$(x+y)^n =x^n+{ } {^nC_1} ~x^{n-1} y+ {^nC_2} ~x^{n-2} ~y^2+…… y^n$

$x^n = T_0$ $\quad$ $^n C_1 = T_1$ $\quad$ $^n C_2 = T_2$ $\quad$ $y^n = T_n$

$\left(\frac{T_1}{T_0}\right)= \frac{{ }^n C_1}{{ }^n C_0} \frac{y}{x}$ $\quad$ $\frac{T_2}{T_1}=\frac{{ }^n C_2}{{ }^n C_1} \frac{y}{x}$$\quad$$\frac{T_{r+1}}{T_r}<1$

Find the numerically greatest term in $(2+3 x)^9, x=3 / 2$

Solution: ${ }^nC_0 ~2^9+{ }^nC_1 ~2^8 . 3 x+{ }^nC_2 . 2^7 .(3 x)^2+….. {^nC_n}(3 x)^9$

$\frac{T_{r+1}}{T_r}=\frac{{ }^9 C_{r+1}}{{ }^9 C_r} \cdot \frac{3 x}{2}$

$=\frac{9 !}{(r+1) !(8-r) !} \cdot \frac{r !(9-r) !}{9 !} \cdot \frac{3 x}{2}$

$=\left(\frac{9-r}{r+1}\right) . \frac{3x}{2}<1$

[$\because x = \frac{3}{2}$]

$=\left(\frac{9-r}{r+1}\right) . \frac{9}{4}<1$

What is $r$ s.t. $\frac{9-r}{r+1} \cdot \frac{9}{4}<1$

$81-9 r <4 r+4$

$77 <13 r$

$r > 77 / 13$

$r = 6, 7, 8, 9$

$\frac{3}{7} \cdot \frac{9}{4}=\frac{27}{28}$

$(3-5 x)^{15} \quad x=1 / 5$

$\frac{T_{r+1}}{T_r}=\frac{{ }^{15} C_{r+1}}{{ }^{15} C_r} : \frac{(-5 x)}{3}$

$=-\frac{15-r}{r+1} \cdot \frac{(-5 x)}{3}$

$\frac{15-r}{r+1}.~5\frac{x}{3} < 1$ [Put x = 1/5]

$\frac{15-r}{r+1}.~\frac{1}{3} < 1$

$15-r<3 r+3$

$ 12<4 r \Rightarrow 3

$r = 3 \rarr$ $\frac{r-15}{r+1}.\frac{1}{3}$

Find the term independent of $x$ in:

$(1+x+2 x^3) (\frac{3 x^2}{2}-\frac{1}{3 x})^9$

Solution $x^0 , \frac{1}{x} , \frac{1}{x^3}$

${ }^n C_r \cdot\left(\frac{3 x^2}{2}\right)^{9-r}\left(\frac{1}{3 x}\right)^r$

$x^{18-2 r-r}=x^{18-3 r}$

$x^0 \Rightarrow r=6$
$x^{-3} \Rightarrow r=7$

$\begin{array}{r}{ }^9 C_6 \cdot\left(\frac{3 x^2}{2}\right)^3\left(-\frac{1}{3 x}\right)^6 + 2 x^3{ }~^9 C_7 \cdot\left(\frac{3 x^2}{2}\right)^2\left(-\frac{1}{3 x}\right)^7\end{array}$

$\begin{aligned} & =\frac{9 \times 8 \times 7}{3 \times 2} \cdot \frac{27 \cdot x^6}{8} \cdot \frac{1}{27 \times 27 \cdot x^6} -2 \times \frac{9 \times 8}{2} x^3 \cdot \frac{9 x^4}{4} \cdot \frac{1}{27 \times 27 \times 3 \times x^7}\end{aligned}$

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

$=\frac{21-4}{54}=\frac{17} {54}$

$f(x) =1-x+x^2-x^3+….+x^{16}-x^{17}$

$=a_0+a_1(1+x)+a_2(1+x)^2+a_3(1+x)^3+\ldots+a_{17}(1+x)^{17}$

What is $a_2 ?$

Solution:
$1 =a_0+a_1+a_2+…..+a_{17}$

$-x =a_1 x+2 a_2 x+3 a_3 x+…..+17 a_{17} x$

$x^2 =a_2 x^2+{ }^3 c_2^{a_3{_ {x^2}}}+{ }^4 c_2^{a_4{_ { x^2}}}+{ }^5 c_2^{a_5{_ {x^2}}}+….{ }^{17} c_2 a_{17} x^2$

$\frac{3 !}{|2 ! 1 !} \frac{4}{2} \frac{4 !}{2 ! 2 !} \frac{5}{3} \frac{5 !}{2 ! 3 !} \rightarrow \frac{6}{4}$

$1=a_0+a_1+a_2+…..+a_{17}$

$-x=\left(a_1+2 a_2+3 a_3+…..+17 a_{17}\right) x$

$x^2=\left(a_2+3 a_3+3 \times \frac{4}{2} a_4+3 \times \frac{4}{2} \times \frac{5}{3} a_5+….+3 \times \frac{4}{2} \times \frac{5}{3} \times …. \frac{17}{15} a_{17}\right) x^2$

$\frac{d f}{d x} =-1+2 x-3 x^2+\cdots+16 x^{15}-17 x^{16}$

$=a_1+2 a_2(1+x)+3 a_3(1+x)^2+4 a_4(1+x)^3 +\ldots+17 a_{17}(1+x)^{16}$

$\ f’’ =\frac{d^2 f}{d x^2}= 2-6 x+4 \times 3 x^2-5 \times 4 x^3+\cdots+16 \times 15 x^{14}-17 \times 16 x^{15}$

$= 2 a_2+3 a_3 \times 2(1+x)+4 a_4 \times 3(1+x)^2+\cdots +17 a_{17} \times 16(1+x)^1$

$1=a_0+a_1+a_2+\ldots+a_{17}$

$-x=(a_1+2 a_2+3 a_3+….+ 17 a_{17}) x$

$x^2=\left(a_2+3 a_3+3 \times \frac{4}{2} a_4+3 \times \frac{4}{2} \times \frac{5}{3} a_5+\ldots+3 \times \frac{4}{2} \times \frac{5}{3} \times \ldots \frac{17}{15} a_{17}\right) x^2$

$2=2 a_2+3 \times 2 a_3+4 \times 3 a_4+5 \times 4 a_5+\cdots +17 \times 16 a_{17}$

$-1=a_1+2 a_2+3 a_3+4 a_4+5 a_5+\cdots +17 a_{17}$

$1=a_0+a_1+a_2+\cdots +a_{17}$

Thank you