Exercise 7.1 - Question 1

Question: Expand each of the expressions: $(1-2x)^5$

Solution:

Using the Binomial Theorem, the expansion of $(a + b)^n$ is given by:

Here, $a = 1$, $b = -2x$, and $n = 5$. Applying the theorem:

Now, compute the binomial coefficients and powers:

• $\binom{5}{0} = 1$, $1^5 = 1$, $(-2x)^0 = 1 \implies 1 \cdot 1 \cdot 1 = 1$
• $\binom{5}{1} = 5$, $1^4 = 1$, $(-2x)^1 = -2x \implies 5 \cdot 1 \cdot (-2x) = -10x$
• $\binom{5}{2} = 10$, $1^3 = 1$, $(-2x)^2 = 4x^2 \implies 10 \cdot 1 \cdot 4x^2 = 40x^2$
• $\binom{5}{3} = 10$, $1^2 = 1$, $(-2x)^3 = -8x^3 \implies 10 \cdot 1 \cdot (-8x^3) = -80x^3$
• $\binom{5}{4} = 5$, $1^1 = 1$, $(-2x)^4 = 16x^4 \implies 5 \cdot 1 \cdot 16x^4 = 80x^4$
• $\binom{5}{5} = 1$, $1^0 = 1$, $(-2x)^5 = -32x^5 \implies 1 \cdot 1 \cdot (-32x^5) = -32x^5$

Combining all the terms, we get the final expanded expression:

Exercise 7.1 - Question 2

Question: Expand each of the expressions: $\left(\frac{2}{x} - \frac{x}{2}\right)^5$

Solution:

Using the Binomial Theorem for $(a + b)^n$ where $a = \frac{2}{x}$, $b = -\frac{x}{2}$, and $n = 5$:

Let us write out each term explicitly for $k = 0, 1, 2, 3, 4, 5$:

• Term for $k = 0$: $$ \binom{5}{0} \left(\frac{2}{x}\right)^5 \left(-\frac{x}{2}\right)^0 = 1 \cdot \frac{32}{x^5} \cdot 1 = \frac{32}{x^5} $$
• Term for $k = 1$: $$ \binom{5}{1} \left(\frac{2}{x}\right)^4 \left(-\frac{x}{2}\right)^1 = 5 \cdot \frac{16}{x^4} \cdot \left(-\frac{x}{2}\right) = -\frac{40}{x^3} $$
• Term for $k = 2$: $$ \binom{5}{2} \left(\frac{2}{x}\right)^3 \left(-\frac{x}{2}\right)^2 = 10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4} = 10 \cdot \frac{2}{x} = \frac{20}{x} $$
• Term for $k = 3$: $$ \binom{5}{3} \left(\frac{2}{x}\right)^2 \left(-\frac{x}{2}\right)^3 = 10 \cdot \frac{4}{x^2} \cdot \left(-\frac{x^3}{8}\right) = 10 \cdot \left(-\frac{x}{2}\right) = -5x $$
• Term for $k = 4$: $$ \binom{5}{4} \left(\frac{2}{x}\right)^1 \left(-\frac{x}{2}\right)^4 = 5 \cdot \left(\frac{2}{x}\right) \cdot \frac{x^4}{16} = 5 \cdot \frac{x^3}{8} = \frac{5}{2}x^3 $$
• Term for $k = 5$: $$ \binom{5}{5} \left(\frac{2}{x}\right)^0 \left(-\frac{x}{2}\right)^5 = 1 \cdot 1 \cdot \left(-\frac{x^5}{32}\right) = -\frac{1}{32}x^5 $$

Combining all the terms, we get:

Exercise 7.1 - Question 3

Question: Expand each of the expressions: $(2x - 3)^6$

Solution:

Using the Binomial Theorem for $(a + b)^n$ where $a = 2x$, $b = -3$, and $n = 6$:

Let us write down each term for $k = 0, 1, 2, 3, 4, 5, 6$:

• Term for $k = 0$: $$ \binom{6}{0} (2x)^6 (-3)^0 = 1 \cdot 64x^6 \cdot 1 = 64x^6 $$
• Term for $k = 1$: $$ \binom{6}{1} (2x)^5 (-3)^1 = 6 \cdot 32x^5 \cdot (-3) = -576x^5 $$
• Term for $k = 2$: $$ \binom{6}{2} (2x)^4 (-3)^2 = 15 \cdot 16x^4 \cdot 9 = 2160x^4 $$
• Term for $k = 3$: $$ \binom{6}{3} (2x)^3 (-3)^3 = 20 \cdot 8x^3 \cdot (-27) = -4320x^3 $$
• Term for $k = 4$: $$ \binom{6}{4} (2x)^2 (-3)^4 = 15 \cdot 4x^2 \cdot 81 = 4860x^2 $$
• Term for $k = 5$: $$ \binom{6}{5} (2x)^1 (-3)^5 = 6 \cdot 2x \cdot (-243) = -2916x $$
• Term for $k = 6$: $$ \binom{6}{6} (2x)^0 (-3)^6 = 1 \cdot 1 \cdot 729 = 729 $$

Combining all the calculated terms, we get the expanded form:

Exercise 7.1 - Question 4

Question: Expand $\left(\frac{x}{3} + \frac{1}{x}\right)^5$ using the binomial theorem.

Solution:
Using the Binomial Theorem, the expansion of $(a + b)^n$ is given by: $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ Here, $a = \frac{x}{3}$, $b = \frac{1}{x}$, and $n = 5$. Applying the formula: $$\left(\frac{x}{3} + \frac{1}{x}\right)^5 = \binom{5}{0}\left(\frac{x}{3}\right)^5 + \binom{5}{1}\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right) + \binom{5}{2}\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2 + \binom{5}{3}\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3 + \binom{5}{4}\left(\frac{x}{3}\right)\left(\frac{1}{x}\right)^4 + \binom{5}{5}\left(\frac{1}{x}\right)^5$$ Now, calculating each binomial coefficient and term:

1. $\binom{5}{0} \left(\frac{x}{3}\right)^5 = 1 \cdot \frac{x^5}{243} = \frac{x^5}{243}$
2. $\binom{5}{1} \left(\frac{x}{3}\right)^4 \left(\frac{1}{x}\right) = 5 \cdot \frac{x^4}{81} \cdot \frac{1}{x} = \frac{5x^3}{81}$
3. $\binom{5}{2} \left(\frac{x}{3}\right)^3 \left(\frac{1}{x}\right)^2 = 10 \cdot \frac{x^3}{27} \cdot \frac{1}{x^2} = \frac{10}{27}x$
4. $\binom{5}{3} \left(\frac{x}{3}\right)^2 \left(\frac{1}{x}\right)^3 = 10 \cdot \frac{x^2}{9} \cdot \frac{1}{x^3} = \frac{10}{9x}$
5. $\binom{5}{4} \left(\frac{x}{3}\right) \left(\frac{1}{x}\right)^4 = 5 \cdot \frac{x}{3} \cdot \frac{1}{x^4} = \frac{5}{3x^3}$
6. $\binom{5}{5} \left(\frac{1}{x}\right)^5 = 1 \cdot \frac{1}{x^5} = \frac{1}{x^5}$

Exercise 7.1 - Question 5

Question: Expand $\left(x + \frac{1}{x}\right)^6$ using the binomial theorem.

Solution:
Using the Binomial Theorem, the expansion of $(a + b)^n$ is given by: $$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$ Here, $a = x$, $b = \frac{1}{x}$, and $n = 6$. Applying the formula: $$\left(x + \frac{1}{x}\right)^6 = \binom{6}{0}x^6 + \binom{6}{1}x^5\left(\frac{1}{x}\right) + \binom{6}{2}x^4\left(\frac{1}{x}\right)^2 + \binom{6}{3}x^3\left(\frac{1}{x}\right)^3 + \binom{6}{4}x^2\left(\frac{1}{x}\right)^4 + \binom{6}{5}x\left(\frac{1}{x}\right)^5 + \binom{6}{6}\left(\frac{1}{x}\right)^6$$ Now, calculating each term: 1. $\binom{6}{0}x^6 = 1 \cdot x^6 = x^6$ 2. $\binom{6}{1}x^5\left(\frac{1}{x}\right) = 6 \cdot x^4 = 6x^4$ 3. $\binom{6}{2}x^4\left(\frac{1}{x}\right)^2 = 15 \cdot x^2 = 15x^2$ 4. $\binom{6}{3}x^3\left(\frac{1}{x}\right)^3 = 20 \cdot 1 = 20$ 5. $\binom{6}{4}x^2\left(\frac{1}{x}\right)^4 = 15 \cdot \frac{1}{x^2} = \frac{15}{x^2}$ 6. $\binom{6}{5}x\left(\frac{1}{x}\right)^5 = 6 \cdot \frac{1}{x^4} = \frac{6}{x^4}$ 7. $\binom{6}{6}\left(\frac{1}{x}\right)^6 = 1 \cdot \frac{1}{x^6} = \frac{1}{x^6}$ Combining all terms, we get: $$x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$$

Exercise 7.1 - Question 6

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

Solution:
Express $96$ as a difference of two numbers whose powers are easy to compute: $$96 = 100 - 4$$ Now, apply the Binomial Theorem to $(100 - 4)^3$: $$(100 - 4)^3 = \binom{3}{0}(100)^3 - \binom{3}{1}(100)^2(4) + \binom{3}{2}(100)(4)^2 - \binom{3}{3}(4)^3$$ Calculating each term step-by-step: 1. $\binom{3}{0}(100)^3 = 1 \cdot 1000000 = 1000000$ 2. $\binom{3}{1}(100)^2(4) = 3 \cdot 10000 \cdot 4 = 120000$ 3. $\binom{3}{2}(100)(4)^2 = 3 \cdot 100 \cdot 16 = 4800$ 4. $\binom{3}{3}(4)^3 = 1 \cdot 64 = 64$ Combining the terms with their respective signs: $$1000000 - 120000 + 4800 - 64$$ $$= 880000 + 4800 - 64$$ $$= 884800 - 64$$ $$= 884736$$

Exercise 7.1 - Question 7

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

Solution:
Express $102$ as a sum of two numbers: $$102 = 100 + 2$$ Now, apply the Binomial Theorem to $(100 + 2)^5$: $$(100 + 2)^5 = \binom{5}{0}(100)^5 + \binom{5}{1}(100)^4(2) + \binom{5}{2}(100)^3(2)^2 + \binom{5}{3}(100)^2(2)^3 + \binom{5}{4}(100)(2)^4 + \binom{5}{5}(2)^5$$ Calculating each term step-by-step: 1. $\binom{5}{0}(100)^5 = 1 \cdot 10000000000 = 10000000000$ 2. $\binom{5}{1}(100)^4(2) = 5 \cdot 100000000 \cdot 2 = 1000000000$ 3. $\binom{5}{2}(100)^3(2)^2 = 10 \cdot 1000000 \cdot 4 = 40000000$ 4. $\binom{5}{3}(100)^2(2)^3 = 10 \cdot 10000 \cdot 8 = 800000$ 5. $\binom{5}{4}(100)(2)^4 = 5 \cdot 100 \cdot 16 = 8000$ 6. $\binom{5}{5}(2)^5 = 1 \cdot 32 = 32$ Summing all the calculated values: $$10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32$$ $$= 11000000000 + 40000000 + 800000 + 8032$$ $$= 11040000000 + 808032$$ $$= 11040808032$$

Exercise 7.1 - Question 8

Question: Evaluate $(101)^4$ using Binomial Theorem.

Solution: We can express $101$ as $(100 + 1)$. Therefore, we need to evaluate $(100 + 1)^4$.
Using the Binomial Theorem expansion for $(x + y)^n$: $$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$$ Substitute $x = 100$, $y = 1$, and $n = 4$: $$(100 + 1)^4 = \binom{4}{0}(100)^4(1)^0 + \binom{4}{1}(100)^3(1)^1 + \binom{4}{2}(100)^2(1)^2 + \binom{4}{3}(100)^1(1)^3 + \binom{4}{4}(100)^0(1)^4$$ Calculating the binomial coefficients: $$\binom{4}{0} = 1, \quad \binom{4}{1} = 4, \quad \binom{4}{2} = 6, \quad \binom{4}{3} = 4, \quad \binom{4}{4} = 1$$ Substituting these values: $$(101)^4 = 1(100000000) + 4(1000000) + 6(10000) + 4(100) + 1(1)$$ $$= 100000000 + 4000000 + 60000 + 400 + 1$$ $$= 104060401$$

Exercise 7.1 - Question 9

Question: Evaluate $(99)^5$ using Binomial Theorem.

Solution: We can express $99$ as $(100 - 1)$. Therefore, we need to evaluate $(100 - 1)^5$.
Using the Binomial Theorem expansion for $(x - y)^n$: $$(x - y)^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{n-k} y^k$$ Substitute $x = 100$, $y = 1$, and $n = 5$: $$(100 - 1)^5 = \binom{5}{0}(100)^5 - \binom{5}{1}(100)^4(1) + \binom{5}{2}(100)^3(1)^2 - \binom{5}{3}(100)^2(1)^3 + \binom{5}{4}(100)(1)^4 - \binom{5}{5}(1)^5$$ Calculating the binomial coefficients: $$\binom{5}{0} = 1, \quad \binom{5}{1} = 5, \quad \binom{5}{2} = 10, \quad \binom{5}{3} = 10, \quad \binom{5}{4} = 5, \quad \binom{5}{5} = 1$$ Substituting these values: $$(99)^5 = 1(10000000000) - 5(100000000) + 10(1000000) - 10(10000) + 5(100) - 1$$ $$= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1$$ $$= 10010000500 - 500100001$$ $$= 9509900499$$

Exercise 7.1 - Question 10

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

Solution: We can write $(1.1)^{10000}$ as $(1 + 0.1)^{10000}$.
Using the Binomial Theorem expansion: $$(1 + 0.1)^{10000} = \binom{10000}{0}(1)^{10000} + \binom{10000}{1}(1)^{9999}(0.1) + \text{remaining positive terms}$$ Expanding the first few terms: $$(1 + 10000 \times 0.1 + \text{other positive terms})$$ $$= 1 + 1000 + \text{other positive terms}$$ Since all terms in the binomial expansion of $(1 + 0.1)^{10000}$ are positive, we have: $$(1.1)^{10000} = 1001 + \text{positive terms}$$ Clearly, $1001 + \text{positive terms} > 1000$.
Therefore, $$(1.1)^{10000} > 1000$$

Exercise 7.1 - Question 11

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

Solution: Using the Binomial Theorem, expand $(a + b)^4$ and $(a - b)^4$:
$$(a + b)^4 = \binom{4}{0}a^4 + \binom{4}{1}a^3b + \binom{4}{2}a^2b^2 + \binom{4}{3}ab^3 + \binom{4}{4}b^4$$ $$= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$ Similarly, for $(a - b)^4$: $$(a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$$ Now, subtracting the second equation from the first: $$(a + b)^4 - (a - b)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) - (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4)$$ $$= 8a^3b + 8ab^3 = 8(a^3b + ab^3) = 8ab(a^2 + b^2)$$ To evaluate $(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4$, substitute $a = \sqrt{3}$ and $b = \sqrt{2}$ into the identity above: $$8(\sqrt{3})(\sqrt{2})\left((\sqrt{3})^2 + (\sqrt{2})^2\right)$$ $$= 8\sqrt{6}(3 + 2)$$ $$= 8\sqrt{6}(5)$$ $$= 40\sqrt{6}$$

Exercise 7.1 - Question 12

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

Solution:

Using the Binomial Theorem, we expand $(x + 1)^6$ and $(x - 1)^6$:

Adding the two expressions, the odd powers of $x$ cancel out:

Evaluating the binomial coefficients:

• ${^6C_0} = 1$
• ${^6C_2} = \frac{6 \times 5}{2 \times 1} = 15$
• ${^6C_4} = {^6C_2} = 15$
• ${^6C_6} = 1$

Substituting these values back:

Hence evaluation: To evaluate $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$, substitute $x = \sqrt{2}$ into the derived result:

Exercise 7.1 - Question 13

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

Solution:

We can write $9^{n+1}$ as $(1 + 8)^{n+1}$. Using the Binomial Theorem:

Since ${^{n+1}C_0} = 1$ and ${^{n+1}C_1} = n + 1$:

Rearranging the terms:

This shows that $9^{n+1} - 8n - 9$ is a multiple of $64$, and hence is divisible by $64$ for all positive integers $n$.

Exercise 7.1 - Question 14

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

Solution:

Consider the binomial expansion of $(1 + x)^n$:

Substitute $x = 3$ into both sides of the identity:

Thus, $\sum_{r=0}^{n} 3^r \, {^nC_r} = 4^n$. Hence proved.

Miscellaneous Exercise - Question 1

Question: 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]

Solution:

Following the hint, let us express $a$ as $(a - b + b)$ and expand $a^n$ using the Binomial Theorem:

Using the binomial expansion for $[(a - b) + b]^n$:

Since ${^nC_0} = 1$ and ${^nC_n} = 1$:

Subtracting $b^n$ from both sides:

Notice that every term on the right-hand side contains $(a - b)$ as a common factor. Factoring out $(a - b)$:

Let the expression inside the bracket be an integer $k$. Then:

This proves that $(a - b)$ is a factor of $a^n - b^n$ for any positive integer $n$.

Miscellaneous Exercise - Question 2

Question: Evaluate $(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6$.

Solution:
Let $x = \sqrt{3}$ and $y = \sqrt{2}$. We need to evaluate: $$(x + y)^6 - (x - y)^6$$ Using the binomial expansion for $(x + y)^6$ and $(x - y)^6$: $$(x + y)^6 = \binom{6}{0}x^6 + \binom{6}{1}x^5y + \binom{6}{2}x^4y^2 + \binom{6}{3}x^3y^3 + \binom{6}{4}x^2y^4 + \binom{6}{5}xy^5 + \binom{6}{6}y^6$$ $$(x - y)^6 = \binom{6}{0}x^6 - \binom{6}{1}x^5y + \binom{6}{2}x^4y^2 - \binom{6}{3}x^3y^3 + \binom{6}{4}x^2y^4 - \binom{6}{5}xy^5 + \binom{6}{6}y^6$$ Subtracting the second expression from the first: $$(x + y)^6 - (x - y)^6 = 2 \left[ \binom{6}{1}x^5y + \binom{6}{3}x^3y^3 + \binom{6}{5}xy^5 \right]$$ Substitute the binomial coefficients:

• $\binom{6}{1} = 6$
• $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$
• $\binom{6}{5} = 6$

• $xy = \sqrt{3} \times \sqrt{2} = \sqrt{6}$
• $x^5 = (\sqrt{3})^5 = 9\sqrt{3}$
• $6x^5y = 6(9\sqrt{3})(\sqrt{2}) = 54\sqrt{6}$
• $x^3 = 3\sqrt{3},\; y^3 = 2\sqrt{2} \implies 20x^3y^3 = 20(3\sqrt{3})(2\sqrt{2}) = 120\sqrt{6}$
• $x = \sqrt{3},\; y^5 = 4\sqrt{2} \implies 6xy^5 = 6(\sqrt{3})(4\sqrt{2}) = 24\sqrt{6}$

Miscellaneous Exercise - Question 3

Question: Find the value of $(a^2 + \sqrt{a^2 - 1})^4 + (a^2 - \sqrt{a^2 - 1})^4$.

Solution: Let $x = a^2$ and $y = \sqrt{a^2 - 1}$. We need to evaluate: $$(x + y)^4 + (x - y)^4$$ Using the binomial expansion: $$(x + y)^4 = \binom{4}{0}x^4 + \binom{4}{1}x^3y + \binom{4}{2}x^2y^2 + \binom{4}{3}xy^3 + \binom{4}{4}y^4$$ $$(x - y)^4 = \binom{4}{0}x^4 - \binom{4}{1}x^3y + \binom{4}{2}x^2y^2 - \binom{4}{3}xy^3 + \binom{4}{4}y^4$$ Adding the two equations: $$(x + y)^4 + (x - y)^4 = 2 \left[ \binom{4}{0}x^4 + \binom{4}{2}x^2y^2 + \binom{4}{4}y^4 \right]$$ Since $\binom{4}{0} = 1$, $\binom{4}{2} = 6$, and $\binom{4}{4} = 1$: $$= 2 \left[ x^4 + 6x^2y^2 + y^4 \right]$$ Now substitute back $x = a^2$ and $y^2 = a^2 - 1$: - $x^4 = (a^2)^4 = a^8$ - $x^2y^2 = (a^2)^2 (a^2 - 1) = a^4(a^2 - 1) = a^6 - a^4$ - $y^4 = (a^2 - 1)^2 = a^4 - 2a^2 + 1$ Substitute these into the expression: $$= 2 \left[ a^8 + 6(a^6 - a^4) + (a^4 - 2a^2 + 1) \right]$$ $$= 2 \left[ a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1 \right]$$ $$= 2 \left[ a^8 + 6a^6 - 5a^4 - 2a^2 + 1 \right]$$ $$= 2a^8 + 12a^6 - 10a^4 - 4a^2 + 2$$

Miscellaneous Exercise - Question 4

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

Solution: We can write $(0.99)^5$ as: $$(1 - 0.01)^5$$ Using the Binomial Theorem for $(1 + x)^n$ where $x = -0.01$ and $n = 5$: $$(1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots$$ Taking the first three terms: $$(1 - 0.01)^5 \approx \binom{5}{0}(1)^5 + \binom{5}{1}(1)^4(-0.01) + \binom{5}{2}(1)^3(-0.01)^2$$ Calculate each term: 1. $\binom{5}{0}(1)^5 = 1 \times 1 = 1$ 2. $\binom{5}{1}(-0.01) = 5 \times (-0.01) = -0.05$ 3. $\binom{5}{2}(-0.01)^2 = 10 \times (0.0001) = 0.0010$ Summing these up: $$(0.99)^5 \approx 1 - 0.05 + 0.0010$$ $$= 0.95 + 0.0010 = 0.9510$$

Miscellaneous Exercise - Question 5

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

Solution: Let $a = 1$ and $b = \left(\frac{x}{2} - \frac{2}{x}\right)$. We apply the binomial expansion for $(a + b)^4$: $$\left(1 + \left(\frac{x}{2} - \frac{2}{x}\right)\right)^4 = \sum_{k=0}^{4} \binom{4}{k} (1)^{4-k} \left(\frac{x}{2} - \frac{2}{x}\right)^k$$ Let's compute each term for $k = 0, 1, 2, 3, 4$: 1. **For $k = 0$:** $$\binom{4}{0} \left(\frac{x}{2} - \frac{2}{x}\right)^0 = 1 \times 1 = 1$$ 2. **For $k = 1$:** $$\binom{4}{1} \left(\frac{x}{2} - \frac{2}{x}\right)^1 = 4 \left(\frac{x}{2} - \frac{2}{x}\right) = 2x - \frac{8}{x}$$ 3. **For $k = 2$:** $$\binom{4}{2} \left(\frac{x}{2} - \frac{2}{x}\right)^2 = 6 \left(\frac{x^2}{4} - 2\left(\frac{x}{2}\right)\left(\frac{2}{x}\right) + \frac{4}{x^2}\right)$$ $$= 6 \left(\frac{x^2}{4} - 2 + \frac{4}{x^2}\right) = \frac{3}{2}x^2 - 12 + \frac{24}{x^2}$$ 4. **For $k = 3$:** $$\binom{4}{3} \left(\frac{x}{2} - \frac{2}{x}\right)^3 = 4 \left[ \left(\frac{x}{2}\right)^3 - 3\left(\frac{x}{2}\right)^2\left(\frac{2}{x}\right) + 3\left(\frac{x}{2}\right)\left(\frac{2}{x}\right)^2 - \left(\frac{2}{x}\right)^3 \right]$$ $$= 4 \left[ \frac{x^3}{8} - \frac{3x}{2} + \frac{6}{x} - \frac{8}{x^3} \right] = \frac{x^3}{2} - 6x + \frac{24}{x} - \frac{32}{x^3}$$ 5. **For $k = 4$:** $$\binom{4}{4} \left(\frac{x}{2} - \frac{2}{x}\right)^4 = 1 \cdot \left( \frac{x^4}{16} - 4\left(\frac{x^3}{8}\right)\left(\frac{2}{x}\right) + 6\left(\frac{x^2}{4}\right)\left(\frac{4}{x^2}\right) - 4\left(\frac{x}{2}\right)\left(\frac{8}{x^3}\right) + \frac{16}{x^4} \right)$$ $$= \frac{x^4}{16} - x^2 + 6 - \frac{16}{x^2} + \frac{16}{x^4}$$ Now, group like powers of $x$: - **Constant terms:** $1 - 12 + 6 = -5$ - **Terms with $x$:** $2x - 6x = -4x$ - **Terms with $x^2$:** $\frac{3}{2}x^2 - x^2 = \frac{x^2}{2}$ - **Terms with $x^3$:** $\frac{x^3}{2}$ - **Terms with $x^4$:** $\frac{x^4}{16}$ - **Terms with $\frac{1}{x}$:** $-\frac{8}{x} + \frac{24}{x} = \frac{16}{x}$ - **Terms with $\frac{1}{x^2}$:** $\frac{24}{x^2} - \frac{16}{x^2} = \frac{8}{x^2}$ - **Terms with $\frac{1}{x^3}$:** $-\frac{32}{x^3}$ - **Terms with $\frac{1}{x^4}$:** $\frac{16}{x^4}$ Combining all terms in standard order: $$\frac{16}{x} + \frac{8}{x^2} - \frac{32}{x^3} + \frac{16}{x^4} - 4x + \frac{x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} - 5$$

Miscellaneous Exercise - Question 6

Question: Find the expansion of $(3x^2 - 2ax + 3a^2)^3$ using binomial theorem.

Solution:

To find the expansion of $(3x^2 - 2ax + 3a^2)^3$, we can group the terms inside the parenthesis to apply the binomial theorem of power $3$.

Let us group the expression as follows:

Now, we apply the binomial expansion formula for $(A - B)^3$ where $A = (3x^2 + 3a^2)$ and $B = 2ax$:

Substitute $A = 3x^2 + 3a^2$ and $B = 2ax$ into the expansion:

Let us simplify each term step by step:

Term 1: $(3x^2 + 3a^2)^3$

Term 2: $-3(3x^2 + 3a^2)^2(2ax)$

Term 3: $3(3x^2 + 3a^2)(2ax)^2$

Term 4: $-(2ax)^3$

Now, combine all the expanded terms together:

Grouping the terms by powers of $x$:

• Degree $6$: $27x^6 + 27a^6$
• Degree $5$: $-54ax^5$
• Degree $4$: $81a^2x^4 + 36a^2x^4 = 117a^2x^4$
• Degree $3$: $-108a^3x^3 - 8a^3x^3 = -116a^3x^3$
• Degree $2$: $81a^4x^2 + 36a^4x^2 = 117a^4x^2$
• Degree $1$: $-54a^5x$

Arranging in descending powers of $x$: