Question 3 Exercise 7.2

Solutions of Question 3 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find the term independent of $x$ in the expansion $(\dfrac{4 x^2}{3}-\dfrac{3}{2 x})$

In the above expansion $n=9, \quad a=\dfrac{4 x^2}{3}$ and $b=-\dfrac{3}{2 x}$.

Let $T_{r+1}$ be the term independent of $x$ in the given expansion.

$T_{r+1}$ of the given expansion is: $$\begin{align}T_{r+1}&=\dfrac{9 !}{(9-r) ! r !}(\dfrac{4 x^2}{3})^{9-r}(-\dfrac{3}{2 x})^r \\ & =\dfrac{9 !}{(9-r) ! r !} \cdot \dfrac{4^{9-r}}{3^{9-r}} \cdot \dfrac{(-3)^r}{2^r} \cdot \dfrac{(x^2)^{9-r}}{x^r} \\ & =\dfrac{9 !}{(9-r) ! r !} \cdot \dfrac{4^{9-r}}{3^{9-r}} \cdot \dfrac{(-3)^r}{2^r} x^{18-2 r r} \\ & =\dfrac{9 !}{(9-r) ! r !} \cdot \dfrac{4^{9-r}}{3^{9-r}} \cdot \dfrac{(-3)^r}{2^r} x^{18-3 r}\end{align}$$ But the term $T_{r+1}$ independent of $x$ is possible only if $x^{18-3 r}=x^0$ $$\Rightarrow 18-3 r=0 \Rightarrow 3 r=18 \Rightarrow r=6$$ Putting $r=6$ in the above $T_{r+1}$ $$\begin{align}T_{6-1}&=\dfrac{9 !}{(9-6) ! 6 !}\cdot \dfrac{4^{9-6}}{3^{9-6}} \cdot \dfrac{(-3)^6}{2^6} \cdot x^{18-16} \\ T_7&=84 \cdot \dfrac{4^7}{3^3} \cdot \dfrac{3^6}{2^6} \because(-3)^6=3^6 \\ \Rightarrow T_7&=84 \cdot \dfrac{2^6}{2^6} \cdot 3^{6-3} \because \quad 4^3=(2^2)^3 \\ \Rightarrow T_7&=84 \times 27=2268\end{align}$$ Hence $T_7$ is independent of $x$ and is $2268.$

Find the term independent of $x$ in the expansion $(x-\dfrac{3}{x^4})^{10}$

In the above expansion $n=10, \quad a=x$ and $b=-\dfrac{3}{x^4}$.

Let $T_{r+1}$ be the term independent of $x$ in the given expansion.

$T_{r+1}$ of the given expansion is: $$\begin{align} T_{r+1}&=\dfrac{10 !}{(10-r) ! r !}(x)^{10 \cdot r}(-\dfrac{3}{x^4})^r \\ & =\dfrac{10 !}{(10-r) ! r !} \cdot(-3)^r x^{10-r} x^{-4 r} \\ & =\dfrac{10 !}{(10-r) ! r !}(-3)^r \cdot x^{10-5 r}\end{align}$$ But the term $T_{r+1}$ independent of $x$ is possible only if $x^{10-5 r}=x^0$ $$\Rightarrow 10-5 r=0 \quad \Rightarrow r=2$$ Putting $r=2$ in the above, we get $$\begin{align} T_{2+1}&=\dfrac{10 !}{(10-2) ! 2 !}(-3)^2 \cdot x^{10} 10 \\ \Rightarrow T_3&=45 \times 9=405\end{align}$$ Hence $T_3$ is independemt of $x$ which is $405$

Find the term independent of $x$ in the expansion $(x-\dfrac{1}{x^2})^{2 !}$

In the above expansion $n=$ 21, $a=x$ and $b=-\dfrac{1}{x^2}$.

Let $T_{r+1}$ be the term independent of $x$ in the given expansion.

$T_{t+1}$ of the given expansion is: $$\begin{align} T_{r-1}&=\dfrac{21 !}{(21-r) ! r !}(x)^{21-r}(-\dfrac{1}{x^2})^r \\ & =\dfrac{21 !}{(21-r) ! r !}(-1)^r \cdot x^{21 \cdots r} \cdot x^{-2 r} \\ & =\dfrac{21 !}{(21-r) ! r !}(-1)^r \cdot x^{21-3 r}\end{align}$$ But the term $T_{r+1}$ independent of $x$ is possible inly if $x^{21-3 t}=x^0$ $$\Rightarrow 21-3 r=0 \Rightarrow r=7$$ Putting $r=7$ in the above, we get $$\begin{align}T_{7+1}&=\dfrac{21 !}{(21-7) ! 7 !}(-1)^7 \cdot x^{21-3.7} \\ \Rightarrow T_8&=-^{21}C_7 \end{align}$$ Hence $T_8$ is independent of $x$ which is $-^{21}C_7$

< Question 2 Question 4 >