====== 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.

=====Question 7(i)=====
Find $(2+\sqrt{3})^5+(2-\sqrt{3})^5$
====Solution====
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}

=====Question 7(ii)=====
$(1+\sqrt{2})^4-(1-\sqrt{2})^{-}$
====Solution====
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}

=====Question 7(iii)=====
Find $(a+b)^5+(a-b)^5$
====Solution====
(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
$$
\begin{aligned}
& \left.=2\left(\begin{array}{l}
5 \\
0
\end{array}\right) a^5+\left(\begin{array}{l}
5 \\
1
\end{array}\right) a^3 b^2+\left(\begin{array}{l}
5 \\
4
\end{array}\right) a b^4\right] \\
& =10 a^4 b+20 a^2 b^3+2 a b^4 .
\end{aligned}
$$
====Go To====
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