# Question 5 & 6 Review Exercise 7

Solutions of Question 5 & 6 of Review Exercise 7 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.

Q5 What is the constant term in the expansion of $\left(\frac{2}{x^2}+\frac{x^2}{2}\right)^{10}$ ? Solution: Here $n=10, a^{\prime}=\frac{2}{x^2}$ and $b=\frac{x^2}{2}$. Let $T_{r+1}$ be the term independent of $x$ that is: \begin{aligned} & T_{r+1}=\frac{10 !}{(10-r) ! r !}\left(\frac{2}{x^2}\right)^{10 r}\left(\frac{x^2}{2}\right)^r \\ & =\frac{10 !}{(10-r) ! r !} \frac{2^{10} r}{2^r} \cdot x^{2 r} \cdot \frac{1}{x^{20-2 r}} \\ & =\frac{10 !}{(10 r) ! r !} 2^{102 r} x^{2 r} 20+2 r \\ & =\frac{10 !}{(10-r) ! r !} 2^{10-2 r} x^{4 r-20} \end{aligned}

But the term $T_{r+1}$ is independent of $x$ is possible only if $x^{4 r-20}=x^0$ $$\Rightarrow 4 r-20=0 \Rightarrow r=5 \text {. }$$

Putting in $T_{r+1}$ we get $$T_{5+1}=\frac{10 !}{5 ! 5 !} \cdot 2^0 \cdot x^0$$

$$\Rightarrow \quad T_6=252 .$$

Hence $T_6$ is constant term in the expansion which is 252.

Q6 Find an approximation of $(0.99)^5$ using the first three terms of its expansion. Solution: We can write $$(0.99)^5=(1-0.01)^5$$

Using binomial theorem, we have \begin{aligned} & (1-0.01)^5 \cong{ }^5 C_0-{ }^5 C_1(0.01) \\ & +{ }^5 C_2(-0.01)^2 \\ & =1-5(0.01)+10(0.0001) \\ & =0.951 . \end{aligned}

Thus $(0.99) 65 \cong 0.951$.