Binomial Theorem

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \frac{\left(1+\frac{2 x}{3}\right)^{-4}(4+5 x)^{1 / 2}}{(9+x)^{3 / 2}} \\ & =\left(1+\frac{2 x}{3}\right)^{-4} \times 2\left(1+\frac{5}{4} x\right)^{1 / 2} \times(9+x)^{-3 / 2} \\ & =\left(1+\frac{2 x}{3}\right)^{-4} \times 2 \times\left(1+\frac{5}{4} x\right)^{1 / 2} \times \frac{1}{27}\left(1+\frac{x}{9}\right)^{-3 / 2} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{2}{27}\left(1-\frac{8 x}{3}\right)\left(1+\frac{5}{8} x\right)\left(1-\frac{1}{6} x\right) \\ & =\frac{2}{27}\left(1-\frac{8}{3} x+\frac{5}{8} x-\frac{1}{6} x\right) \\ & =\frac{2}{27}\left(1-\frac{53}{24} x\right) \end{aligned}\\ &\text { Put, } x=\frac{6}{371}\\ &\begin{aligned} & =\frac{2}{27}\left(1-\frac{53 \times 6}{24 \times 371}\right) \\ & =\frac{2}{27}\left(1-\frac{53}{1484}\right) \\ & =\frac{2}{27} \times \frac{1431}{1484}=\frac{1}{14} \end{aligned} \end{aligned} $

We have, $\sqrt{3+x}+\sqrt{5+x}$

Here, $3 < x < 5$

$ \begin{aligned} \therefore & x^{1 / 2}\left(1+\frac{3}{x}\right)^{1 / 2}+5^{1 / 2}\left(1+\frac{x}{5}\right)^{1 / 2} \\ & x^{1 / 2}\left[1+\frac{1}{2}\left(\frac{3}{x}\right)+\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{9}{x^2}\right) \ldots\right] \end{aligned} $

$ +5^{1 / 2}\left[\begin{array}{l} 1+\frac{1}{2}\left(\frac{x}{5}\right)+\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{x^2}{5^2}\right) \\ +\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\left(\frac{x^3}{5^3}\right) \end{array}\right] $

∴ Coefficient of $x^{-3 / 2}$ is,

$ \frac{1}{2}\left(\frac{1}{2}-1\right) 9=\frac{1}{2} \times-\frac{1}{2} \times 9=-\frac{9}{4} $

and coefficient of $x^3$ is,

$ \begin{aligned} 5^{1 / 2} & \times \frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right) \frac{1}{5^3} \\ & =5^{\frac{1}{2}-3} \times \frac{1}{2} \times-\frac{1}{2} \times-\frac{3}{2}=5^{\frac{-5}{2}} \times \frac{3}{8} \end{aligned} $

Sum of coefficient of $x^{-3 / 2}$ and $\quad x^3=\frac{-9}{4}+\frac{3}{8} 5^{-5 / 2}$

$ =\frac{-18+3(5)^{-5 / 2}}{8} $

Given, 9th and 10th terms are numerically greatest terms in the expansion of $(5 x-6 y)^n$

$ 9 \leq \frac{(n+1)\left(\frac{6 y}{5 x}\right)}{1+\left(\frac{6 y}{5 x}\right)} $

Here, $y=\frac{1}{2}$ and $x=\frac{2}{5}$

$ \begin{array}{ll} \because & \frac{y}{x}=\frac{5}{4} \Rightarrow \frac{6 y}{5 x}=\frac{3}{2} \\ & 9 \leq \frac{(n+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}} \\ & 9 \leq \frac{3(n+1)}{5} \\ & n+1 \geq 15 \\ & n \geq 14 \\\because & n=14 \end{array} $

$ \begin{aligned} & \text { Middle term of }(5 x-6 y)^{14} \\ & \quad=\left|{ }^{14} C_7(5 x)^7(-6 y)^7\right| \\ & \quad=\left|{ }^{14} C_7\left(5 \times \frac{2}{5}\right)^7\left(-6 \times \frac{1}{2}\right)^7\right| \\ & \quad={ }^{14} C_7 \times 6^7 \end{aligned} $

$ \begin{aligned} &\text {We have, }\\ &1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^2-\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^3+\ldots= \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} (1-x)^n=1-n x+ & \frac{n(n-1)}{2!} x^2 \\ & -\frac{n(n-1)(n-2)}{3!} x^3 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let }(1-x)^n=1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^2 \\ & \qquad \begin{aligned} \quad & -\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^3 \ldots \\ \therefore \quad n x=\frac{3}{16} & \text { and } \frac{n(n-1)}{2!} x^2=\frac{1 \cdot 4}{2!}\left(\frac{3}{16}\right)^2 \\ \quad n x & =\frac{3}{16} \text { and } \\ n(n-1) x^2 & =4\left(\frac{3}{16}\right)^2 \end{aligned} \end{aligned} $

$ \begin{gathered} n^2 \cdot x^2=\frac{3^2}{16} \text { and } n(n-1) x^2=4\left(\frac{3}{16}\right)^2 \\ n(n-1) x^2=4 n^2 x^2 \\ n^2-n=4 n^2 \\ n-1=4 n \Rightarrow n=-\frac{1}{3} \\ x=\frac{3}{16} \times-3=\frac{-9}{16} \\ \because \quad(1-x)^n=\left(1+\frac{9}{16}\right)^{-1 / 3} \\ =\left(\frac{25}{16}\right)^{-1 / 3}=\left(\frac{4}{5}\right)^{2 / 3} \end{gathered} $

We have,

$(1-x)^3=1-3 x+3 x^2$ [neglecting

higher term]

$ (2+x)^6={ }^6 C_0 2^6 x^0+{ }^6 C_1 2^5 x^1+{ }^6 C_2 2^4 x^2 $

[neglecting higher term]

$ =64+192 x+240 x^2 $

Now, $(1-x)^3(2+x)^6=\left(1-3 x+3 x^2\right)$

$ \begin{array}{r} \left(64+192 x+240 x^2\right) \\ =64+192 x+240 x^2-192 x \\ -576 x^2+192 x^2 \end{array} $

(neglecting higher term)

$ \begin{aligned} & =-144 x^2+64 \\ \therefore \quad & a=64, b=0, c=-144 \\ \therefore & a+b+c=64+0-144=-80 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, } \\ & \left(1+\frac{1}{x}\right)^{1 / 2} \\ & =1+\frac{1}{2}\left(\frac{1}{x}\right)+\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)}{2!}\left(\frac{1}{x}\right)^2 \\ & +\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}\left(\frac{1}{x}\right)^3 \\ & +\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\left(\frac{1}{2}-3\right)}{4!}\left(\frac{1}{x}\right)^4+\ldots \\ & =1+\frac{1}{2 x}-\frac{1}{2!}\left(\frac{1}{2 x}\right)^2+\frac{1 \cdot 3}{3!}\left(\frac{1}{2 x}\right)^3 \\ & -\frac{1 \cdot 3 \cdot 5}{4!}\left(\frac{1}{2 x}\right)^4+\ldots . \infty \end{aligned} \end{aligned} $