📚 NCERT Solutions Step-by-Step Logic 🎯 MCQ’s Quiz Master 📄 Assignments DPPs/Assignments

Class 11 NCERT Solutions

Chapter 7: Binomial Theorem

Master the expansion of algebraic powers, the properties of general terms, and the logic of Pascal’s patterns with our step-by-step logic.

Exercise 7.1

1. Expand $(1-2x)^5$

Apply the Binomial Theorem. Since the expression has the form $(x – y)^n$, the terms alternate between positive and negative:

$ (1-2x)^5 = {}^5C_0(1)^5 – {}^5C_1(1)^4(2x) + {}^5C_2(1)^3(2x)^2 – {}^5C_3(1)^2(2x)^3 + {}^5C_4(1)(2x)^4 – {}^5C_5(2x)^5 $ $ \begin{aligned} &= 1 – 5(2x) + 10(4x^2) – 10(8x^3) + 5(16x^4) – 32x^5 \\ &\left[\because {}^5C_5 = {}^5C_0 = 1,\ {}^5C_4 = {}^5C_1 = 5,\ {}^5C_3 = {}^5C_2 = \frac{5\times4}{2\times1} = 10\right] \\ &= 1 – 10x + 40x^2 – 80x^3 + 80x^4 – 32x^5 \end{aligned} $

2. Expand $\left(\dfrac{2}{x} – \dfrac{x}{2}\right)^5$

Use the Binomial Theorem, treating $\dfrac{2}{x}$ as the first term and $\dfrac{x}{2}$ as the second:

$ \left(\frac{2}{x}-\frac{x}{2}\right)^5 = {}^5C_0\left(\frac{2}{x}\right)^5 – {}^5C_1\left(\frac{2}{x}\right)^4\left(\frac{x}{2}\right) + {}^5C_2\left(\frac{2}{x}\right)^3\left(\frac{x}{2}\right)^2 $ $ \begin{aligned} &\quad – {}^5C_3\left(\frac{2}{x}\right)^2\left(\frac{x}{2}\right)^3 + {}^5C_4\left(\frac{2}{x}\right)\left(\frac{x}{2}\right)^4 – {}^5C_5\left(\frac{x}{2}\right)^5 \\ &= \frac{32}{x^5} – 5\left(\frac{16}{x^4}\right)\left(\frac{x}{2}\right) + 10\left(\frac{8}{x^3}\right)\left(\frac{x^2}{4}\right) \\ &\quad – 10\left(\frac{4}{x^2}\right)\left(\frac{x^3}{8}\right) + 5\left(\frac{2}{x}\right)\left(\frac{x^4}{16}\right) – \frac{x^5}{32} \\ &\quad \left[\because {}^5C_5 = {}^5C_0 = 1,\ {}^5C_4 = {}^5C_1 = 5,\ {}^5C_3 = {}^5C_2 = \frac{5\times4}{2\times1} = 10\right] \\ &= \frac{32}{x^5} – \frac{40}{x^3} + \frac{20}{x} – 5x + \frac{5}{8}x^3 – \frac{1}{32}x^5 \end{aligned} $

3. Expand $(2x-3)^6$

Expand using the Binomial Theorem with $n = 6$, taking $2x$ and $3$ as the two terms:

$ (2x-3)^6 = {}^6C_0(2x)^6 – {}^6C_1(2x)^5(3) + {}^6C_2(2x)^4(3)^2 $ $ \begin{aligned} &\quad – {}^6C_3(2x)^3(3)^3 + {}^6C_4(2x)^2(3)^4 – {}^6C_5(2x)(3)^5 + {}^6C_6(3)^6 \\ &= 64x^6 – 6(32x^5)(3) + 15(16x^4)(9) – 20(8x^3)(27) \\ &\quad + 15(4x^2)(81) – 6(2x)(243) + 729 \\ &\left[\because {}^6C_6 = {}^6C_0 = 1,\ {}^6C_5 = {}^6C_1 = 6,\ {}^6C_4 = {}^6C_2 = 15,\ {}^6C_3 = 20\right] \\ &= 64x^6 – 576x^5 + 2160x^4 – 4320x^3 + 4860x^2 – 2916x + 729 \end{aligned} $

4. Expand $\left(\dfrac{x}{3} + \dfrac{1}{x}\right)^5$

Apply the Binomial Theorem with the two terms $\dfrac{x}{3}$ and $\dfrac{1}{x}$:

$ \left(\frac{x}{3}+\frac{1}{x}\right)^5 = {}^5C_0\left(\frac{x}{3}\right)^5 + {}^5C_1\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right) + {}^5C_2\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2 $ $ + {}^5C_3\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3 + {}^5C_4\left(\frac{x}{3}\right)\left(\frac{1}{x}\right)^4 + {}^5C_5\left(\frac{1}{x}\right)^5 $ $ \begin{aligned} &= \frac{x^5}{243} + 5\left(\frac{x^4}{81}\right)\left(\frac{1}{x}\right) + 10\left(\frac{x^3}{27}\right)\left(\frac{1}{x^2}\right) \\ &\quad + 10\left(\frac{x^2}{9}\right)\left(\frac{1}{x^3}\right) + 5\left(\frac{x}{3}\right)\left(\frac{1}{x^4}\right) + \frac{1}{x^5} \\ &\left[\because {}^5C_5 = {}^5C_0 = 1,\ {}^5C_4 = {}^5C_1 = 5,\ {}^5C_3 = {}^5C_2 = 10\right] \\ &= \frac{1}{243}x^5 + \frac{5}{81}x^3 + \frac{10}{27}x + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5} \end{aligned} $

5. Expand $\left(x + \dfrac{1}{x}\right)^6$

Expand using the Binomial Theorem with $n = 6$:

$ \left(x+\frac{1}{x}\right)^6 = {}^6C_0 x^6 + {}^6C_1 x^5\left(\frac{1}{x}\right) + {}^6C_2 x^4\left(\frac{1}{x}\right)^2 + {}^6C_3 x^3\left(\frac{1}{x}\right)^3 $ $ \begin{aligned} &+ {}^6C_4 x^2\left(\frac{1}{x}\right)^4 + {}^6C_5 x\left(\frac{1}{x}\right)^5 + {}^6C_6\left(\frac{1}{x}\right)^6 \\ &= x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6} \end{aligned} $

6. Using binomial theorem, evaluate $(96)^3$.

Rewrite 96 as a sum/difference of a round number to make computation easier:

$ (96)^3 = (100-4)^3 $ $ \begin{aligned} &= {}^3C_0(100)^3 – {}^3C_1(100)^2(4) + {}^3C_2(100)(4)^2 – {}^3C_3(4)^3 \\ &= 1000000 – 3(40000) + 3(1600) – 64 \\ &\left[\because {}^3C_3 = {}^3C_0 = 1,\ {}^3C_2 = {}^3C_1 = 3\right] \\ &= 1000000 – 120000 + 4800 – 64 \\ &= 884736 \end{aligned} $

7. Using binomial theorem, evaluate $(102)^5$.

Express 102 as $100 + 2$ to apply the Binomial Theorem conveniently:

$ (102)^5 = (100+2)^5 $ $ \begin{gathered} = {}^5C_0(100)^5 + {}^5C_1(100)^4(2) + {}^5C_2(100)^3(2)^2 \\ + {}^5C_3(100)^2(2)^3 + {}^5C_4(100)(2)^4 + {}^5C_5(2)^5 \\ = 10000000000 + 5(200000000) + 10(4000000) \\ + 10(80000) + 5(1600) + 32 \end{gathered} $ $ \begin{aligned} &\left[\because {}^5C_5 = {}^5C_0 = 1,\ {}^5C_4 = {}^5C_1 = 5,\ {}^5C_3 = {}^5C_2 = 10\right] \\ &= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 \\ &= 11040808032 \end{aligned} $

8. Using binomial theorem, evaluate $(101)^4$.

Write 101 as $100 + 1$ and apply the Binomial Theorem:

$ (101)^4 = (100+1)^4 $ $ \begin{aligned} &= {}^4C_0(100)^4 + {}^4C_1(100)^3 + {}^4C_2(100)^2 + {}^4C_3(100) + {}^4C_4 \\ &= 100000000 + 4(1000000) + 6(10000) + 4(100) + 1 \\ &\left[\because {}^4C_4 = {}^4C_0 = 1,\ {}^4C_3 = {}^4C_1 = 4,\ {}^4C_2 = 6\right] \\ &= 100000000 + 4000000 + 60000 + 400 + 1 \\ &= 104060401 \end{aligned} $

9. Using binomial theorem, evaluate $(99)^5$.

Write 99 as $100 – 1$ to make use of the Binomial Theorem:

$ (99)^5 = (100-1)^5 $ $ \begin{aligned} &= {}^5C_0(100)^5 – {}^5C_1(100)^4 + {}^5C_2(100)^3 – {}^5C_3(100)^2 \\ &\quad + {}^5C_4(100) – {}^5C_5 \\ &= 10000000000 – 5(100000000) + 10(1000000) \\ &\quad – 10(10000) + 5(100) – 1 \\ &\left[\because {}^5C_5 = {}^5C_0 = 1,\ {}^5C_4 = {}^5C_1 = 5,\ {}^5C_3 = {}^5C_2 = 10\right] \\ &= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1 \\ &= 9509900499 \end{aligned} $

10. Using Binomial Theorem, indicate which number is larger $(1.1)^{10000}$ or $1000$.

Rewrite $(1.1)^{10000}$ as $(1 + 0.1)^{10000}$ and expand the first few terms. Since all remaining terms are positive, we can establish a lower bound:

$ (1.1)^{10000} = (1+0.1)^{10000} $ $ \begin{aligned} &= {}^{10000}C_0 + {}^{10000}C_1(0.1) + \text{other positive terms} \\ &= 1 + 10000 \times 0.1 + \text{other positive terms} \\ &= 1 + 1000 + \text{other positive terms} \\ &> 1000 \end{aligned} $ $\Rightarrow \quad (1.1)^{10000} > 1000$

Therefore, $(1.1)^{10000}$ is the larger number.

11. Find $(a+b)^4 – (a-b)^4$. Hence, evaluate $(\sqrt{3}+\sqrt{2})^4 – (\sqrt{3}-\sqrt{2})^4$.

Expand both expressions using the Binomial Theorem, then subtract. Notice that when we subtract, all even-powered terms cancel:

$ \begin{aligned} (a+b)^4 – (a-b)^4 &= \left({}^4C_0 a^4 + {}^4C_1 a^3 b + {}^4C_2 a^2b^2 + {}^4C_3 ab^3 + {}^4C_4 b^4\right) \\ &\quad – \left({}^4C_0 a^4 – {}^4C_1 a^3 b + {}^4C_2 a^2b^2 – {}^4C_3 ab^3 + {}^4C_4 b^4\right) \end{aligned} $ $ = 2\cdot{}^4C_1 a^3b + 2\cdot{}^4C_3 ab^3 = 2(4a^3b) + 2(4ab^3) $ $ (a+b)^4 – (a-b)^4 = 8ab(a^2+b^2) \quad \left[\because {}^4C_3 = {}^4C_1 = 4\right] $

Now substitute $a = \sqrt{3}$ and $b = \sqrt{2}$:

$ \begin{aligned} (\sqrt{3}+\sqrt{2})^4 – (\sqrt{3}-\sqrt{2})^4 &= 8\sqrt{3}\sqrt{2}(3+2) \\ &= 40\sqrt{6} \end{aligned} $

12. Find $(x+1)^6 + (x-1)^6$. Hence or otherwise evaluate $(\sqrt{2}+1)^6 + (\sqrt{2}-1)^6$.

Expand both terms using the Binomial Theorem. When we add the two expansions, all odd-powered terms cancel out:

$ \begin{aligned} (x+1)^6+(x-1)^6 &= \left[{}^6C_0 x^6 + {}^6C_1 x^5 + {}^6C_2 x^4 + {}^6C_3 x^3 + {}^6C_4 x^2 + {}^6C_5 x + {}^6C_6\right] \\ &\quad + \left[{}^6C_0 x^6 – {}^6C_1 x^5 + {}^6C_2 x^4 – {}^6C_3 x^3 + {}^6C_4 x^2 – {}^6C_5 x + {}^6C_6\right] \end{aligned} $ $ = 2\left({}^6C_0 x^6 + {}^6C_2 x^4 + {}^6C_4 x^2 + {}^6C_6\right) = 2\left(x^6 + 15x^4 + 15x^2 + 1\right) $ $ \left[\because {}^6C_6 = {}^6C_0 = 1,\ {}^6C_4 = {}^6C_2 = 15\right] $ $ \therefore (x+1)^6 + (x-1)^6 = 2\left(x^6 + 15x^4 + 15x^2 + 1\right) $

Substitute $x = \sqrt{2}$ to get the required value:

$ \begin{aligned} (\sqrt{2}+1)^6 + (\sqrt{2}-1)^6 &= 2\left[(\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1\right] \\ &= 2[8 + 15(4) + 15(2) + 1] \\ &= 2(99) = 198 \end{aligned} $

13. Show that $9^{n+1} – 8n – 9$ is divisible by 64, whenever $n$ is a positive integer.

Rewrite 9 as $1 + 8$ and expand $9^{n+1} = (1+8)^{n+1}$ using the Binomial Theorem:

$ 9^{n+1} = (1+8)^{n+1} $ $ = {}^{n+1}C_0 + {}^{n+1}C_1\cdot 8 + {}^{n+1}C_2\cdot 8^2 + {}^{n+1}C_3\cdot 8^3 + \ldots + {}^{n+1}C_{n+1}\cdot 8^{n+1} $ $ \left[\because (1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + {}^nC_3 x^3 \ldots + {}^nC_n x^n\right] $ $ \text{or} \quad 9^{n+1} = 1 + (n+1)\cdot 8 + {}^{n+1}C_2\cdot 8^2 + {}^{n+1}C_3\cdot 8^3 + \ldots + {}^{n+1}C_{n+1}\cdot 8^{n+1} $

Move the first two terms of the right-hand side to the left:

$ \begin{aligned} 9^{n+1} – 8n – 9 &= 8^2\left[{}^{n+1}C_2 + {}^{n+1}C_3\cdot 8 + \ldots + 8^{n-1}\right] \\ &= 64 \times \text{a positive integer} \end{aligned} $

$\therefore 9^{n+1} – 8n – 9$ is divisible by 64.

14. Prove that $\displaystyle\sum_{r=0}^n 3^r\, {}^nC_r = 4^n$.

Start with the standard binomial expansion for $(1+x)^n$ and substitute a specific value of $x$:

$ (1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + \ldots + {}^nC_r x^r + \ldots + {}^nC_n x^n $

Now put $x = 3$ in both sides of equation (i):

$ \begin{aligned} (1+3)^n = 4^n &= {}^nC_0 + {}^nC_1\cdot 3 + {}^nC_2\cdot 3^2 + \ldots + {}^nC_r\cdot 3^r + \ldots + {}^nC_n\cdot 3^n \\ \text{or}\quad 4^n &= \sum_{r=0}^n 3^r\cdot {}^nC_r \\ \text{or}\quad \sum_{r=0}^n {}^nC_r\cdot 3^r &= 4^n \end{aligned} $

Miscellaneous Exercise on Chapter 7

1. Find $a$, $b$ and $n$ in the expansion of $(a+b)^n$ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Set up equations using the general term formula. The first three terms give us:

$ \begin{aligned} T_1 = 729 &\Rightarrow {}^nC_0\, a^n = 729 \Rightarrow a^n = 729 \\ T_2 = 7290 &\Rightarrow {}^nC_1\, a^{n-1}b = 7290 \Rightarrow na^{n-1}b = 7290 \\ T_3 = 30375 &\Rightarrow {}^nC_2\, a^{n-2}b^2 = 30375 \Rightarrow \frac{n(n-1)}{2}a^{n-2}b^2 = 30375 \end{aligned} $ $\Rightarrow n(n-1)a^{n-2}b^2 = 60750$

Divide equation (ii) by equation (i) to find the ratio $\dfrac{b}{a}$:

$ \frac{na^{n-1}b}{a^n} = \frac{7290}{729} \Rightarrow \frac{nb}{a} = 10 \Rightarrow \frac{b}{a} = \frac{10}{n} \quad \cdots(iv) $

Divide equation (iii) by equation (ii):

$ \frac{n(n-1)a^{n-2}b^2}{na^{n-1}b} = \frac{60750}{7290} \Rightarrow \frac{b}{a} = \frac{25}{3(n-1)} \quad \cdots(v) $

Equate (iv) and (v) to find $n$:

$ \frac{10}{n} = \frac{25}{3(n-1)} \Rightarrow \frac{2}{n} = \frac{5}{3(n-1)} \Rightarrow 6(n-1) = 5n \Rightarrow n = 6 $

Substitute $n = 6$ in equation (i) to find $a$:

$ a^6 = 729 = 3^6 \Rightarrow a = 3 $

Substitute $n = 6$ and $a = 3$ in equation (ii) to find $b$:

$ 6 \times 3^5 \times b = 3^6 \times 10 = 3^5 \times 30 \Rightarrow b = 5 $

Hence $a = 3$, $b = 5$, $n = 6$.

2. Find $a$ if the coefficients of $x^2$ and $x^3$ in the expansion of $(3+ax)^9$ are equal.

Write the general term of the expansion and identify the coefficients of $x^2$ and $x^3$:

$ T_{r+1} = {}^9C_r \cdot 3^{9-r} \cdot (ax)^r = {}^9C_r \cdot 3^{9-r} a^r x^r $

The coefficient of $x^r$ is ${}^9C_r \cdot 3^{9-r} \cdot a^r$. Substituting $r = 2$ and $r = 3$:

$ \text{Coefficient of } x^2 = {}^9C_2 \cdot 3^7 a^2 $ $ \text{Coefficient of } x^3 = {}^9C_3 \cdot 3^6 a^3 $

Setting these equal and dividing both sides by $3^6 a^2$:

$ {}^9C_2 \cdot 3^7 a^2 = {}^9C_3 \cdot 3^6 a^3 $ $ \begin{aligned} &\Rightarrow \frac{9\times8}{2\times1}\times 3 = \frac{9\times8\times7}{3\times2\times1}\times a \\ &\Rightarrow 108 = 84a \Rightarrow a = \frac{108}{84} = \frac{9}{7} \end{aligned} $

3. Find the coefficient of $x^5$ in the product $(1+2x)^6(1-x)^7$ using binomial theorem.

First expand each factor completely using the Binomial Theorem:

$ \begin{gathered} (1+2x)^6 = \left[{}^6C_0 + {}^6C_1(2x) + {}^6C_2(2x)^2 + {}^6C_3(2x)^3 + {}^6C_4(2x)^4 + {}^6C_5(2x)^5 + {}^6C_6(2x)^6\right] \\ (1-x)^7 = \left[{}^7C_0 – {}^7C_1 x + {}^7C_2 x^2 – {}^7C_3 x^3 + {}^7C_4 x^4 – {}^7C_5 x^5 + {}^7C_6 x^6 – {}^7C_7 x^7\right] \end{gathered} $ $ \begin{gathered} \left[\because {}^6C_0=1,\ {}^6C_1=6,\ {}^6C_2=15,\ {}^6C_3=20,\ {}^7C_0=1,\ {}^7C_1=7,\ {}^7C_2=21,\ {}^7C_3=35\right] \\ = (1+12x+60x^2+160x^3+240x^4+192x^5+64x^6) \\ \times (1-7x+21x^2-35x^3+35x^4-21x^5+7x^6-x^7) \end{gathered} $

Now collect all terms whose product gives $x^5$:

$ \begin{aligned} &= (1)(-21x^5) + (12x)(35x^4) + (60x^2)(-35x^3) + (160x^3)(21x^2) \\ &\quad + (240x^4)(-7x) + (192x^5)(1) \\ &= (-21 + 420 – 2100 + 3360 – 1680 + 192)x^5 \\ &= 171x^5 \end{aligned} $

$\therefore$ The coefficient of $x^5$ is $\mathbf{171}$.

4. 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.

The key idea is to write $a = (a – b) + b$, so that $a^n$ becomes a binomial expression:

$ a = a – b + b \Rightarrow a^n = [(a-b)+b]^n $

Expand the right-hand side using the Binomial Theorem:

$ = {}^nC_0(a-b)^n + {}^nC_1(a-b)^{n-1}b + \ldots + {}^nC_{n-1}(a-b)b^{n-1} + {}^nC_n b^n $

Move the last term $b^n$ to the left-hand side:

$ \begin{aligned} a^n – b^n &= (a-b)^n + {}^nC_1(a-b)^{n-1}b + \ldots + {}^nC_{n-1}(a-b)b^{n-1} \\ &= (a-b)\left[(a-b)^{n-1} + {}^nC_1(a-b)^{n-2}b + \ldots + {}^nC_{n-1}b^{n-1}\right] \\ &= (a-b)(\text{an integer}) \end{aligned} $

$\therefore a^n – b^n$ is divisible by $a – b$. $\blacksquare$

5. Evaluate $(\sqrt{3}+\sqrt{2})^6 – (\sqrt{3}-\sqrt{2})^6$.

Substitute $a = \sqrt{3}$ and $b = \sqrt{2}$ so the expression becomes $(a+b)^6 – (a-b)^6$. Expand both terms: the even-powered terms cancel and we keep only the odd-powered terms:

$ (a+b)^6 – (a-b)^6 = 2\left({}^6C_1 a^5b + {}^6C_3 a^3b^3 + {}^6C_5 ab^5\right) $ $ = 2(6a^5b + 20a^3b^3 + 6ab^5) \quad \left[\because {}^6C_5 = {}^6C_1 = 6,\ {}^6C_3 = 20\right] $ $ = 4ab(3a^4 + 10a^2b^2 + 3b^4) $ $ \therefore (a+b)^6 – (a-b)^6 = 4ab(3a^4 + 10a^2b^2 + 3b^4) \quad \cdots(i) $

Now substitute $a = \sqrt{3}$ and $b = \sqrt{2}$ back into (i):

$ \begin{aligned} (\sqrt{3}+\sqrt{2})^6 – (\sqrt{3}-\sqrt{2})^6 &= 4\sqrt{3}\sqrt{2}\left[3(9) + 10(3)(2) + 3(4)\right] \\ &= 4\sqrt{6}(27 + 60 + 12) \\ &= 4\sqrt{6}(99) = 396\sqrt{6} \end{aligned} $

6. Find the value of $\left(a^2+\sqrt{a^2-1}\right)^4 + \left(a^2-\sqrt{a^2-1}\right)^4$.

Simplify by substituting $x = a^2$ and $y = \sqrt{a^2-1}$, so the expression becomes $(x+y)^4 + (x-y)^4$. Expand using the Binomial Theorem: the odd-powered terms cancel:

$ \begin{aligned} (x+y)^4 + (x-y)^4 &= 2\left[{}^4C_0 x^4 + {}^4C_2 x^2y^2 + {}^4C_4 y^4\right] \\ &= 2(x^4 + 6x^2y^2 + y^4) \end{aligned} $ $ \left[\because {}^4C_4 = {}^4C_0 = 1,\ {}^4C_2 = 6\right] $ $ \therefore (x+y)^4 + (x-y)^4 = 2(x^4 + 6x^2y^2 + y^4) \quad \cdots(i) $

Substitute back $x = a^2$ and $y = \sqrt{a^2-1}$:

$ \begin{aligned} \left(a^2+\sqrt{a^2-1}\right)^4 + \left(a^2-\sqrt{a^2-1}\right)^4 &= 2\left[a^8 + 6a^4(a^2-1) + (a^2-1)^2\right] \\ &= 2\left(a^8 + 6a^6 – 6a^4 + a^4 – 2a^2 + 1\right) \\ &= 2\left(a^8 + 6a^6 – 5a^4 – 2a^2 + 1\right) \\ &= 2a^8 + 12a^6 – 10a^4 – 4a^2 + 2 \end{aligned} $

7. Find an approximation of $(0.99)^5$ using the first three terms of its expansion.

Write $0.99 = 1 – 0.01$ so the expression becomes $(1-0.01)^5$. Keep only the first three terms since higher-order terms are negligible:

$ (0.99)^5 = (1-0.01)^5 $ $ \begin{aligned} &\approx {}^5C_0 – {}^5C_1(0.01) + {}^5C_2(0.01)^2 \\ &= 1 – 5(0.01) + 10(0.0001) \\ &\left[\because {}^5C_0 = 1,\ {}^5C_1 = 5,\ {}^5C_2 = 10\right] \\ &= 1 – 0.05 + 0.001 \\ &= 0.951 \end{aligned} $

$\therefore (0.99)^5$ is approximately equal to $\mathbf{0.951}$.

8. Find $n$, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\left(\sqrt[4]{2}+\dfrac{1}{\sqrt[4]{3}}\right)^n$ is $\sqrt{6}:1$.

Set $x = 2^{1/4}$ and $y = \dfrac{1}{3^{1/4}}$, so the expansion becomes $(x+y)^n$. Use the key fact that the $p$-th term from the end equals the $p$-th term from the beginning in $(y+x)^n$:

$ \frac{T_5 \text{ of }(x+y)^n}{T_5 \text{ of }(y+x)^n} = \sqrt{6} $ $ \begin{aligned} &\Rightarrow \frac{{}^nC_4\, x^{n-4}\, y^4}{{}^nC_4\, y^{n-4}\, x^4} = \sqrt{6} \\ &\Rightarrow \frac{x^{n-4-4}}{y^{n-4-4}} = \sqrt{6} \Rightarrow \left(\frac{x}{y}\right)^{n-8} = \sqrt{6} \end{aligned} $

Substitute back $x = 2^{1/4}$ and $y = \dfrac{1}{3^{1/4}}$:

$ \left(2^{1/4}\cdot 3^{1/4}\right)^{n-8} = \sqrt{6} \Rightarrow \left(6^{1/4}\right)^{n-8} = 6^{1/2} $ $ \Rightarrow 6^{\frac{n-8}{4}} = 6^{1/2} \Rightarrow \frac{n-8}{4} = \frac{1}{2} $

Cross-multiplying: $2n – 16 = 4 \Rightarrow 2n = 20 \Rightarrow \mathbf{n = 10}$.

9. Expand using Binomial Theorem $\left(1+\dfrac{x}{2}-\dfrac{2}{x}\right)^4$, $x \neq 0$.

Group the expression as $\left[1 + \left(\dfrac{x}{2}-\dfrac{2}{x}\right)\right]^4$ and treat $y = \dfrac{x}{2}-\dfrac{2}{x}$ as a single entity. Apply the Binomial Theorem $(1+y)^4$, then expand each power of $y$ using the Binomial Theorem again:

$ = {}^4C_0 + {}^4C_1\left(\frac{x}{2}-\frac{2}{x}\right) + {}^4C_2\left(\frac{x}{2}-\frac{2}{x}\right)^2 + {}^4C_3\left(\frac{x}{2}-\frac{2}{x}\right)^3 + {}^4C_4\left(\frac{x}{2}-\frac{2}{x}\right)^4 $ $ = 1 + 4\left(\frac{x}{2}-\frac{2}{x}\right) + 6\left[\frac{x^2}{4} – 2 + \frac{4}{x^2}\right] $ $ + 4\left[\frac{x^3}{8} – 6\left(\frac{x^2}{4}\right)\left(\frac{2}{x}\right) + 3\left(\frac{x}{2}\right)\left(\frac{4}{x^2}\right) – \frac{8}{x^3}\right] $ $ + \left[\frac{x^4}{16} – 4\left(\frac{x^3}{8}\right)\left(\frac{2}{x}\right) + 6\left(\frac{x^2}{4}\right)\left(\frac{4}{x^2}\right) – 4\left(\frac{x}{2}\right)\left(\frac{8}{x^3}\right) + \frac{16}{x^4}\right] $ $ = 1 + 2x – \frac{8}{x} + \frac{3}{2}x^2 – 12 + \frac{24}{x^2} + \frac{x^3}{2} – 6x + \frac{24}{x} – \frac{32}{x^3} + \frac{x^4}{16} – x^2 + 6 – \frac{16}{x^2} + \frac{16}{x^4} $

10. Find the expansion of $\left(3x^2 – 2ax + 3a^2\right)^3$ using binomial theorem.

Regroup the expression as $\left[(3x^2 – 2ax) + 3a^2\right]^3$ and apply the Binomial Theorem with $n = 3$, treating $(3x^2 – 2ax)$ as the first part:

$ = {}^3C_0(3x^2-2ax)^3 + {}^3C_1(3x^2-2ax)^2(3a^2)^1 + {}^3C_2(3x^2-2ax)^1(3a^2)^2 + {}^3C_3(3a^2)^3 $

Expand $(3x^2 – 2ax)^3$ using the Binomial Theorem, then collect all terms:

$ \begin{aligned} &= \left[27x^6 – 3(9x^4)(2ax) + 3(3x^2)(4a^2x^2) – 8a^3x^3\right] \\ &\quad + 3\left(9x^4 – 12ax^3 + 4a^2x^2\right)(3a^2) + 3\left(3x^2 – 2ax\right)(9a^4) + 27a^6 \\ &= 27x^6 – 54ax^5 + 36a^2x^4 – 8a^3x^3 \\ &\quad + 9a^2(9x^4 – 12ax^3 + 4a^2x^2) + 27a^4(3x^2 – 2ax) + 27a^6 \\ &= 27x^6 – 54ax^5 + 36a^2x^4 – 8a^3x^3 + 81a^2x^4 \\ &\quad – 108a^3x^3 + 36a^4x^2 + 81a^4x^2 – 54a^5x + 27a^6 \end{aligned} $ $ = 27x^6 – 54ax^5 + 117a^2x^4 – 116a^3x^3 + 117a^4x^2 – 54a^5x + 27a^6 $