Chapter 7 Binomial Theorem Miscellaneous Exercise
Miscellaneous Exercise on Chapter 7
1. If $a$ and $b$ are distinct integers, prove that $a-b$ is a factor of $a^{n}-b^{n}$, whenever $n$ is a positive integer.
[Hint write $a^{n}=(a-b+b)^{n}$ and expand]
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Answer :
In order to prove that (a-$b$ ) is a factor of $(a^{n} - b^{n})$, it has to be proved that $a^{n} - b^{n}=k(a - b)$,
where $k$ is some natural number It can be written that, $a=a - b+b$
$ \begin{aligned} \therefore a^{n} & =(a-b+b)^{n}=[(a-b)+b]^{n} \\ & ={ }^{n} C_0(a-b)^{n}+{ }^{n} C_1(a-b)^{n-1} b+\ldots+{ }^{n} C _{n-1}(a-b) b^{n-1}+{ }^{n} C_n b^{n} \\ & =(a-b)^{n}+{ }^{n} C_1(a-b)^{n-1} b+\ldots+{ }^{n} C _{n-1}(a-b) b^{n-1}+b^{n} \\ \Rightarrow & a^{n}-b^{n}=(a-b)[(a-b)^{n-1}+{ }^{n} C_1(a-b)^{n-2} b+\ldots+{ }^{n} C _{n-1} b^{n-1}] \\ \Rightarrow & a^{n}-b^{n}=k(a-b) \end{aligned} $
where, $k=[(a-b)^{n-1}+{ }^{n} C_1(a-b)^{n-2} b+\ldots+{ }^{n} C _{n-1} b^{n-1}]$ is a natural number
This shows that (a-$b$ ) is a factor of $(a^{n} - b^{n})$, where $n$ is a positive integer.
2. Evaluate $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$
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Answer :
Firstly, the expression $(a+b)^{6} -(a - b)^{6}$ is simplified by using Binomial Theorem.
This can be done as
$ \begin{aligned} &(a+b)^{6}={ }^{6} C_0 a^{6}+{ }^{6} C_1 a^{5} b+{ }^{6} C_2 a^{4} b^{2}+{ }^{6} C_3 a^{3} b^{3}+{ }^{6} C_4 a^{2} b^{4}+{ }^{6} C_5 a^{1} b^{5}+{ }^{6} C_6 b^{6} \\ &=a^{6}+6 a^{5} b+15 a^{4} b^{2}+20 a^{3} b^{3}+15 a^{2} b^{4}+6 a b^{5}+b^{6} \\ &(a-b)^{6}={ }^{6} C_0 a^{6}-{ }^{6} C_1 a^{5} b+{ }^{6} C_2 a^{4} b^{2}-{ }^{6} C_3 a^{3} b^{3}+{ }^{6} C_4 a^{2} b^{4}-{ }^{6} C_5 a^{1} b^{5}+{ }^{6} C_6 b^{6} \\ &=a^{6}-6 a^{5} b+15 a^{4} b^{2}-20 a^{3} b^{3}+15 a^{2} b^{4}-6 a b^{5}+b^{6} \\ & \therefore(a+b)^{6}-(a-b)^{6}=2[6 a^{5} b+20 a^{3} b^{3}+6 a b^{5}] \end{aligned} $
Putting $a=\sqrt{3}$ and $b=\sqrt{2}$, we obtain
$ \begin{aligned} (\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6} & =2[6(\sqrt{3})^{5}(\sqrt{2})+20(\sqrt{3})^{3}(\sqrt{2})^{3}+6(\sqrt{3})(\sqrt{2})^{5}] \\ & =2[54 \sqrt{6}+120 \sqrt{6}+24 \sqrt{6}] \\ & =2 \times 198 \sqrt{6} \\ & =396 \sqrt{6} \end{aligned} $
3. Find the value of $(a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}$.
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Answer :
Firstly, the expression $(x+y)^{4}+(x - y)^{4}$ is simplified by using Binomial Theorem.
This can be done as
$ \begin{aligned} &(x+y)^{4}={ }^{4} C_0 x^{4}+{ }^{4} C_1 x^{3} y+{ }^{4} C_2 x^{2} y^{2}+{ }^{4} C_3 x^{3}+{ }^{4} C_4 y^{4} \\ &=x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4} \\ &(x-y)^{4}={ }^{4} C_0 x^{4}-{ }^{4} C_1 x^{3} y+{ }^{4} C_2 x^{2} y^{2}-{ }^{4} C_3 x^{3}+{ }^{4} C_4 y^{4} \\ &=x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4} \\ & \therefore(x+y)^{4}+(x-y)^{4}=2(x^{4}+6 x^{2} y^{2}+y^{4}) \end{aligned} $
Putting $x=a^{2}$ and $y=\sqrt{a^{2}-1}$, we obtain
$ \begin{aligned} (a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4} & =2[(a^{2})^{4}+6(a^{2})^{2}(\sqrt{a^{2}-1})^{2}+(\sqrt{a^{2}-1})^{4}] \\ & =2[a^{8}+6 a^{4}(a^{2}-1)+(a^{2}-1)^{2}] \\ & =2[a^{8}+6 a^{6}-6 a^{4}+a^{4}-2 a^{2}+1] \\ & =2[a^{8}+6 a^{6}-5 a^{4}-2 a^{2}+1] \\ & =2 a^{8}+12 a^{6}-10 a^{4}-4 a^{2}+2 \end{aligned} $
4. Find an approximation of $(0.99)^{5}$ using the first three terms of its expansion.
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Answer :
$0.99=1$-0.01
$ \begin{aligned} \therefore(0.99)^{5} & =(1-0.01)^{5} \\ & ={ }^{5} C_0(1)^{5}-{ }^{5} C_1(1)^{4}(0.01)+{ }^{5} C_2(1)^{3}(0.01)^{2} \\ & =1-5(0.01)+10(0.01)^{2} \\ & =1-0.05+0.001 \\ & =1.001-0.05 \\ & =0.951 \end{aligned} $
(Approximately)
Thus, the value of $(0.99)^{5}$ is approximately 0.951 .
5. Expand using Binomial Theorem $(1+\dfrac{x}{2}-\dfrac{2}{x})^{4}, x \neq 0$.
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Answer :
Using Binomial Theorem, the given expression $(1+\dfrac{x}{2}-\dfrac{2}{x})^{4}$ can be expanded as
$$ \begin{align*} & {[(1+\dfrac{x}{2})-\dfrac{2}{x}]^{4}} \\ & ={ }^{4} C_0(1+\dfrac{x}{2})^{4}-{ }^{4} C_1(1+\dfrac{x}{2})^{3}(\dfrac{2}{x})+{ }^{4} C_2(1+\dfrac{x}{2})^{2}(\dfrac{2}{x})^{2}-{ }^{4} C_3(1+\dfrac{x}{2})(\dfrac{2}{x})^{3}+{ }^{4} C_4(\dfrac{2}{x})^{4} \\ & =(1+\dfrac{x}{2})^{4}-4(1+\dfrac{x}{2})^{3}(\dfrac{2}{x})+6(1+x+\dfrac{x^{2}}{4})(\dfrac{4}{x^{2}})-4(1+\dfrac{x}{2})(\dfrac{8}{x^{3}})+\dfrac{16}{x^{4}} \\ & =(1+\dfrac{x}{2})^{4}-\dfrac{8}{x}(1+\dfrac{x}{2})^{3}+\dfrac{24}{x^{2}}+\dfrac{24}{x}+6-\dfrac{32}{x^{3}}-\dfrac{16}{x^{2}}+\dfrac{16}{x^{4}} \\ & =(1+\dfrac{x}{2})^{4}-\dfrac{8}{x}(1+\dfrac{x}{2})^{3}+\dfrac{8}{x^{2}}+\dfrac{24}{x}+6-\dfrac{32}{x^{3}}+\dfrac{16}{x^{4}} \tag{1} \end{align*} $$
Again by using Binomial Theorem, we obtain
$$ \begin{align*} (1+\dfrac{x}{2})^{4} & ={ }^{4} C_0(1)^{4}+{ }^{4} C_1(1)^{3}(\dfrac{x}{2})+{ }^{4} C_2(1)^{2}(\dfrac{x}{2})^{2}+{ }^{4} C_3(1)^{1}(\dfrac{x}{2})^{3}+{ }^{4} C_4(\dfrac{x}{2})^{4} \\ & =1+4 \times \dfrac{x}{2}+6 \times \dfrac{x^{2}}{4}+4 \times \dfrac{x^{3}}{8}+\dfrac{x^{4}}{16} \\ & =1+2 x+\dfrac{3 x^{2}}{2}+\dfrac{x^{3}}{2}+\dfrac{x^{4}}{16} \tag{2}\\ (1+\dfrac{x}{2})^{3} & ={ }^{3} C_0(1)^{3}+{ }^{3} C_1(1)^{2}(\dfrac{x}{2})+{ }^{3} C_2(1)(\dfrac{x}{2})^{2}+{ }^{3} C_3(\dfrac{x}{2})^{3} \\ & =1+\dfrac{3 x}{2}+\dfrac{3 x^{2}}{4}+\dfrac{x^{3}}{8} \tag{3} \end{align*} $$
From(1), (2), and (3), we obtain
$ \begin{aligned} & {[(1+\dfrac{x}{2})-\dfrac{2}{x}]^{4}} \\ & =1+2 x+\dfrac{3 x^{2}}{2}+\dfrac{x^{3}}{2}+\dfrac{x^{4}}{16}-\dfrac{8}{x}(1+\dfrac{3 x}{2}+\dfrac{3 x^{2}}{4}+\dfrac{x^{3}}{8})+\dfrac{8}{x^{2}}+\dfrac{24}{x}+6-\dfrac{32}{x^{3}}+\dfrac{16}{x^{4}} \\ & =1+2 x+\dfrac{3}{2} x^{2}+\dfrac{x^{3}}{2}+\dfrac{x^{4}}{16}-\dfrac{8}{x}-12-6 x-x^{2}+\dfrac{8}{x^{2}}+\dfrac{24}{x}+6-\dfrac{32}{x^{3}}+\dfrac{16}{x^{4}} \\ & =\dfrac{16}{x}+\dfrac{8}{x^{2}}-\dfrac{32}{x^{3}}+\dfrac{16}{x^{4}}-4 x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{2}+\dfrac{x^{4}}{16}-5 \end{aligned} $
6. Find the expansion of $(3 x^{2}-2 a x+3 a^{2})^{3}$ using binomial theorem.
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Answer :
Using Binomial Theorem, the given expression $(3 x^{2}-2 a x+3 a^{2})^{3}$ can be expanded as
$[(3 x^{2}-2 ax)+3 a^{2}]^{3}$
$={ }^{3} C_0(3 x^{2}-2 ax)^{3}+{ }^{3} C_1(3 x^{2}-2 ax)^{2}(3 a^{2})+{ }^{3} C_2(3 x^{2}-2 ax)(3 a^{2})^{2}+{ }^{3} C_3(3 a^{2})^{3}$
$=(3 x^{2}-2 ax)^{3}+3(9 x^{4}-12 ax^{3}+4 a^{2} x^{2})(3 a^{2})+3(3 x^{2}-2 ax)(9 a^{4})+27 a^{6}$
$=(3 x^{2}-2 ax)^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+36 a^{4} x^{2}+81 a^{4} x^{2}-54 a^{5} x+27 a^{6}$
$=(3 x^{2}-2 ax)^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}$
Again by using Binomial Theorem, we obtain
$$ \begin{align*} & (3 x^{2}-2 a x)^{3} \\ & ={ }^{3} C_0(3 x^{2})^{3}-{ }^{3} C_1(3 x^{2})^{2}(2 a x)+{ }^{3} C_2(3 x^{2})(2 a x)^{2}-{ }^{3} C_3(2 a x)^{3} \\ & =27 x^{6}-3(9 x^{4})(2 a x)+3(3 x^{2})(4 a^{2} x^{2})-8 a^{3} x^{3} \\ & =27 x^{6}-54 a x^{5}+36 a^{2} x^{4}-8 a^{3} x^{3} \tag{2} \end{align*} $$
From (1) and (2), we obtain
$$ \begin{aligned} & (3 x^{2}-2 a x+3 a^{2})^{3} \\ & =27 x^{6}-54 a x^{5}+36 a^{2} x^{4}-8 a^{3} x^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6} \\ & =27 x^{6}-54 a x^{5}+117 a^{2} x^{4}-116 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6} \end{aligned} $$
