Question 7 Exercise 7.2

Solutions of Question 7 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 $(2+\sqrt{3})^5+(2-\sqrt{3})^5$

Using binomial formula $$\begin{align}(2+\sqrt{3})^5+(2 \cdot \sqrt{3})^5& =[(2)^5+{ }^5 C_1 \cdot 2^4 \cdot \sqrt{3}+{ }^5 C_2 \cdot 2^3 \cdot(\sqrt{3})^2 \\ & +^5 C_3 \cdot 2^2 \cdot(\sqrt{3})^4+{ }^5 C_4 \cdot 2 \cdot(\sqrt{3})^4 \\ & +{ }^5 C_5 \cdot(\sqrt{3})^5+(2)^5 +{ }^5 C_1 \cdot 2^4 \cdot \sqrt{3} \\ & +^5 C_2 \cdot 2^2 \cdot(\sqrt{3})^2 +^5 C_3 \cdot 2^2(\sqrt{3})^3 \\ & +{ }^5 C_4 \cdot 2 \cdot(\sqrt{3})^4-{ }^5 C_5 \cdot(\sqrt{3})^5]\\ &=2(2)^5+{ }^5 C_1 \cdot 2^5 \cdot \sqrt{3}+{ }^5 C_2 \cdot 2^4 \cdot(\sqrt{3})^2 \\ & +^5 C_3 \cdot 2^3 \cdot(\sqrt{3})^4+{ }^5 C_4 \cdot 2^2 \cdot(\sqrt{3})^4+2{ }^5 C_5 \cdot(\sqrt{3})^5\end{align}$$ simplifing, we get $$\begin{align} & =2 \cdot 2^5+2^5 C_2 \cdot 2^3 \cdot(\sqrt{3})^2+2^5 C_4 \cdot 2 \cdot(\sqrt{3})^4 \\ & =2 \cdot 32+2 \cdot 10 \cdot 8 \cdot 3+2 \cdot 5 \cdot 2 \cdot 9 \\ & =64+480+180=692 \\ & \text { Thus }(2+\sqrt{3})^5+(2-\sqrt{3})^5=724\end{align}$$

$(1+\sqrt{2})^4-(1-\sqrt{2})^{-}$

Using binomial formula $$\begin{align} (1+\sqrt{2})^4-(1-\sqrt{2})^4 & =[1+{ }^4 C_1 \cdot \sqrt{2}+{ }^4 C_2 \cdot(\sqrt{2})^2 \\ & +{ }^4 C_3 \cdot(\sqrt{2})^3+{ }^4 C_4 \cdot(\sqrt{2})^4]- \\ & {[1-{ }^4 C_1 \cdot \sqrt{2}+{ }^4 C_2 \cdot(\sqrt{2})^2} \\ & -{ }^4 C_3 \cdot(\sqrt{2})^3 +{ }^4 C_4 \cdot(\sqrt{2})^4]\end{align}$$ simplifing, we get $$\begin{align} & =2^4 C_1 \cdot \sqrt{2}+2^4 C_3 \cdot(\sqrt{2})^3 \\ & =2 \cdot 4 \cdot \sqrt{2}+2 \cdot 4 \cdot(\sqrt{2})^3 \\ & =8 \sqrt{2}[1+(\sqrt{2})^2] \\ & =8 \sqrt{2}[1+2] \\ & =24 \sqrt{2} . \text { Thus } \\ (1+\sqrt{2})^4-(1-\sqrt{2})^4&=24 \sqrt{2}\end{align}$$

Find $(a+b)^5+(a-b)^5$

(iii) $(a+b)^5+(a-b)^5$

Solution: Using binomial theorem $$\begin{aligned} (a+b)^5+(a-b)^5&=\left[\left(\begin{array}{l} 5 \\ 0 \end{array}\right) a^5+\left(\begin{array}{l} 5 \\ 1 \end{array}\right) a^4 b\right. \\ & +\left(\begin{array}{l} 5 \\ 2 \end{array}\right) a^3 b^2+\left(\begin{array}{l} 5 \\ 3 \end{array}\right) a^2 b^3+\left(\begin{array}{l} 5 \\ 4 \end{array}\right) a^1 b^4+ \\ & \left.\left(\begin{array}{l} 5 \\ 5 \end{array}\right) b^5\right]+\left[\left(\begin{array}{l} 5 \\ 0 \end{array}\right) a^5-\left(\begin{array}{l} 5 \\ 1 \end{array}\right) a^4 b+\left(\begin{array}{l} 5 \\ 2 \end{array}\right) a^3 b^2-\right. \\ & \left.\left(\begin{array}{l} 5 \\ 3 \end{array}\right) a^2 b^3+\left(\begin{array}{l} 5 \\ 4 \end{array}\right) a^1 b^4-\left(\begin{array}{l} 5 \\ 5 \end{array}\right) b^5\right] \end{aligned}$$

Simplifying, we get $$\begin{aligned} & \left.=2\left(\begin{array}{l} 5 \\ 0 \end{array}\right) a^5+\left(\begin{array}{l} 5 \\ 1 \end{array}\right) a^3 b^2+\left(\begin{array}{l} 5 \\ 4 \end{array}\right) a b^4\right] \\ & =10 a^4 b+20 a^2 b^3+2 a b^4 . \end{aligned}$$

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