Question 1

Q1: Expand each of the expressions in Exercises 1 to 5. 1. $(1-2 x)^{5}$

Accepted Answer

Given: The expression $(1-2x)^5$. To Find: The expansion of the given expression. Formula: The binomial theorem for $(a-b)^n$ is: $$(a-b)^n = \sum_{r=0}^{n} (-1)^r \,^nC_r a^{n-r} b^r$$ $$(a-b)^n = \,^nC_0 a^n - \,^nC_1 a^{n-1}b + \,^nC_2 a^{n-2}b^2 - \dots + (-1)^n \,^nC_n b^n$$ Solution: Here, $a=...

Question 2

Q10: Using Binomial Theorem, indicate which number is larger $(1.1)^{10000}$ or 1000.

Accepted Answer

Given: Two numbers, $(1.1)^{10000}$ and $1000$. To Find: Which of the two numbers is larger. Solution: We can write $1.1$ as $(1 + 0.1)$. So, $(1.1)^{10000} = (1 + 0.1)^{10000}$. Using the binomial theorem for $(1+x)^n = \,^nC_0 + \,^nC_1 x + \,^nC_2 x^2 + \dots + \,^nC_n x^n$. Here, $x=0.1$ and $n=...

Question 3

Q11: Find $(a+b)^{4}-(a-b)^{4}$. Hence, evaluate $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$.

Accepted Answer

Part 1: Find $(a+b)^{4}-(a-b)^{4}$ Solution: First, we expand $(a+b)^4$ and $(a-b)^4$ using the binomial theorem. $$(a+b)^4 = \,^4C_0 a^4 + \,^4C_1 a^3b + \,^4C_2 a^2b^2 + \,^4C_3 ab^3 + \,^4C_4 b^4$$ $$= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$ $$(a-b)^4 = \,^4C_0 a^4 - \,^4C_1 a^3b + \,^4C_2 a^2b^2 -...

Question 4

Q12: Find $(x+1)^{6}+(x-1)^{6}$. Hence or otherwise evaluate $(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}$.

Accepted Answer

Part 1: Find $(x+1)^{6}+(x-1)^{6}$ Solution: First, we expand $(x+1)^6$ and $(x-1)^6$ using the binomial theorem. $$(x+1)^6 = \,^6C_0 x^6 + \,^6C_1 x^5 + \,^6C_2 x^4 + \,^6C_3 x^3 + \,^6C_4 x^2 + \,^6C_5 x + \,^6C_6$$ $$(x-1)^6 = \,^6C_0 x^6 - \,^6C_1 x^5 + \,^6C_2 x^4 - \,^6C_3 x^3 + \,^6C_4 x^2 -...

Question 5

Q13: Show that $9^{n+1}-8 n-9$ is divisible by 64, whenever $n$ is a positive integer.

Accepted Answer

To Prove: $9^{n+1}-8n-9$ is divisible by 64 for any positive integer $n$. Proof: We can write $9$ as $(1+8)$. Therefore, $9^{n+1} = (1+8)^{n+1}$. Using the binomial theorem to expand $(1+8)^{n+1}$: $$(1+8)^{n+1} = \,^{n+1}C_0 (1)^{n+1} + \,^{n+1}C_1 (1)^n (8)^1 + \,^{n+1}C_2 (1)^{n-1} (8)^2 + \,^{n+...

Question 6

Q14: Prove that $\sum_{r=0}^{n} 3^{r}{ }^{n} \mathrm{C}_{r}=4^{n}$.

Accepted Answer

To Prove: $\sum_{r=0}^{n} 3^{r} \,^nC_r = 4^n$. Proof: We start with the general formula for the binomial theorem: $$(a+b)^n = \sum_{r=0}^{n} \,^nC_r a^{n-r} b^r$$ Let's compare the given summation with the general formula. The given summation is $\sum_{r=0}^{n} \,^nC_r 3^r$. We can write this as $\...

Question 7

Q2: $\left(\frac{2}{x}-\frac{x}{2}\right)^{5}$

Accepted Answer

Given: The expression $\left(\frac{2}{x}-\frac{x}{2}\right)^{5}$. To Find: The expansion of the given expression. Formula: The binomial theorem for $(a-b)^n$ is: $$(a-b)^n = \,^nC_0 a^n - \,^nC_1 a^{n-1}b + \,^nC_2 a^{n-2}b^2 - \dots + (-1)^n \,^nC_n b^n$$ Solution: Here, $a=\frac{2}{x}$, $b=\frac{x...

Question 8

Q3: $(2 x-3)^{6}$

Accepted Answer

Given: The expression $(2x-3)^6$. To Find: The expansion of the given expression. Formula: The binomial theorem for $(a-b)^n$ is: $$(a-b)^n = \sum_{r=0}^{n} (-1)^r \,^nC_r a^{n-r} b^r$$ Solution: Here, $a=2x$, $b=3$, and $n=6$. $$(2x-3)^6 = \,^6C_0(2x)^6 - \,^6C_1(2x)^5(3) + \,^6C_2(2x)^4(3)^2 - \,^...

Question 9

Q4: $\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$

Accepted Answer

Given: The expression $\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$. To Find: The expansion of the given expression. Formula: The binomial theorem for $(a+b)^n$ is: $$(a+b)^n = \sum_{r=0}^{n} \,^nC_r a^{n-r} b^r$$ Solution: Here, $a=\frac{x}{3}$, $b=\frac{1}{x}$, and $n=5$. $$\left(\frac{x}{3}+\frac{1}...

Question 10

Q5: $\left(x+\frac{1}{x}\right)^{6}$

Accepted Answer

Given: The expression $\left(x+\frac{1}{x}\right)^{6}$. To Find: The expansion of the given expression. Formula: The binomial theorem for $(a+b)^n$ is: $$(a+b)^n = \sum_{r=0}^{n} \,^nC_r a^{n-r} b^r$$ Solution: Here, $a=x$, $b=\frac{1}{x}$, and $n=6$. $$\left(x+\frac{1}{x}\right)^{6} = \,^6C_0(x)^6...

Question 11

Q6: Using binomial theorem, evaluate each of the following: 6. $(96)^{3}$

Accepted Answer

Given: The expression $(96)^3$. To Find: The value of the expression using the binomial theorem. Solution: First, we express 96 as a difference of two numbers whose powers are easy to calculate. We can write $96 = 100 - 4$. So, $(96)^3 = (100-4)^3$. Using the binomial expansion for $(a-b)^3 = \,^3C_...

Question 12

Q7: $(102)^{5}$

Accepted Answer

Given: The expression $(102)^5$. To Find: The value of the expression using the binomial theorem. Solution: First, we express 102 as a sum of two numbers. We can write $102 = 100 + 2$. So, $(102)^5 = (100+2)^5$. Using the binomial expansion for $(a+b)^5$: $$(a+b)^5 = \,^5C_0 a^5 + \,^5C_1 a^4b + \,^...

Question 13

Q8: $(101)^{4}$

Accepted Answer

Given: The expression $(101)^4$. To Find: The value of the expression using the binomial theorem. Solution: First, we express 101 as a sum. We can write $101 = 100 + 1$. So, $(101)^4 = (100+1)^4$. Using the binomial expansion for $(a+b)^4$: $$(a+b)^4 = \,^4C_0 a^4 + \,^4C_1 a^3b + \,^4C_2 a^2b^2 + \...

Question 14

Q9: $(99)^{5}$

Accepted Answer

Given: The expression $(99)^5$. To Find: The value of the expression using the binomial theorem. Solution: First, we express 99 as a difference. We can write $99 = 100 - 1$. So, $(99)^5 = (100-1)^5$. Using the binomial expansion for $(a-b)^5$: $$(a-b)^5 = \,^5C_0 a^5 - \,^5C_1 a^4b + \,^5C_2 a^3b^2...

Question 15

Q1: 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...

Accepted Answer

To Prove: $a-b$ is a factor of $a^n - b^n$ for any positive integer $n$. Proof: Following the hint, we write $a$ as $a = (a-b) + b$. Then, $a^n = ((a-b) + b)^n$. We expand this using the binomial theorem, treating $(a-b)$ as the first term and $b$ as the second term. $$a^n = \,^nC_0 (a-b)^n b^0 + \,...

Question 16

Q2: Evaluate $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$.

Accepted Answer

Given: The expression $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$. To Find: The value of the expression. Solution: Let $a = \sqrt{3}$ and $b = \sqrt{2}$. The expression becomes $(a+b)^6 - (a-b)^6$. Let's expand $(a+b)^6$ and $(a-b)^6$ using the binomial theorem. $$(a+b)^6 = \,^6C_0a^6 + \,^6C_...

Question 17

Q3: Find the value of $\left(a^{2}+\sqrt{a^{2}-1}\right)^{4}+\left(a^{2}-\sqrt{a^{2}-1}\right)^{4}$.

Accepted Answer

Given: The expression $\left(a^{2}+\sqrt{a^{2}-1}\right)^{4}+\left(a^{2}-\sqrt{a^{2}-1}\right)^{4}$. To Find: The value of the expression. Solution: Let $x = a^2$ and $y = \sqrt{a^2-1}$. The expression becomes $(x+y)^4 + (x-y)^4$. Let's expand $(x+y)^4$ and $(x-y)^4$ using the binomial theorem. $$(x...

Question 18

Q4: Find an approximation of $(0.99)^{5}$ using the first three terms of its expansion.

Accepted Answer

Given: The expression $(0.99)^5$. To Find: An approximation of its value using the first three terms of its binomial expansion. Solution: We can write $0.99$ as $(1 - 0.01)$. So, $(0.99)^5 = (1 - 0.01)^5$. Using the binomial theorem for $(1-x)^n = \,^nC_0 - \,^nC_1 x + \,^nC_2 x^2 - \dots$ Here, $x=...

Question 19

Q5: Expand using Binomial Theorem $\left(1+\frac{x}{2}-\frac{2}{x}ight)^{4}, x 
eq 0$.

Accepted Answer

Given: The expression $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}$. To Find: The expansion of the expression using the binomial theorem. Solution: This is a trinomial expansion. We can group the terms to apply the binomial theorem. Let $a = 1+\frac{x}{2}$ and $b = \frac{2}{x}$. The expression become...

Question 20

Q6: Find the expansion of $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$ using binomial theorem.

Accepted Answer

Given: The expression $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$. To Find: The expansion of the expression using the binomial theorem. Solution: We group the terms to apply the binomial theorem. Let $A = 3x^2 + 3a^2$ and $B = 2ax$. The expression becomes $(A-B)^3$. Using the formula for $(A-B)^3$: $$(...

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