# BINOMIAL THEOREM (Exercise Problems with Solutions)

$\LARGE{\color {purple} {PART- A}}$
$1. \ \color {red}{Find\ the\ general\ term\ of\ the\ expansion\ of}\ (x\ +\ \frac{1}{x})^{10}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x,\ a\ =\ \frac{1}{x},\ n\ =\ 10\ \hspace{14cm}$
$T_{r + 1} = 10C_r\ x^{10-r}\ (\frac{1}{x})^r \hspace{15cm}$
$T_{r + 1} = 10C_r\ x^{10-r}\ 1^r\ x^{-r}\ \hspace{15cm}$
$= 10C_r\ 1^r\ x^{10-r-r}\ \hspace{15cm}$
$T_{r + 1} = 10C_r\ 1^r\ x^{10-2r}\ \hspace{15cm}$
$2.\ \color {red}{which\ term\ in\ (3\ x\ -\ y)^6\ is\ the\ middle\ term\ ?}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Here\ n\ is\ even\ number,\ so\ there\ is\ one\ middle\ term\ \hspace{18cm}$
$(\frac{n+2}{2})^{th}\ term\ =\ (\frac{6+2}{2})^{th}\ term\ =\ (\frac{8}{2})^{th}\ term\ =\ 4^{th}\ term\ \hspace{16cm}$
$3.\ \color {red}{Write\ down\ the\ first\ three\ terms\ in\ the\ expansion\ of}\ (1\ +\ x)^{-4}\ when\ \mid x\ \mid \lt\ 1\ \hspace{15cm}$
$\color {blue}{Solution:}\ W.\ K.\ T\ (1\ +\ x)^n\ =\ 1\ +\ nx\ +\ \frac{n(n-1)}{1.\ 2}\ x^2\ +\ \frac{n(n-1)(n-2)}{1.\ 2.\ 3}\ x^3\ +\ …….\ \hspace{5cm}$
$Here\ X\ =\ x,\ n\ =\ – 4\ \hspace{14cm}$
$(1\ +\ x)^{-4}\ =\ 1\ +\ (-4)(x)\ +\ \frac{-4(-4-1)}{1.\ 2}\ (x)^2\ +\ \frac{-4(-4-1)(-4-2)}{1.\ 2.\ 3}\ (x)^3\ …….$
$(1\ +\ x)^{-2}\ =\ 1\ +\ 2x\ +\ (-1) (-3) x^2\ +\ (-2) (-5)(-2) x^3\ …….$
$(1\ +\ x)^{-4}\ =\ 1\ -\ 4x\ +\ 10 x^2\ -\ 20\ x^3\ …….$
$\LARGE{\color {purple} {PART- B}}$
$4.\ \color {red}{Find\ the\ 4^{th}\ term\ in\ the\ expansion\ of}\ (x^2\ +\ \frac{1}{x})^8\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x^2,\ a\ =\ \frac{1}{x},\ n\ =\ 8,\ r\ =\ 3\ \hspace{14cm}$
$T_{3 + 1} = 8C_3\ {(x^2)}^{8 – 3}\ (\frac{1}{x})^3 \hspace{15cm}$
$T_{4} = 8C_3\ {(x^2)}^{5}\ \frac{1^3}{x^3} \hspace{15cm}$
$T_4 = 8C_3\ x^{10}\ 1^3\ x^{-3}\ \hspace{15cm}$
$= 8C_3\ x^{10 – 3}\ \hspace{15cm}$
$T_4 = 8C_3\ x^7\ \hspace{15cm}$
$5.\ \color {red}{Find\ the\ 5^{th}\ term\ in\ the\ expansion\ of\ (x^2\ +\ \frac{1}{x^2})^{10}}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x^2,\ a\ =\ \frac{1}{x^2},\ n\ =\ 10,\ r\ =\ 4\ \hspace{14cm}$
$T_{4 + 1} = 10C_4\ {(x^2)}^{10-4}\ (\frac{1}{x^2})^4 \hspace{15cm}$
$T_{5} = 10C_4\ {(x^2)}^{6}\ \frac{1^4}{x^8} \hspace{15cm}$
$T_5 = 10C_4\ x^{12}\ 1^4\ x^{-8}\ \hspace{15cm}$
$= 10C_4\ x^{12 – 8}\ \hspace{15cm}$
$T_5 = 10C_4\ x^4\ \hspace{15cm}$
$6.\ \color {red}{Find\ the\ 6^{th}\ term\ in\ the\ expansion\ of}\ (x^3\ -\ \frac{1}{x^2})^{14}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x^3,\ a\ =\ -\frac{1}{x^2},\ n\ =\ 14,\ r\ =\ 5\ \hspace{14cm}$
$T_{5 + 1} = 14C_5\ {(x^3)}^{14-5}\ (-\frac{1}{x^2})^5 \hspace{15cm}$
$T_{6} = 14C_5\ {(x^3)}^{9}\ \frac{{-1}^5}{x^{10}} \hspace{15cm}$
$T_6 = 14C_5\ x^{27}\ {(-1)}^5\ x^{-10}\ \hspace{15cm}$
$= -\ 14C_5\ x^{27 – 10}\ \hspace{15cm}$
$T_6 =\ -\ 14C_5\ x^{17}\ \hspace{15cm}$
$\LARGE{\color {purple} {PART- C}}$
$7 .\ \color {red}{Find\ the\ middle\ terms\ in\ the\ expansion}\ of\ (x\ +\ \frac{2}{x^3})^{10}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Here\ n\ is\ even\ number,\ so\ there\ is\ one\ middle\ term\ \hspace{18cm}$
$(\frac{n+2}{2})^{th}\ term\ =\ (\frac{10+2}{2})^{th}\ term\ =\ (\frac{12}{2})^{th}\ term\ =\ 6^{th}\ term\ \hspace{16cm}$
$T_{r + 1} = nC_rx^{n-r} a^r \hspace{19cm}$
$Here\ X\ =\ x,\ a\ =\ \frac{2}{x^3},\ n\ =\ 10,\ r\ =\ 5\ \hspace{14cm}$
$T_{5 + 1} = 10C_5\ (x)^{10-5}\ (\frac{2}{x^3})^5 \hspace{15cm}$
$T_6 = 10C_5\ (x)^5\ \frac{2^5}{x^15}\ \hspace{15cm}$
$= 10C_5\ \ x^5\ 2^5\ x^{-15}\ \hspace{15cm}$
$= 10C_5\ \ 2^5\ \ x^5\ x^{-15}\ \hspace{15cm}$
$= 10C_5\ \ 2^5\ \ x^{5-15}\ \hspace{15cm}$
$\boxed{T_6 = 10C_5\ 2^5\ x^{-10}}\ \hspace{15cm}$

$8 .\ \color {red}{Find\ the\ middle\ terms\ in\ the\ expansion}\ of\ (x^3\ +\ \frac{2}{x^3})^{11}\ \hspace{15cm}$
$\color {blue}{Solution:}\ Here\ n\ is\ odd\ number,\ so\ there\ are\ two\ middle\ terms\ \hspace{18cm}$
$(\frac{n+1}{2})^{th}\ term\ =\ (\frac{11+1}{2})^{th}\ term\ =\ (\frac{12}{2})^{th}\ term\ =\ 6^{th}\ term\ \hspace{16cm}$
$(\frac{n+3}{2})^{th}\ term\ =\ (\frac{11+3}{2})^{th}\ term\ =\ (\frac{14}{2})^{th}\ term\ =\ 7^{th}\ term\ \hspace{16cm}$
$\underline{To\ find\ T_6\ :}\ \hspace{15cm}$
$T_{r + 1} = nC_rx^{n-r} a^r \hspace{19cm}$
$Here\ X\ =\ x^3,\ a\ =\ \frac{2}{x^3},\ n\ =\ 11,\ r\ =\ 5\ \hspace{14cm}$
$T_{5 + 1} = 11C_5\ (x^3)^{11-5}\ (\frac{2}{x^3})^5 \hspace{15cm}$
$T_6 = 11C_5\ (x^3)^6\ \frac{(2)^5}{(x^3)^5}\ \hspace{15cm}$
$= 11C_5\ \ 2^5\ \ x^{18}\ x^{-15}\ \hspace{15cm}$
$= 11C_5\ \ 2^5\ \ x^{18\ -\ 15}\ \hspace{15cm}$
$T_6 = 11C_5\ 2^5\ x^3\ \hspace{15cm}$
$\underline{To\ find\ T_7\ :}\ \hspace{15cm}$
$T_{r + 1} = nC_rx^{n-r} a^r \hspace{19cm}$
$Here\ X\ =\ x^3,\ a\ =\ \frac{2}{x^3},\ n\ =\ 11,\ r\ =\ 6\ \hspace{14cm}$
$T_{6 + 1} = 11C_6\ (x^3)^{11-6}\ (\frac{2}{x^3})^6 \hspace{15cm}$
$T_7 = 11C_6\ (x^3)^5\ \frac{(2)^6}{(x^3)^6}\ \hspace{15cm}$
$= 11C_6\ \ 2^6\ \ x^{15}\ x^{-18}\ \hspace{15cm}$
$= 11C_6\ \ 2^6\ \ x^{15\ -\ 18}\ \hspace{15cm}$
$T_7 = 11C_6\ 2^6\ x^{-3}\ \hspace{15cm}$
$9.\ \color {red}{Find\ the\ term\ independent\ of\ x\ in\ the\ expansion\ of}\ (x^2\ +\ \frac{1}{x})^{12}\ \hspace{15cm}$
$\color {blue}{Solution:}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}$
$Here\ X\ =\ x^2,\ a\ =\ \frac{1}{x},\ n\ =\ 12,\ r\ =\ r\ \hspace{14cm}$
$T_{r + 1} = 12C_r\ (x^2)^{12-r}\ (\frac{1}{x})^r \hspace{15cm}$
$= 12C_r\ (x^2)^{12 – r}\ \frac{1^r}{x^{r}} \hspace{15cm}$
$= 12C_r\ x^{24\ -\ 2r}\ 1^r\ x^{-r}\ \hspace{15cm}$
$= 12C_r\ 1^r\ x^{24 – 2r – r} \hspace{15cm}$
$T_{r + 1} = 12C_r\ 1^r\ x^{24 – 3r}\ ———————— (1)\ \hspace{15cm}$
$To\ find\ the\ independent\ term\ of\ x,\ find\ the\ coefficient\ of\ x^0.\ \hspace{15cm}$
$\therefore\ x^{24\ -\ 3r}\ =\ x^0\ \hspace{15cm}$
$24 – 3r\ =\ 0\ \hspace{15cm}$
$-3r\ = -\ 24\ \hspace{15cm}$
$r\ = 8\ \hspace{15cm}$
$Put\ r\ =\ 8\ in\ (1)\ \hspace{15cm}$
$\therefore\ independent\ term\ of\ x =\ 12C_8\ 1^8\ \hspace{15cm}$
$=\ 12C_8\ \hspace{15cm}$