# BINOMIAL THEOREM (Text)

$\color {royalblue} {Definition\ of\ Factorial\ Notation}:\ \hspace{20cm}$
$The\ continued\ product\ of\ first\ ‘n’ \ natural\ numbers\ is\ called\ n\ factorial$
$and\ is\ denoted\ by\ n!\ and\ defined\ by\ n!\ = 1×2×3×………..×n$
$Ex:\ 5!\ =\ 1×2×3×4×5\ = 120\ \hspace{10cm}$
$Note:\ 0!\ =\ 1\ \hspace{14cm}$
$\color {royalblue} {Binomial\ theorem\ for\ a\ positive\ integral\ index}:\ \hspace{20cm}$
$If\ ‘n’\ is\ any\ positive\ integer,\ then\ \hspace{15cm}$
$(X\ +\ a)^n\ =\ X^n\ +\ nc_1 X^{n-1}a\ +\ \ nc_2 X^{n-2}a^2\ +\ …………………..\ +\ \ nc_r X^{n-r}a^r\ +\ …….\ a^n$
$Note:\ nc_r\ =\ \frac{n!}{(n-r)!\ r!}\ \hspace{15cm}$
$\color {royalblue} {General\ term\ of\ the\ expansion\ of\ (X\ +\ a)^n}:\ \hspace{20cm}$
$In\ the\ binomial\ expansion\ for\ (X\ +\ a)^n ,\ we\ observe\ that\ the\ first\ term\ is\ nC_0x^n,\ the\ second\ term\ is\ nC_1 x^{n-1} a,\ the\ third\ term\ is\ nC_2x^{n-2},\ and\ so\ on.$
$Looking\ at\ the\ pattern\ of\ the\ successive\ terms\ we\ can\ say\ that\ the\ (r +1 )^{th}\ term\ is\ nC_rx^{n-r} a^r.$
$The\ (r +1 )^{th}\ term\ is\ also\ called\ the\ general\ term\ of\ the\ expansion\ ( x +a )^n .\ It\ is\ denoted\ by\ T_{r + 1}$
$T_{r + 1} = nC_rx^{n-r} a^r.\ \hspace{15cm}$
$\color {purple} {Example\ 1:}\ \color{red}{Write\ down\ the\ general\ term\ of\ the\ expansion\ of}\ (x\ +\ \frac{2}{x})^{12}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x,\ a\ =\ \frac{2}{x},\ n\ =\ 12\ \hspace{14cm}$
$T_{r + 1} = 12C_r\ x^{12-r}\ (\frac{2}{x})^r \hspace{15cm}$
$T_{r + 1} = 12C_r\ x^{12-r}\ 2^r\ x^{-r}\ \hspace{15cm}$
$= 12C_r\ 2^r\ x^{12-r-r}\ \hspace{15cm}$
$\boxed{T_{r + 1} = 12C_r\ 2^r\ x^{12-2r}}\ \hspace{15cm}$
$\color {purple} {Example\ 2:}\ \color {red}{Write\ down\ the\ general\ term\ of\ the\ expansion\ of}\ (x^2\ -\ \frac{2}{x})^{9}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x^2,\ a\ =\ -\frac{2}{x},\ n\ =\ 9\ \hspace{14cm}$
$T_{r + 1} = 9C_r\ (x^2)^{9-r}\ (-\frac{2}{x})^r \hspace{15cm}$
$= 9C_r\ x^{18-2r}\ (-2)^r\ x^{-r}\ \hspace{15cm}$
$= 9C_r\ (-2)^r\ x^{18-2r-r}\ \hspace{15cm}$
$\boxed{T_{r + 1} = 9C_r\ (-2)^r\ x^{18-3r}}\ \hspace{15cm}$
$\color {purple} {Example\ 3:}\ \color {red}{Find\ the\ 6^{th}\ term\ in\ the\ expansion\ of}\ (x\ +\ \frac{1}{x})^{10}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x,\ a\ =\ \frac{1}{x},\ n\ =\ 10,\ r\ =\ 5\ \hspace{14cm}$
$T_{5 + 1} = 10C_5\ x^{10-5}\ (\frac{1}{x})^5 \hspace{15cm}$
$= 10C_5\ x^5\ 1^5\ x^{-5}\ \hspace{15cm}$
$= 10C_5\ x^{5 – 5}\ \hspace{15cm}$
$T_6 = 10C_5\ x^0\ \hspace{15cm}$
$\therefore\ \boxed{The\ 6^{th}\ term\ is\ 10C_5}\ \hspace{15cm}$
$\color {royalblue} {To\ find\ the\ Middle\ term\ of\ the\ expansion\ of\ (X\ +\ a)^n}:\ \hspace{20cm}$
$\color {brown} {cases}:\ \hspace{18cm}$
$1.\ If\ n\ is\ an\ even\ number,\ there\ is\ one\ middle\ term\ =\ (\frac{n+2}{2})^{th}\ term$
$2.\ If\ n\ is\ odd\ number,\ there\ are\ two\ middle\ terms\ (\frac{n+1}{2})^{th}\ and\ \ (\frac{n+1}{2})^{th}\ term$
$\color {purple} {Example\ 4\ .}\ \color {red}{Find\ the\ middle\ term\ in\ the\ expansion\ of}\ (2x^4\ +\ \frac{3}{x})^{16}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Here\ n\ is\ even\ number,\ so\ there\ is\ one\ middle\ term\ \hspace{18cm}$
$(\frac{n+2}{2})^{th}\ term\ =\ (\frac{16+2}{2})^{th}\ term\ =\ (\frac{18}{2})^{th}\ term\ =\ 9^{th}\ term\ \hspace{16cm}$
$T_{r + 1} = nC_rx^{n-r} a^r \hspace{19cm}$
$Here\ X\ =\ 2x^4,\ a\ =\ \frac{3}{x},\ n\ =\ 16,\ r\ =\ 8\ \hspace{14cm}$
$T_{8 + 1} = 16C_8\ (2x^4)^{16-8}\ (\frac{3}{x})^8 \hspace{15cm}$
$T_9 = 16C_8\ (2x^4)^8\ \frac{3^8}{x^8}\ \hspace{15cm}$
$= 16C_8\ \ 2^8\ \ 3^8\ x^{32}\ x^{-8}\ \hspace{15cm}$
$= 16C_8\ \ 2^8\ \ 3^8\ x^{32 – 8}\ \hspace{15cm}$
$\boxed{T_9 = 16C_8\ 2^8\ 3^8\ x^{24}}\ \hspace{15cm}$
$\color {royalblue} {To\ find\ the\ coefficient\ of\ x^n\ for\ particular\ value\ of\ n\ and\ the\ term\ independent\ of\ x\ in\ the\ expansion\ of\ (X\ +\ a)^n}$
$\color {purple} {Example\ 5 :}\ \color {red}{Find\ the\ coefficient\ of\ x^{32}\ in\ the\ expansion\ of}\ (x^4\ +\ \frac{1}{x^3})^{15}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x^4,\ a\ =\ \frac{1}{x^3},\ n\ =\ 15,\ r\ =\ r\ \hspace{14cm}$
$T_{r + 1} = 15C_r\ (x^4)^{15-r}\ (\frac{1}{x^3})^r \hspace{15cm}$
$= 15C_r\ x^{60 – 4r}\ \frac{1^r}{x^{3r}} \hspace{15cm}$
$= 15C_r\ x^{60 – 4r – 3r} \hspace{15cm}$
$T_{r + 1} = 15C_r\ x^{60 – 7r}\ ———————— (1)\ \hspace{15cm}$
$We\ assume\ that\ x^{32}\ occurs\ in\ 15C_r\ x^{60 – 7r}\ \hspace{15cm}$
$\therefore\ x^{60 – 7r}\ =\ x^{32}\ \hspace{15cm}$
$Equating\ the\ indices\ 60 – 7r\ =\ 32\ \hspace{15cm}$
$-7r\ =\ 32\ -\ 60\ \hspace{15cm}$
$-7r\ = -\ 28\ \hspace{15cm}$
$\boxed{r\ = 4}\ \hspace{15cm}$
$Put\ r\ =\ 4\ in\ (1)\ \hspace{15cm}$
$T_{4 + 1} = 15C_4\ 1^4\ x^{60\ -\ 28}\ \hspace{15cm}$
$T_5 = 15C_4\ x^{32}\ \hspace{15cm}$
$\boxed{The\ coefficient\ of\ x^{32}\ =\ 15\ C_4}\ \hspace{15cm}$
$\color {purple} {Example\ 6 :}\ \color {red}{Find\ the\ coefficient\ of\ x^5\ in\ the\ expansion\ of}\ (ax\ +\ \frac{b}{x})^{11}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ ax,\ a\ =\ \frac{b}{x},\ n\ =\ 11,\ r\ =\ r\ \hspace{14cm}$
$T_{r + 1} = 11C_r\ (ax)^{11-r}\ (\frac{b}{x})^r \hspace{15cm}$
$= 11C_r\ a^{11\ -\ r}\ x^{11 – r}\ b^r\ x^{-r}\ \hspace{15cm}$
$= 11C_r\ a^{11\ -\ r}\ b^r\ x^{11\ -\ r}\ x^{-r}\ \hspace{15cm}$
$=\ 11C_r\ a^{11\ -\ r}\ b^r\ x^{11\ -\ 2r}\ \hspace{15cm}$
$T_{r + 1}\ =\ =\ 11C_r\ a^{11\ -\ r}\ b^r\ x^{11\ -\ 2r}\ ———————— (1)\ \hspace{15cm}$
$We\ assume\ that\ x^5\ occurs\ in\ 11C_r\ a^{11\ -\ r}\ b^r\ x^{11\ -\ 2r}\ \hspace{15cm}$
$\therefore\ x^{11\ -\ 2r}\ =\ x^5\ \hspace{15cm}$
$Equating\ the\ indices\ 11\ -\ 2r\ =\ 5\ \hspace{15cm}$
$-2r\ =\ 5\ -\ 11\ \hspace{15cm}$
$-2r\ = -\ 6\ \hspace{15cm}$
$\boxed{r\ = 3}\ \hspace{15cm}$
$Put\ r\ =\ 3\ in\ (1)\ \hspace{15cm}$
$T_{3 + 1} = 11C_3\ a^{11\ -\ 3}\ b^3\ x^{11\ -\ 6}\ \hspace{15cm}$
$T_4 =\ 11C_3\ a^8\ b^3\ x^5\ \hspace{15cm}$
$\boxed{The\ coefficient\ of\ x^5\ =\ 11C_3\ a^8\ b^3}\ \hspace{15cm}$
$\color {purple} {Example\ 7 :}\ \color {red}{Find\ the\ term\ independent\ of\ x\ in\ the\ expansion\ of}\ (\sqrt{x} – \frac{1}{x^2})^{20}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ \sqrt{x},\ a\ =\ \frac{-1}{x^2},\ n\ =\ 20,\ r\ =\ r\ \hspace{14cm}$
$T_{r + 1} = 20C_r\ (\sqrt{x})^{20-r}\ (-\frac{1}{x^2})^r \hspace{15cm}$
$= 20C_r\ x^(\frac{20-r}{2})\ \frac{-1^r}{x^{2r}} \hspace{15cm}$
$= 20C_r\ x^(\frac{20-r}{2})\ {-1}^r\ \ x^{-2r}\ \hspace{15cm}$
$= 20C_r\ x^(\frac{20-r}{2}\ -\ 2r)\ {-1}^r\ \hspace{15cm}$
$= 20C_r\ x^(\frac{20-r – 4r}{2})\ {-1}^r\ \hspace{15cm}$
$T_{r + 1} = 20C_r\ x^(\frac{20-5r}{2})\ {-1}^r\ ———————— (1)\ \hspace{15cm}$
$To\ find\ the\ independent\ term\ of\ x,\ find\ the\ coefficient\ of\ x^0.\ \hspace{15cm}$
$\therefore\ x^{\frac{20-5r}{2}}\ =\ x^0\ \hspace{15cm}$
$\frac{20-5r}{2}\ =\ 0\ \hspace{15cm}$
$20 – 5r\ =\ 0\ \hspace{15cm}$
$-5r\ = -\ 20\ \hspace{15cm}$
$\boxed{r\ = 4}\ \hspace{15cm}$
$Put\ r\ =\ 4\ in\ (1)\ \hspace{15cm}$
$\therefore\ independent\ term\ of\ x =\ 20C_4\ (-1)^4\ \hspace{15cm}$
$=\ 20C_4\ \hspace{15cm}$
$\color {royalblue} {Binomial\ theorem\ for\ rational\ index}:\ \hspace{20cm}$
$If\ ‘n’\ is\ any\ rational\ number\ then\ \hspace{15cm}$
$(1\ +\ x)^n\ =\ 1\ +\ nx\ +\ \frac{n(n-1)}{1.\ 2}\ x^2\ +\ \frac{n(n-1)(n-2)}{1.\ 2.\ 3}\ x^3\ +\ …….$
$\color {purple} {Example\ 8:}\ \color {red}{Write\ down\ the\ first\ three\ terms\ in\ the\ expansion\ of\ (1\ -\ x)^{-2}\ when\ \mid x\ \mid \lt\ 1}\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ (1\ +\ x)^n\ =\ 1\ +\ nx\ +\ \frac{n(n-1)}{1.\ 2}\ x^2\ +\ \frac{n(n-1)(n-2)}{1.\ 2.\ 3}\ x^3\ +\ …….\ \hspace{5cm}$
$Here\ X\ =\ – x,\ n\ =\ – 2\ \hspace{14cm}$
$(1\ +\ x)^{-2}\ =\ 1\ +\ (-2)(-x)\ +\ \frac{-2(-2-1)}{1.\ 2}\ (-x)^2\ +\ …….$
$(1\ +\ x)^{-2}\ =\ 1\ +\ 2x\ +\ (-1) (-3) x^2\ +\ …….$
$(1\ +\ x)^{-2}\ =\ 1\ +\ 2x\ +\ 3 x^2\ +\ …….$
$\color {purple} {Example\ 9:}\ \color {red}{Expand\ using\ (1\ -\ 2x)^{-3}\ Binomial\ Theorem}\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ (1\ +\ x)^n\ =\ 1\ +\ nx\ +\ \frac{n(n-1)}{1.\ 2}\ x^2\ +\ \frac{n(n-1)(n-2)}{1.\ 2.\ 3}\ x^3\ +\ …….\ \hspace{5cm}$
$Here\ X\ =\ – 2x,\ n\ =\ – 3\ \hspace{14cm}$
$W.\ K.\ T\ (1\ -\ 2x)^{-3}\ =\ 1\ +\ (-3)(-2x)\ +\ \frac{-3(-3-1)}{1.\ 2}\ (-2x)^2\ +\ \frac{-3(-3-1)(-3-2)}{1.\ 2.\ 3}\ (-2x)^3\ +\ …….$
$=\ 1\ +\ 6x\ +\ 6(4x^2)\ +\ 8(10x^3) +\ …….$
$=\ 1\ +\ 6x\ +\ 24x^2\ +\ 80x^3 +\ …….$

### Exercise Problems

$\LARGE{\color {purple} {PART- A}}$
$1. \ \color {red}{Find\ the\ general\ term\ of\ the\ expansion\ of}\ (x\ +\ \frac{1}{x})^{10}\ \hspace{15cm}$
$2.\ \color {red}{which\ term\ in\ (3\ x\ -\ y)^6\ is\ the\ middle\ term\ ?}\ \hspace{15cm}$
$3.\ \color {red}{Write\ down\ the\ first\ three\ terms\ in\ the\ expansion\ of\ (1\ +\ x)^{-4}\ when\ \mid x\ \mid \lt\ 1}\ \hspace{15cm}$
$4.\ \color {red}{Write\ down\ the\ first\ two\ terms\ in\ the\ expansion\ of\ (1\ -\ 2x)^{-3}\ when\ \mid x\ \mid \lt\ 1}\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- B}}$
$5.\ \color {red}{Find\ the\ 4^{th}\ term\ in\ the\ expansion\ of}\ (x^2\ +\ \frac{1}{x})^8\ \hspace{15cm}$
$6.\ \color {red}{Find\ the\ 5^{th}\ term\ in\ the\ expansion\ of}\ (x^2\ +\ \frac{1}{x^2})^{10}\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- C}}$
$7 .\ \color {red}{Find\ the\ middle\ terms\ in\ the\ expansion}\ of\ (x\ +\ \frac{2}{x^3})^{10}\ \hspace{15cm}$
$8 .\ \color {red}{Find\ the\ middle\ terms\ in\ the\ expansion\ of}\ (x^3\ +\ \frac{2}{x^3})^{11}\ \hspace{15cm}$
$9.\ \color {red}{Find\ the\ term\ independent\ of\ x\ in\ the\ expansion\ of}\ (x^2\ +\ \frac{1}{x})^{12}\ \hspace{15cm}$