Binomial Theorem

$ \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} $

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

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} $

Let 3 consecutive terms in the expansion of $(1+x)^{23}$ be

$ T_{r+1}, T_{r+2} \text { and } T_{r+3} $

Coefficient of $T_{r+1}={ }^{23} C_r$

Coefficient of $T_{r+2}={ }^{23} C_{r+1}$

Coefficient of $T_{r+3}={ }^{23} C_{r+2}$

According to the question,

$ \begin{array}{ll} { }^{23} C_{r+1}-{ }^{23} C_r & ={ }^{23} C_{r+2}-{ }^{23} C_{r+1} \\\\ \Rightarrow \quad & \frac{23!}{(r+1)!(22-r)!}-\frac{23!}{r!(23-r)!} \\\\ & =\frac{23!}{(r+2)!(21-r)!}-\frac{23!}{(r+1)!(22-r)!} \\\\ \Rightarrow \quad & \frac{23-r}{(r+1)!(23-r)!}-\frac{(r+1)}{(r+1)!(23-r)!} \\\\ & =\frac{22-r}{(r+2)!(22-r)!}-\frac{r+2}{(r+2)!(22-r)!} \\\\ \Rightarrow \quad & (r+2)[(23-r)-(r+1)] \\\\ \Rightarrow \quad & =(23-r)[(22-r)-(r+2)] \\\\ \Rightarrow \quad & (r+2)(22-2 r)=(23-r)(20-2 r) \\\\ \Rightarrow \quad & r^2-84 r+416=0 \\\\ \Rightarrow \quad & (r-8)(r-13)=0 \\\\ \Rightarrow \quad r=8,13 \end{array} $

Hence, 3 consecutive terms are either $T_9, T_{10}$ and $T_{11}$ or $T_{14}, T_{15}$ and $T_{16}$.

Let the numerically greatest term in the expansion of $(3 x-16 y)^{15}$ when $x=\frac{2}{3}$ and $y=\frac{3}{2}$ is $T_{r+1}$

$ \begin{aligned} T_{r+1} & ={ }^{15} C_r\left(3\left(\frac{2}{3}\right)\right)^{15-r}\left(-16\left(\frac{3}{2}\right)\right)^r \\\\ & ={ }^{15} C_r(2)^{15-r}(-24)^r \end{aligned} $

Clearly, the value the expression will be maximum only when the value $r$ is largest possible even integer.

$ \text { So, } r=14 \quad[\because 0 \leq r \leq 15 \text { and } r \in Z] $

$\therefore 15$ th term will be the numerically greatest term.

To find largest positive integer $K$, that divides $81^n+20 n-1$ for $n \in N$, we start evaluating the expression for small values of $n$

For $n=1: 81^1+20.1-1=100$

For $n=2: 81^2+20.2-1=6600$

Now,

The greatest common divisor of 100 and 6600 is 100 . Hence, $K=100$ is the largest integer that divides

$81^n+20 n-1$ for all $n \in N$

and the divisior of 100 are

$ \begin{aligned} & 1,2,4,5,10,20,25,50,100 \\\\ & \begin{aligned} \Rightarrow S=1+2+4+5+10+20+25+50+100=217 \end{aligned} \end{aligned} $

Hence, $S-K=117$

$ \begin{aligned} &\\ &\begin{aligned} & \begin{aligned} (1 & \left.-3 x+2 x^3\right)\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9 \\ T_{r+1} & ={ }^9 C_r\left(\frac{3 x^2}{2}\right)^{9-r}\left(\frac{-1}{3 x}\right)^r \\ & ={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(\frac{-1}{3}\right)^r x^{18-3 r} \\ & 1\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9-3 x\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9 \\ & +2 x^3\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9 \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &r=6 \quad r=\frac{19}{3} \text { not possible } \quad r=7\\ &\Rightarrow r=6\\ &\text { We get, } 1\left(T_{6+1}\right)+2 x^3\left(T_{7+1}\right) \end{aligned} $

$ \begin{aligned} & ={ }^9 C_6\left(\frac{3 x^2}{2}\right)^3\left(\frac{-1}{3 x}\right)^6+2 x^3 \quad\left({ }^9 C_7\left(\frac{3 x^2}{2}\right)^2\left(\frac{-1}{3 x}\right)^7\right) \end{aligned} $

$ =\frac{9 \times 8 \times 7}{3!} \times \frac{(3)^3}{(2)^3}\left(\frac{1}{3}\right)^6+2 \times \frac{9 \times 8}{2}\times \frac{3^2}{2^2}\left(\frac{-1}{3}\right)^7 $

$ =\frac{7}{18}-\frac{2}{27}=\frac{17}{54} $

$ \begin{aligned} & { }^{20} C_0+{ }^{21} C_1+{ }^{22} C_2 \ldots{ }^{40} C_{20}=\frac{p}{q}{ }^{40} C_{20} \\ & \Rightarrow{ }^{41} C_{21}=\frac{p}{q}{ }^{40} C_{20} \\ & \Rightarrow \quad \frac{41}{21}{ }^{40} C_{20}=\frac{p}{q}{ }^{40} C_{20} \\ & \Rightarrow \quad p^2-q^2=(41)^2-(21)^2 \\ & \quad=(41+21)(41-21)=1240 \end{aligned} $

$ \begin{aligned} & \text { } x=\frac{2 \cdot 5}{2!3}+\frac{2 \cdot 5 \cdot 7}{3!3^2}+\frac{2 \cdot 5 \cdot 7 \cdot 9}{4!3^3}+\ldots \\ & \begin{aligned} \Rightarrow \frac{x}{2}+1+\frac{1}{1!}= & 1+\frac{3}{1!}\left(\frac{1}{3}\right) \\ & +\frac{3 \cdot 5}{2!}\left(\frac{1}{3}\right)^2+\frac{3 \cdot 5 \cdot 7}{3!}\left(\frac{1}{3}\right)^3 \ldots \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{x}{2}+2=\left(1-\frac{2}{3}\right)^{-\frac{3}{2}} \\ & \Rightarrow \frac{x}{2}+2=3 \sqrt{3} \Rightarrow \frac{x^2}{4}+4+2 x=27 \\ & \Rightarrow x^2+8 x+8=27 \times 4-8=100 \end{aligned} $

$ \text { } \begin{aligned} & \frac{x}{(x-1)^2(x-2)} \\ & =x(x-1)^{-2}(x-2)^{-1} \\ & =-\frac{x}{2}(1-x)^2\left(1-\frac{x}{2}\right)^{-1} \end{aligned} $

Expansion is possible only, when $|x|<1$ which is not mentions in question.

The given expression $(7-5 x)^{-2 / 3}$ can be written as, $7^{-2 / 3}\left(1-\frac{5}{7} x\right)^{-2 / 3}$ Now, the expansion of $\left(1-\frac{5}{7} x\right)^{-2 / 3}$ is valid, if

$ \begin{aligned} & \left|\frac{5}{7} x\right| < 1 \Rightarrow \frac{5}{7}|x| < 1 \\ & \Rightarrow|x| < \frac{7}{5} \Rightarrow-\frac{7}{5} < x < \frac{7}{5} \end{aligned} $

Now, it can be concluded that $c=7 / 5$

$ \therefore \quad 5 c+7=5\left(\frac{7}{5}\right)+7=7+7=14 $

Given, $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n \therefore f(2023)$ be the coefficient of $x^{2023}$ in the expansion of $(1+x)(1-x)^{2023}$

$ \begin{aligned} & \text { Now, }(1+x)(1-x)^{2023} \\ & \begin{array}{l} =(1+x)(1-2023 x+\ldots+2023 x^{2022}-x^{2023} \\ \begin{aligned} f(2023) & \left.=1 \times(-1)+1 \times 2023 x^{2022}-x^{2023}\right) \\ & =-1+2023=2022 \end{aligned} \end{array} \end{aligned} $

$ \begin{aligned} & \text { Given, } y=\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \\ & \begin{array}{r} \Rightarrow 1+y=1+\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}+\ldots \\ \Rightarrow 1+y=1+\frac{3}{1}\left(\frac{1}{4}\right)^1 \\ +\frac{3 \cdot 5}{2!}\left(\frac{1}{4}\right)^2+\frac{3 \cdot 5 \cdot 7}{3!}\left(\frac{1}{4}\right)^3+\ldots \end{array} \\ & \Rightarrow 1+y=\left(1-\frac{1}{2}\right)^{-3 / 2} \\ & {\left[\begin{array}{r} \because(1-x)^{-p / q}=1+\frac{p}{1!}\left(\frac{x}{q}\right)^1+\frac{p(p+q)}{2!}\left(\frac{x}{q}\right)^2 \\ +\frac{p(p+q)(p+2 q)}{3!}+\ldots \end{array}\right]} \end{aligned} $

$ \begin{aligned} & \Rightarrow 1+y=\left(\frac{1}{2}\right)^{-3 / 2} \\ & \Rightarrow 1+y=2^{3 / 2} \Rightarrow(1+y)^2=\left(2^{3 / 2}\right)^2 \\ & \Rightarrow y^2+2 y+1=8 \Rightarrow y^2+2 y-7=0 \end{aligned} $

Given, binomial expansion

$ \begin{aligned} (2 x-3 y)^5 & =(2 x)^5\left(1-\frac{3 y}{2 x}\right)^5 \\ & =\left(2 \times \frac{3}{2}\right)^5\left(1-\frac{3 y}{2 \times 3 / 2}\right)^5 \\ & =3^5(1-y)^5 \end{aligned} $

Now, $\frac{T r+1}{T r}=\frac{n-r+1}{r}|-y|$

$ \begin{aligned} & \Rightarrow \frac{5-r+1}{r}\left|-\frac{2}{3}\right| \Rightarrow \frac{6-r}{r}\left(\frac{2}{3}\right) \\ & \Rightarrow \frac{12-2 r}{3 r} \\ & \therefore \frac{T r+1}{T r} \geq 1 \Rightarrow \frac{12-2 r}{3 r} \geq 1 \\ & \Rightarrow 12-2 r \geq 3 r \Rightarrow 12 \geq 5 r \\ & \Rightarrow r \leq \frac{12}{5} \Rightarrow r \leq 2 \frac{2}{5} \end{aligned} $

Greatest term $=T_3$

$ \begin{aligned} & =3^5 \times{ }^5 C_2(-1)^2(y)^2 \\ & =3^5 \times \frac{5 \times 4}{2 \times 1} \times\left(\frac{2}{3}\right)^2=3^3 \times 10 \times 4 \\ & =27 \times 40=1080 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, } \frac{2 x^3+3 x^2+3 x+5}{\left(x^2+1\right)\left(x^2+2\right)} \\ & =\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1}\left(2+x^2\right)^{-1} \\ & =\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1} \\ & \quad\left[2\left(1+\frac{x^2}{2}\right)\right]^{-1} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{1}{2}\left(2 x^3+3 x^2+3 x+5\right)\left(1+x^2\right)^{-1} \\ & =\frac{1}{2}\left(1+\frac{x^2}{2}\right)^{-1} \\ & \left(1-x^2+x^4-x^6+\ldots\right) \\ & \left(1-\frac{x^2}{2}+\frac{x^4}{4}-\frac{x^6}{6}+\ldots\right) \end{aligned}\\ &\text { Now, coefficient of } x^5\\ &\begin{aligned} & =\frac{1}{2}\left[\begin{array}{l} 2 \times 1 \times\left(-\frac{1}{2}\right)+2 \times(-1) \times 1+3 \\ \times 1 \times \frac{1}{4}+3 \times 1 \times 1+3 \times(-1) \times\left(-\frac{1}{2}\right) \end{array}\right] \\ & =\frac{1}{2}\left[-1-2+\frac{3}{4}+3+\frac{3}{2}\right]=\frac{1}{2}\left(\frac{9}{4}\right)=\frac{9}{8} \end{aligned} \end{aligned} $

Number of terms in expansion of

$ \begin{aligned} (x-2 y+3 z)^5 & ={ }^{5+3-1} C_{3-1} \\ P & ={ }^7 C_2 \Rightarrow P=21 \end{aligned} $

Coefficient of $x^p y^q z^r$ in $(a x+b y+c z)^n$ is

$ \frac{n!}{p!q!r!}\left(a^p \cdot b^q \cdot c^r\right) $

$q=$ Coefficient of $x^2 y z^2$ in $(x-2 y+3 z)^5$

$ =\frac{5!}{2!\times 1!\times 2!}\left\{1^2 \cdot(-2)^1(3)^2\right\}=-540 $

Now, $\frac{q}{p}=-\frac{540}{21}=-\frac{180}{7}$

$ \begin{aligned} & \text { } \begin{aligned} S_{n+1}=5 \cdot C_0+8 \cdot & C_1+11 \cdot C_2 \\ & +\ldots+(n+1) \text { terms } \\ = & \sum_{r=0}^n(3 r+5) \cdot C_r \\ = & \sum_{r=0}^n\left(3 \cdot r \cdot C_r\right)+\sum_{r=0}^n 5 \cdot C_r \\ = & 3 \sum_{r=0}^n r C_r+5 \sum_{r=0}^n C_r \\ S_{n+1}= & 3\left(n 2^{n-1}\right)+5 \cdot 2^n \end{aligned} \end{aligned} $

On putting $n=10$

$ S_{11}=3\left(10 \cdot 2^9\right)+5 \cdot 2^{10}=20480 $

$\frac{1}{\sqrt{4-x}(2+x)^3}$

Using binomial expansion for negative index

$ \begin{aligned} & \frac{1}{16}\left(1-\frac{x}{4}\right)^{-1 / 2}\left(1+\frac{x}{2}\right)^{-3} \\ = & \frac{1}{16}\left(1+\frac{x}{8}+\frac{3}{128} x^2\right)\left(1-\frac{3 x}{2}+\frac{3}{2} x^2\right) \\ = & \frac{1}{16}\left(1-\frac{3 x}{2}+\frac{3}{2} x^2+\frac{x}{8}-\frac{3 x^2}{16}+\frac{3 x^2}{128}\right) \\ = & \frac{1}{16}\left(1-\frac{11 x}{8}+\frac{171}{128} x^2\right) \end{aligned} $

To determine the number of integral terms in the expansion of $(\sqrt{3}+\sqrt[8]{5})^{256}$, consider the general term in the expansion:

$ T_{r+1} = \binom{256}{r} (\sqrt{3})^{256-r} (\sqrt[8]{5})^r $

This expression simplifies to:

$ T_{r+1} = \binom{256}{r} (3)^{\frac{256-r}{2}} (5)^{\frac{r}{8}} $

For the term to be an integer, both $\frac{256-r}{2}$ and $\frac{r}{8}$ must be integers. Therefore:

$256 - r$ must be even, implying $r$ must be even.

$\frac{r}{8}$ also needs to be an integer, meaning $r$ must be a multiple of 8.

Thus, $r$ can be 0, 8, 16, 24, …, up to 256. This sequence forms an arithmetic progression (AP) where $r = 0, 8, 16, \ldots, 256$.

To find the total number of terms in this sequence:

The first term ($a$) is 0.

The common difference ($d$) is 8.

The last term ($l$) is 256.

Using the formula for the nth term of an AP, $l = a + (n-1) \cdot d$:

$ 256 = 0 + (n-1) \cdot 8 $

Solving for $n$:

$ 256 = 8(n-1) $

$ n-1 = \frac{256}{8} = 32 $

$ n = 32 + 1 = 33 $

Therefore, the number of integral terms in the expansion is 33.

We know that the expansion $(1+x)^n$ in power of $x$ is valid if $|x|<1$, where $n$ is a rational number.

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

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

Now, the expansion of $\left(1+x+x^2\right)^{-\frac{3}{2}}$ in power of $x$ is valid, if

$ \begin{aligned} & \left|\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\right|<1 \\ \Rightarrow & -1<\left(x+\frac{1}{2}\right)^2-\frac{1}{4}<1 \\ \Rightarrow & -\frac{3}{4}<\left(x+\frac{1}{2}\right)^2<\frac{5}{4} \\ \Rightarrow & 0 \leq\left(x+\frac{1}{2}\right)^2<\frac{5}{4} \\ & \quad\left[\because\left(x+\frac{1}{2}\right)^2 \geq 0 \forall x \in R\right] \\ \Rightarrow & \quad\left|x+\frac{1}{2}\right|<\frac{\sqrt{5}}{2} \end{aligned} $

$(1+x)^n=c_0+c_1 x+c_2 x^2+\ldots c_n x^n$.

By integrating on both sides, we get

$ \begin{aligned} &\begin{aligned} \frac{(1+x)^{n+1}}{n+1}=c_0 x+\frac{c_1 x^2}{2}+\frac{c_2 x^3}{3} & +\frac{c_n x^{n+1}}{n+1} \end{aligned}\\ &\begin{aligned} & \frac{2^{n+1}}{n+1}=c_0+\frac{c_1}{2}+\frac{c_2}{3}+\ldots \frac{c_n}{n+1} \\ & \therefore \quad c_0+\frac{c_1}{2}+\frac{c_2}{3}+\ldots \frac{c_n}{n+1}=\frac{2^{n+1}}{n+1} \end{aligned} \end{aligned} $

$ \begin{aligned} &\left(\sqrt{x}-\frac{k}{x^2}\right)^{10} \\ & T_{r+1}={ }^{10} C_r\left(x^{\frac{1}{2}}\right)^{10-r}\left(\frac{-k}{x^2}\right)^r \\ &={ }^{10} C_r \frac{x^{\frac{10-r}{2}}}{x^{2 r}}(-k)^r \\ &={ }^{10} C_r x^{\frac{10-r}{2}-2 r}(-k)^r \\ &={ }^{10} C_r x^{\frac{10-5 r}{2}}(-k)^r \end{aligned} $

To be independent of $x, r=2$

$ \begin{array}{ll} \therefore & 405={ }^{10} C_2(-k)^2 \\ \Rightarrow & 405=\frac{10 \times 9}{2}(-k)^2 \\ \Rightarrow & 405=45 k^2 \\ \Rightarrow & 9=k^2 \\ \therefore & k= \pm 3 \end{array} $

To determine the number of rational terms in the binomial expansion $(\sqrt[4]{5} + \sqrt[5]{4})^{100}$, consider the term:

$ \sum\limits_{r=0}^{100} \binom{100}{r} \cdot 5^{\frac{100-r}{4}} \cdot 4^{\frac{r}{5}} $

For a term to be rational, both $\frac{r}{5}$ and $\frac{100-r}{4}$ must be integers.

Condition 1: $\frac{r}{5}$ must be an integer. This implies that $r$ must be a multiple of 5. The possible values of $r$ are:

$0, 5, 10, 15, 20, \ldots, 100$.

Condition 2: $\frac{100-r}{4}$ must also be an integer. For this condition, check which values from the list above also satisfy this.

For $r = 0$, $\frac{100-0}{4} = 25$ (integer)

For $r = 20$, $\frac{100-20}{4} = 20$ (integer)

For $r = 40$, $\frac{100-40}{4} = 15$ (integer)

For $r = 60$, $\frac{100-60}{4} = 10$ (integer)

For $r = 80$, $\frac{100-80}{4} = 5$ (integer)

For $r = 100$, $\frac{100-100}{4} = 0$ (integer)

The values of $r$ that make both $\frac{r}{5}$ and $\frac{100-r}{4}$ integers are $0, 20, 40, 60, 80, 100$.

Therefore, there are 6 rational terms in the expansion.

Here, $(1+x)^{101}\left(1-x+x^2\right)^{100}$

$ \begin{aligned} &=(1+x)(1+x)^{100}\left(1-x+x^2\right)^{100} \\ &=(1+x)\left(1+x^3\right)^{100} \\ &=\left(1+x^3\right)^{100}+x\left(1+x^3\right)^{100} \\ &=\Sigma^{100} C_r x^{3 r}+\Sigma^{100} C_r x^{3 r+1} \\ & 3 r=50 ; 3 r+1=50 \\ & r=\frac{50}{3} ; \quad r=\frac{49}{3} \\ & \Rightarrow \text { Coefficient of } x^{50}=0 \end{aligned} $

For $(a+b)^n$

$ T_{r+1}={ }^n C_r a^r b^{(n-r)} $

For numerically greatest term,

$ \begin{aligned} \left|\frac{T_{r+1}}{T_r}\right| & =\left|\frac{{ }^n C_r a^r b^{n-r}}{{ }^n C_{r-1} a^{r-1} b^{n-r+1}}\right| \\ & =\left|\frac{{ }^n C_r}{{ }^n C_{r-1}}\right|\left|\frac{a}{b}\right|=\left|\frac{n-r+1}{r}\right|\left|\frac{a}{b}\right| \end{aligned} $

Here given, $a=2 x=\frac{2}{3}, b=-3 y=\frac{-3}{2}$,

$ \begin{aligned} & n=11 \\ & \therefore \quad\left|\frac{T_{r+1}}{T_r}\right|=\left|\frac{11-r+1}{r}\right|\left|\frac{2 / 3}{-3 / 2}\right| \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad=\left|\frac{12-r}{r}\right|\left|\frac{4}{9}\right|>1 \end{aligned} $

$ \begin{array}{rlrl} \Rightarrow & & 48-4 r & >9 r \\ \Rightarrow & & 13 r & <48 \Rightarrow r<\frac{48}{13} \\ \Rightarrow & & r & =3 \\&& T_{r+1} & ={ }^n C_r a^r b^{n-r} \\ & & & ={ }^{11} C_3\left(\frac{2}{3}\right)^3\left(\frac{-3}{2}\right)^8 \\ & & & ={ }^{11} C_3\left(\frac{3}{2}\right)^5 \end{array} $

Given,

$ \begin{aligned} & \left(\frac{1}{8}-\frac{7}{8 \times 12}+\frac{7 \times 10}{8 \times 12 \times 16}-\ldots\right) \\ & =\left(\frac{3}{4}+\frac{1}{8}-\frac{7}{8 \times 12}+\frac{7 \times 10}{8 \times 12 \times 16}-\ldots\right)-\frac{3}{4} \\ & =\left(1-\frac{1}{4}+\frac{1}{8}-\frac{7}{8 \times 12}+\frac{7 \times 10}{8 \times 12 \times 16}-\ldots\right)-\frac{3}{4} \end{aligned} $

$ =\left(\begin{array}{r} 1-\frac{1}{3} \times \frac{3}{4}+\frac{\frac{1}{3} \times \frac{4}{3} \times\left(\frac{3}{4}\right)^2}{2!} \\ -\frac{\frac{1}{3} \times \frac{4}{3} \times \frac{7}{3} \times\left(\frac{3}{4}\right)^3}{3!} \end{array}\right. $

$ \left.+\frac{\frac{1}{3} \times \frac{4}{3} \times \frac{7}{3} \times \frac{10}{3} \times\left(\frac{3}{4}\right)^4}{4!}-\ldots\right)-\frac{3}{4} $

$ =\left(\begin{array}{l} 1-\frac{1}{3} \times \frac{3}{4}+\frac{\frac{1}{3} \times\left(\frac{1}{3}+1\right) \times\left(\frac{3}{4}\right)^2}{2!} \\ -\frac{\frac{1}{3} \times\left(\frac{1}{3}+1\right)\left(\frac{1}{3}+2\right) \times\left(\frac{3}{4}\right)^3}{3!} \end{array}\right. $

$ \left.+\frac{\frac{1}{3}\left(\frac{1}{3}+1\right)\left(\frac{1}{3}+2\right)\left(\frac{1}{3}+3\right) \times\left(\frac{3}{4}\right)^4}{4!}-\ldots\right)-\frac{3}{4} $

$ \begin{aligned} & =\left(1+\frac{3}{4}\right)^{-1 / 3}-\frac{3}{4} \\ & =\left(\frac{7}{4}\right)^{-1 / 3}-\frac{3}{4}=\sqrt[3]{\frac{4}{7}}-\frac{3}{4} \end{aligned} $

Given, the expansion $(a+x)^n$ contains 15 terms $\Rightarrow n=14$

Middle term of expansion $(a+1)^n$ is

$ T_8={ }^{14} C_7 a^7 $

Ratio of the neighbouring term in the expansion $(a+x)^n=16$

$ \begin{array}{ll} \therefore & \frac{T_8}{T_6}=16 \Rightarrow \frac{{ }^{14} C_8 a^8}{{ }^{14} C_6 a^6}=16 \\ \Rightarrow & a^2=16 \,\,\,\,\,\,\,\left[\because{ }^{14} C_8={ }^{14} C_6\right]\\ \Rightarrow & a=4 \end{array} $

$ \begin{aligned} T_{r+1} & ={ }^n C_r\left(2 x^2\right)^{n-r}\left(\frac{-1}{3 x^3}\right)^r \\ T_{r+1} & ={ }^5 C_r(2)^{5-r}\left(\frac{-1}{3}\right)^r x^{10-2 r-3 r} \\ & ={ }^5 C_r(2)^{5-r}\left(\frac{-1}{3}\right)^r x^{10-5 r} \end{aligned} $

$ \begin{aligned} &\begin{aligned} 10-5 r & =5 \Rightarrow 5 r=5 \Rightarrow r=1 \\ T_2 & ={ }^5 C_1(2)^4\left(\frac{-1}{3}\right) x^5 \end{aligned}\\ &\begin{aligned} & \text { Coefficient of } x^5=5(2)^4\left(\frac{-1}{3}\right) \\ & \therefore \quad k=\frac{-16 \times 5}{3} \Rightarrow \frac{3 k}{2}=-40 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } T_4=T_{3+1}=-20 \\ & \Rightarrow{ }^6 C_3\left(\frac{x}{2}\right)^3\left(\frac{-2 y}{3}\right)^3=-20 \\ & \Rightarrow 20 \times \frac{x^3}{8} \times\left(\frac{-8 y^3}{27}\right)=-20 \end{aligned} $

$ \begin{array}{ll} \Rightarrow & x^3 y^3=27 \\ \therefore & x y=3 \end{array} $

General term of $\left(a x+\frac{b}{x^2}\right)^{11}$

$ T_{r+1}={ }^{11} C_r a^r b^{11-r} x^{3 r-22} $

For $x^{-7} \quad 3 r-22=-7$

$ \begin{aligned} \Rightarrow \quad r & =5 \\ L & =\text { Coefficient of } x^{-7}={ }^{11} C_5 a^5 b^6 \end{aligned} $

General term of $\left(b x^2+\frac{a}{x}\right)^{11}$

$ T_{r+1}={ }^{11} C_r a^{11-r} b^r x^{3 r-11} $

For $x^7$

$ \begin{aligned} & 3 r-11=7 \quad \Rightarrow \quad r=6 \\ & M=\text { Coefficient of } x^7={ }^{11} C_6 a^5 b^6 \end{aligned} $

$ \begin{aligned} &\begin{aligned} L+M & =\left({ }^{11} C_5+{ }^{11} C_6\right) a^5 b^6 \\ & ={ }^{12} C_6 a^5 b^6 \end{aligned}\\ &\text { Now, coefficient of } x^6 \text { in }\\ &\begin{aligned} & \left(a x^2+\frac{b}{x}\right)^{12}={ }^{12} C_0 a^6 b^6 \\ & \therefore \quad L+M=\frac{1}{a} \\ & {\left[\text { coefficient of } x^{-6} \text { in }\left(a x^2+\frac{b}{x}\right)^{12}\right]} \end{aligned} \end{aligned} $

Given $n=10$, then $\sum_{r=1}^{10}{ }^n C_r r(r-4)$

$ \begin{aligned} & =\sum_{r=1}^{10}{ }^n C_r r((r-1)-3) \\ & =\sum_{r=1}^{10}\left(r(r-1){ }^n C_r-3 r{ }^n C_r\right) \\ & =\sum_{r=1}^{10}\left(r(r-1) \frac{n(n-1)}{r(r-1)}{ }^{n-2} C_{r-2}-3 \cdot r \frac{n}{r} \cdot{ }^{n-1} C_{r-1}\right) \\ & =\sum_{r=1}^{10}\left(n(n-1){ }^{n-2} C_{r-2}-3 n \cdot{ }^{n-1} C_{r-1}\right) \\ & =n(n-1) \sum_{r=1}^{10}{ }^{n-2} C_{r-2}-3 n \sum_{r=1}^{10}{ }^{n-1} C_{r-1} \end{aligned} $

Given, $n=10$ and $\sum_{r=1}^{10}{ }^n C_r=2^n$

$ \begin{aligned} & \therefore 10(10-1) \cdot 2^{10-2}-3 \times 10 \cdot 2^{10-1} \\ & =10 \times 9 \times 2^8-30 \times 2^9 \\ & =(90-60) \cdot 2^8 \\ & =30 \times 256=7680 \end{aligned} $

Given sum of coefficient in expansion of $\left(1+3 x-2 x^2\right)^n$ is 128

∴ For sum of coefficient, putting $n=1$ in the expression $\left(1+3 x-2 x^2\right)^n$

$ \begin{array}{lc} \Rightarrow & (1+3 \cdot 1-2 \cdot 1)^n=128 \\ \Rightarrow & 2^n=2^7 \\ \Rightarrow & n=7 \end{array} $

Now $\sum_{r=1}^{2 n} \frac{r \cdot{ }^{2 n} C_r}{{ }^{2 n} C_{r-1}}=\sum_{r=1}^{2 n} \frac{r \cdot \frac{2 n!}{r!(2 n-r)!}}{\frac{2 n!}{(r-1)!(2 n-r+1)!}}$

$ \begin{aligned} & =\sum_{r=1}^{2 n} \frac{r \cdot(2 n-r+1)}{r}=\sum_{r=1}^{2 n}(2 n-r+1) \,\,\,\,\,\,\,[\because n=7]\\ & =(2 n+1)(2 n)-\frac{2 n(2 n+1)}{2} \\ & =(14+1)(14)-7(14+1) \\ & =15 \times 7=105 \end{aligned} $

$ \begin{aligned} & &\begin{aligned} & \text { } \operatorname{Let}\left(3 \sqrt{126}+\sin 61^{\circ}\right)=A+B \\ & \begin{aligned} & A=3 \sqrt{126}=3 \sqrt{125+1}=\left(5^3+1\right)^{1 / 3} \\ &=5\left(1+\frac{1}{5^3}\right)^{1 / 3} \\ &=5\left(1+\frac{1}{3} \cdot \frac{1}{5^3}-\frac{1}{9} \cdot \frac{1}{5^6}+\ldots\right) \\ &\left\{\because(1+x)^n=1+n(x)+\frac{n(n-1)}{2!} x^2+\ldots\right\} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & =5+\frac{1}{3} \cdot \frac{1}{5^2}-\frac{1}{9} \cdot \frac{1}{5^5}+\ldots \\ & =5+\frac{1}{3} \frac{2^2}{10^2}-\frac{1}{9} \frac{2^5}{10^5}+\ldots \\ & =5+\frac{0.04}{3}-\frac{0.00032}{9}+\ldots \\ & =5+0.0133-0.000035+\ldots \end{aligned} $

$ \begin{aligned} & A=5.0132 \Rightarrow B=\sin 61^{\circ}=\sin \left(60^{\circ}+1^{\circ}\right) \\ & B=\sin 60^{\circ} \cos 1^{\circ}+\cos 60^{\circ} \sin 1^{\circ} \\ & {[\because \sin (A+B)=\sin A \cos B+\cos A \sin B]} \\ & \quad=\frac{\sqrt{3}}{2} \cos 1^{\circ}+\frac{1}{2} \sin 1^{\circ} \\ & \quad \simeq \frac{\sqrt{3}}{2} \cos 0^{\circ}+\frac{1}{2}\left(\frac{\pi}{180}\right) \\ & {\left[\because \cos 1^{\circ}=\cos 0^{\circ} \text { and } \sin 1^{\circ}=1 \text { or } \frac{\pi}{180}\right]} \\ & \quad \simeq 0.866+\frac{1}{2} \times 0.0174 \\ & B \simeq 0.8747 \\ & \text { So, }\left(3 \sqrt{126}+\sin 61^{\circ}\right)=5.0132+0.8747 \\ & \quad=5.8879 \end{aligned} $

$\left(3 \sqrt{126}+\sin 61^{\circ}\right) \simeq 5.888$ to three places of decimals.