplifying, we get:

$ 1 + 8(n+1) + 8^2 {^{n+1}C_2} + {^{n+1}C_3} \cdot 8^3 + \ldots $

Subtracting $8n + 9$ from this sequence, we have:

$ 1 + 8n + 8 + 8^2 {^{n+1}C_2} + \ldots - 8n - 9 $

Which simplifies to:

$ 8^2 ({^{n+1}C_2} + 8 {^{n+1}C_3} + \ldots) $

The expression $8^2$ can be further evaluated as $2^6$. Thus, the binomial expansion shows that after simplification, the given expression is a multiple of $2^6$.

Therefore, the maximum value of $p$ such that the expression is divisible by $2^p$ is:

$ \boxed{6} $

Given the polynomial expression $(1 + x + x^2 + x^3)^{100}$, we are interested in the coefficient of $x^r$, denoted by $a_r$. To find the sum of the products $r \cdot a_r$, we perform the following steps:

First, evaluate the polynomial at $x = 1$:

$ (1 + 1 + 1^2 + 1^3)^{100} = (4)^{100} $

This shows that the sum of all coefficients, $\sum_{r=0}^{300} a_r$, equals $4^{100}$.

Next, differentiate $(1 + x + x^2 + x^3)^{100}$ with respect to $x$:

$ \frac{d}{dx}\left((1 + x + x^2 + x^3)^{100}\right) = 100(1 + x + x^2 + x^3)^{99}(1 + 2x + 3x^2) $

Substitute $x = 1$ into the derivative:

$ \begin{aligned} \sum_{r=0}^{300} r \cdot a_r &= 100(4)^{99}(1 + 2 \cdot 1 + 3 \cdot 1^2) \\ &= 100(4^{99}) \cdot 6 \end{aligned} $

Simplify the expression to find:

$ \sum_{r=0}^{300} r \cdot a_r = 600(4^{99}) = 150 \times 4^{100} $

Since $4^{100} = \sum_{r=0}^{300} a_r = S$, we have:

$ \sum_{r=0}^{300} r \cdot a_r = 150S $

Thus, $\sum_{r=0}^{300} r \cdot a_r$ is equal to $150S$.

Given the binomial distribution $ X \sim B(6, p) $ and the condition $ \frac{P(X=4)}{P(X=2)} = \frac{1}{9} $, let's solve for $ p $.

Start with the equation:

$ 9 P(X=4) = P(X=2) $

For a binomial distribution:

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

Using this formula, we set up the probabilities:

$ P(X=4) = \binom{6}{4} p^4 (1-p)^2 $

$ P(X=2) = \binom{6}{2} p^2 (1-p)^4 $

We substitute these into our equation:

$ 9 \times \binom{6}{4} p^4 (1-p)^2 = \binom{6}{2} p^2 (1-p)^4 $

Calculate the binomial coefficients:

$ \binom{6}{4} = 15 $

$ \binom{6}{2} = 15 $

Substitute these values:

$ 9 \times 15 p^4 (1-p)^2 = 15 p^2 (1-p)^4 $

Cancel out the common term $ 15 $:

$ 9 p^4 (1-p)^2 = p^2 (1-p)^4 $

Divide both sides by $ p^2 (1-p)^2 $ (assuming $ p \neq 0 $ and $ p \neq 1 $):

$ 9 p^2 = (1-p)^2 $

Take the square root of both sides:

$ 3 p = 1-p $

Solve for $ p $:

$ 3p + p = 1 $

$ 4p = 1 $

$ p = \frac{1}{4} $

Therefore, the value of $ p $ is $\frac{1}{4}$.

We know that $ T_{r+1}={ }^{n} C_{r} x^{n-r} a^{r} $

$ \begin{aligned} T_{6+1} & ={ }^{8} C_{6}\left(\frac{5}{p^{3}}\right)^{8-6}\left(-\frac{3 q}{7}\right)^{6}=700 \\\\ & =28 \times \frac{25}{p^{6}} \times \frac{729 q^{6}}{7^{6}}=700 \\\\ \Rightarrow(7 p)^{6} & =\frac{28 \times 25 \times 729}{700} q^{6} \\\\ \Rightarrow(7 p)^{6} & =(3 q)^{6} \Rightarrow 7 p= \pm 3 q \Rightarrow 49 p^{2}=9 q^{2} \end{aligned} $

Given,

$ \left(5 x+\frac{7}{x}\right)^{-3 / 2}=(5 x)^{-\frac{3}{2}}\left(1+\frac{7}{5 x^{2}}\right)^{-3 / 2} $

We know that

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

$ \begin{array}{l} \therefore T_{4}=\frac{1}{5 x \sqrt{5 x}}\left[\frac{-3}{2} \times \frac{5}{2} \times \frac{7}{2} \times \frac{1}{6} \times\left(\frac{7}{5 x^{2}}\right)^{3}\right] \\\\ \begin{aligned} \left(x^{7} \sqrt{5 x}\right) T_{4} & =-\frac{1}{5} \times \frac{3}{2} \times \frac{5}{2} \times \frac{7}{2} \times \frac{1}{6} \times \frac{7^{3}}{5^{3}} \\\\ & =-\frac{7^{4}}{2^{4} 5^{3}} \end{aligned} \end{array} $