# BINOMIAL THEOREM (REVISION)

$1\ Find\ the\ middle\ terms\ in\ the\ expansion\ of\ (x^3\ +\ \frac{2}{x^3})^{11}\ \hspace{15cm}$
$\color {black}{Solution:}\ Here\ n\ is\ odd\ number,\ so\ there\ are\ two\ middle\ terms\ \hspace{18cm}$
$(\frac{n+1}{2})^{th}\ term\ =\ (\frac{11+2}{2})^{th}\ term\ =\ (\frac{12}{2})^{th}\ term\ =\ 6^{th}\ term\ \hspace{16cm}$
$(\frac{n+3}{2})^{th}\ term\ =\ (\frac{11+3}{2})^{th}\ term\ =\ (\frac{14}{2})^{th}\ term\ =\ 7^{th}\ term\ \hspace{16cm}$
$To\ find\ T_6:\ \hspace{18cm}$
$T_{r + 1} = nC_rx^{n-r} a^r \hspace{19cm}$
$Here\ X\ =\ x^3,\ a\ =\ \frac{2}{x^3},\ n\ =\ 11,\ r\ =\ 5\ \hspace{14cm}$
$T_{5 + 1} = 11C_5\ (x^3)^{11-5}\ (\frac{2}{x^3})^5 \hspace{15cm}$
$T_6 = 11C_5\ (x^3)^6\ \frac{2^5}{x^{15}}\ \hspace{15cm}$
$= 11C_5\ \ 2^5\ x^{18}\ x^{-15}\ \hspace{15cm}$
$= 11C_5\ \ 2^5\ x^{18 – 15}\ \hspace{15cm}$
$T_6 = 11C_5\ 2^5\ x^{3}\ \hspace{15cm}$
$To\ find\ T_7:\ \hspace{18cm}$
$T_{r + 1} = nC_rx^{n-r} a^r \hspace{19cm}$
$Here\ X\ =\ x^3,\ a\ =\ \frac{2}{x^3},\ n\ =\ 11,\ r\ =\ 6\ \hspace{14cm}$
$T_{6 + 1} = 11C_6\ (x^3)^{11-6}\ (\frac{2}{x^3})^6 \hspace{15cm}$
$T_7 = 11C_6\ (x^3)^5\ \frac{2^6}{x^{18}}\ \hspace{15cm}$
$= 11C_6\ \ 2^6\ x^{15}\ x^{-18}\ \hspace{15cm}$
$= 11C_6\ \ 2^6\ x^{15 – 18}\ \hspace{15cm}$
$T_7 = 11C_6\ 2^6\ x^{-3}\ \hspace{15cm}$
$2\ Find\ the\ term\ independent\ of\ x\ in\ the\ expansion\ of\ (4\ x^3 + \frac{3}{x^2})^{20}\ \hspace{15cm}$
$\color {black}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ 4\ x^3,\ a\ =\ \frac{3}{x^2},\ n\ =\ 20,\ r\ =\ r\ \hspace{14cm}$
$T_{r + 1} = 20C_r\ (4\ x^3)^{20-r}\ (\frac{3}{x^2})^r \hspace{15cm}$
$= 20C_r\ 4^{20\ -\ r}\ x^{60\ -\ 3\ r}\ \frac{3^r}{x^{2r}} \hspace{15cm}$
$= 20C_r\ 4^{20\ -\ r}\ x^{60\ -\ 3\ r}\ 3^r\ \ x^{-2r}\ \hspace{15cm}$
$= 20C_r\ 3^r\ 4^{20\ -\ r}\ x^{60\ -\ 3\ r\ -\ 2\ r}\ \hspace{15cm}$
$T_{r + 1} = 20C_r\ 3^r\ 4^{20\ -\ r}\ x^{60\ -\ 5\ r}\ ———————— (1)\ \hspace{15cm}$
$To\ find\ the\ independent\ term\ of\ x,\ find\ the\ coefficient\ of\ x^0.\ \hspace{15cm}$
$\therefore\ x^{60\ -\ 5\ r}\ =\ x^0\ \hspace{15cm}$
$60\ -\ 5\ r\ =\ 0\ \hspace{10cm}$
$-5r\ = -\ 60\ \hspace{15cm}$
$r\ = 12\ \hspace{15cm}$
$Put\ r\ =\ 12\ in\ (1)\ \hspace{15cm}$
$\therefore\ independent\ term\ of\ x =\ 20C_{12}\ 3^{12}\ 4^{20\ -\ {12}}\ \hspace{15cm}$
$=\ 20C_{12}\ 3^{12}\ 4^8\ \hspace{15cm}$