Question 2 Exercise 7.2

Solutions of Question 2 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 indicate term in the expansion $4^{th}$ term in $(2+a)^7$.

$\ln$ the above $n=7$, $a=2$ and $b=a$.

Thus the general term of the given expansion is: $$T_{r+1}=\frac{7 !}{(7-r) ! r !}(2)^{7-r } a^r$$ For $4^{\text {th }}$ term, putting $r=3$ $$\begin{align} & T_{3+1}=\dfrac{7 !}{(7-3) ! 3 !} 2^{7-3} a^3 \\ & \Rightarrow T_4=\dfrac{7 !}{4 ! 3 !} \cdot 2^4 a^3 \\ & \Rightarrow T_4=35 \times 16 a^3 \\ & \Rightarrow T_4=560 a ^3 \end{align}$$

Find the indicate term in the expansion $8^{\text {th }}$ term in $(\dfrac{x}{2}-\dfrac{3}{y})^{10}$

In the above $n=10$, $a=\dfrac{x}{2}$ and $b=-\dfrac{3}{y}$.

Thus the general term of the given expansion is: $$T_{r+1}=\dfrac{10 !}{(10-r) ! r !}(\dfrac{x}{2})^{10-r}(-\dfrac{3}{y})^r$$ For $8^{\text {th }}$ term, putting $r=7$ $$\begin{align}T_{7+1}&=\dfrac{10 !}{(10-7) ! 7 !}(\dfrac{x}{2})^{10-7}(-\dfrac{3}{y})^7 \\ & =-{}^{10}C_7 \,2^{-3}\cdot 3^7 x^3 \cdot y^{-7} \end{align}$$

Find the indicate term in the expansion $8^{\text {th }}$ term in $3^{\text {rd }}$ term in $(x^2+\dfrac{1}{\sqrt{x}})^4$.

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.

$T_{r+1}$ of the given expansion is: $$\begin{align}T_{r-1}&=\dfrac{21 !}{(21-r) ! r !}(x)^{21-r}(-\dfrac{1}{x^2})^r \\ & =\dfrac{21 !}{(21-r) ! r !}(-1)^r \cdot x^{21 \cdots r} \cdot x^{-2 r} \\ & =\dfrac{21 !}{(21-r) ! r !}(-1)^r \cdot x^{21 \cdot 3 r}\end{align}$$ But the term $T_{r+1}$ independent of $x$ is possible conly 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

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