OP Malhotra Class 11 Maths Solutions Chapter 13 Binomial Theorem Exercise 13 (A)

S Chand Class 11 ICSE Maths Solutions Chapter 13 Binomial Theorem Ex 13(a)

Question 1. What is the number of terms in the expansion of each of the following?
(i) \([(x-2y)^3]^3\)
(ii) \((5a + 7b)^8\)
(iii) \(\left(6 x-\frac{1}{x^3}\right)^{17}\)
(iv) \((4x^2 + 12xy + 9y^2)^9\)
(v) \((3 + 2\sqrt{5})^{18} - (3 – 2\sqrt{5})^{18}\)
(vi) \((5 + 7x)^{15} + (5 – 7x)^{15}\)
(vii) \((a + bx)^{17} – (a – bx)^{17}\)
Answer:
(i) For \([(x-2y)^3]^3\), which simplifies to \((x-2y)^9\), the power is 9. The number of terms in an expansion of \((A+B)^n\) is \(n+1\).
So, the number of terms is \(9 + 1 = 10\).
(ii) For \((5a + 7b)^8\), the power is 8.
So, the number of terms is \(8 + 1 = 9\).
(iii) For \(\left(6 x-\frac{1}{x^3}\right)^{17}\), the power is 17.
So, the number of terms is \(17 + 1 = 18\).
(iv) For \((4x^2 + 12xy + 9y^2)^9\), this can be rewritten. We know that \(4x^2 + 12xy + 9y^2 = (2x + 3y)^2\).
So, the expression becomes \([(2x + 3y)^2]^9 = (2x + 3y)^{18}\). The power is 18.
Therefore, the number of terms is \(18 + 1 = 19\).
(v) For \((A+B)^n - (A-B)^n\), if 'n' is even, the number of terms is \(\frac{n}{2}\). Here, \(n=18\) (even).
So, the number of terms in \((3 + 2\sqrt{5})^{18} - (3 – 2\sqrt{5})^{18}\) is \(\frac{18}{2} = 9\).
(vi) For \((A+B)^n + (A-B)^n\), if 'n' is odd, the number of terms is \(\frac{n+1}{2}\). Here, \(n=15\) (odd).
So, the number of terms in \((5 + 7x)^{15} + (5 – 7x)^{15}\) is \(\frac{15+1}{2} = \frac{16}{2} = 8\).
(vii) For \((A+B)^n - (A-B)^n\), if 'n' is odd, the number of terms is \(\frac{n+1}{2}\). Here, \(n=17\) (odd).
So, the number of terms in \((a + bx)^{17} – (a – bx)^{17}\) is \(\frac{17+1}{2} = \frac{18}{2} = 9\).
In simple words: When you expand a binomial like \((a+b)^n\), there are always \(n+1\) terms. If you have a sum or difference of two binomials like \((A+B)^n + (A-B)^n\) or \((A+B)^n - (A-B)^n\), the number of terms changes based on whether 'n' is odd or even. For example, if 'n' is odd and you have a sum, it's \(\frac{n+1}{2}\) terms.

🎯 Exam Tip: Remember to simplify the base of the binomial first if it's a perfect square (like in part iv) before applying the formula for the number of terms. Also, pay close attention to whether the exponent 'n' is odd or even for sum/difference expansions.

Question 2. Write out the expansions of the following:
(a) \((3x - y)^4\)
(b) \((3 + 2x^2)^4\)
(c) \(\left(x-\frac{y}{2}\right)^4\)
(d) \(\left(2 x+\frac{y}{2}\right)^5\)
(e) \((1 + 2x)^7\)
(f) \(\left(\frac{2}{x}-\frac{x}{2}\right)^5, x ≠ 0\)
Answer: We use the binomial theorem: \((x + a)^n = ^nC_0 x^n a^0 + ^nC_1 x^{n-1} a^1 + ^nC_2 x^{n-2} a^2 + ...... + ^nC_n x^0 a^n\). This formula helps us expand expressions with a power 'n'.
(a) \((3x-y)^4 = ^4C_0 (3x)^4 (-y)^0 + ^4C_1 (3x)^3 (-y)^1 + ^4C_2 (3x)^2 (-y)^2 + ^4C_3 (3x)^1 (-y)^3 + ^4C_4 (3x)^0 (-y)^4\)
\( = 1 \times 81x^4 \times 1 + 4 \times 27x^3 (-y) + \frac{4 \times 3}{2} \times 9x^2 y^2 + 4 \times 3x (-y^3) + 1 \times 1 \times y^4\)
\( = 81x^4 - 108x^3y + 54x^2y^2 - 12xy^3 + y^4\)
(b) \((3 + 2x^2)^4 = ^4C_0 3^4 (2x^2)^0 + ^4C_1 3^3 (2x^2)^1 + ^4C_2 3^2 (2x^2)^2 + ^4C_3 3^1 (2x^2)^3 + ^4C_4 3^0 (2x^2)^4\)
\( = 1 \times 81 \times 1 + 4 \times 27 \times 2x^2 + 6 \times 9 \times 4x^4 + 4 \times 3 \times 8x^6 + 1 \times 1 \times 16x^8\)
\( = 81 + 216x^2 + 216x^4 + 96x^6 + 16x^8\)
(c) \(\left(x-\frac{y}{2}\right)^4 = ^4C_0 x^4 \left(-\frac{y}{2}\right)^0 + ^4C_1 x^3 \left(-\frac{y}{2}\right)^1 + ^4C_2 x^2 \left(-\frac{y}{2}\right)^2 + ^4C_3 x^1 \left(-\frac{y}{2}\right)^3 + ^4C_4 x^0 \left(-\frac{y}{2}\right)^4\)
\( = x^4 \times 1 + 4x^3 \left(-\frac{y}{2}\right) + 6x^2 \left(\frac{y^2}{4}\right) + 4x \left(-\frac{y^3}{8}\right) + 1 \times 1 \times \frac{y^4}{16}\)
\( = x^4 - 2x^3y + \frac{3}{2}x^2y^2 - \frac{1}{2}xy^3 + \frac{1}{16}y^4\)
(d) \(\left(2 x+\frac{y}{2}\right)^5 = ^5C_0 (2x)^5 \left(\frac{y}{2}\right)^0 + ^5C_1 (2x)^4 \left(\frac{y}{2}\right)^1 + ^5C_2 (2x)^3 \left(\frac{y}{2}\right)^2 + ^5C_3 (2x)^2 \left(\frac{y}{2}\right)^3 + ^5C_4 (2x)^1 \left(\frac{y}{2}\right)^4 + ^5C_5 (2x)^0 \left(\frac{y}{2}\right)^5\)
\( = 32x^5 + 5 \times 16x^4 \times \frac{y}{2} + 10 \times 8x^3 \times \frac{y^2}{4} + 10 \times 4x^2 \times \frac{y^3}{8} + 5 \times 2x \times \frac{y^4}{16} + 1 \times 1 \times \frac{y^5}{32}\)
\( = 32x^5 + 40x^4y + 20x^3y^2 + 5x^2y^3 + \frac{5}{8}xy^4 + \frac{1}{32}y^5\)
(e) \((1 + 2x)^7 = ^7C_0 1^7 (2x)^0 + ^7C_1 1^6 (2x)^1 + ^7C_2 1^5 (2x)^2 + ^7C_3 1^4 (2x)^3 + ^7C_4 1^3 (2x)^4 + ^7C_5 1^2 (2x)^5 + ^7C_6 1^1 (2x)^6 + ^7C_7 1^0 (2x)^7\)
\( = 1 + 7 \times 1 \times 2x + \frac{7 \times 6}{2} \times 1 \times 4x^2 + \frac{7 \times 6 \times 5}{6} \times 1 \times 8x^3 + \frac{7 \times 6 \times 5}{6} \times 1 \times 16x^4 + \frac{7 \times 6}{2} \times 1 \times 32x^5 + 7 \times 1 \times 64x^6 + 1 \times 1 \times 128x^7\)
\( = 1 + 14x + 84x^2 + 280x^3 + 560x^4 + 682x^5 + 448x^6 + 128x^7\)
(f) \(\left(\frac{2}{x}-\frac{x}{2}\right)^5 = ^5C_0 \left(\frac{2}{x}\right)^5 \left(-\frac{x}{2}\right)^0 + ^5C_1 \left(\frac{2}{x}\right)^4 \left(-\frac{x}{2}\right)^1 + ^5C_2 \left(\frac{2}{x}\right)^3 \left(-\frac{x}{2}\right)^2 + ^5C_3 \left(\frac{2}{x}\right)^2 \left(-\frac{x}{2}\right)^3 + ^5C_4 \left(\frac{2}{x}\right)^1 \left(-\frac{x}{2}\right)^4 + ^5C_5 \left(\frac{2}{x}\right)^0 \left(-\frac{x}{2}\right)^5\)
\( = \frac{32}{x^5} - 5 \times \frac{16}{x^4} \times \frac{x}{2} + 10 \times \frac{8}{x^3} \times \frac{x^2}{4} - 10 \times \frac{4}{x^2} \times \frac{x^3}{8} + 5 \times \frac{2}{x} \times \frac{x^4}{16} - \frac{x^5}{32}\)
\( = \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5}{8}x^3 - \frac{1}{32}x^5\)
In simple words: To expand these, we use the binomial theorem. It breaks down the power into a sum of terms. Each term follows a pattern using combinations \((^nC_r)\), the first part of the binomial, and the second part. The powers of the first part decrease while the powers of the second part increase. Remember to handle negative signs and fractions carefully.

🎯 Exam Tip: When expanding, always calculate the binomial coefficients \((^nC_r)\) correctly first. Also, pay close attention to the signs, especially when one of the terms in the binomial is negative, as the signs will alternate.

Question 4. Expand \([(x + y)^5 + (x - y)^5]\) and hence find the value of \([(\sqrt{3} + 1)^5 – (\sqrt{3} – 1)^5]\).
Answer: We use the binomial theorem to expand \((x+y)^5\) and \((x-y)^5\).
\((x + y)^5 = ^5C_0 x^5 y^0 + ^5C_1 x^4 y^1 + ^5C_2 x^3 y^2 + ^5C_3 x^2 y^3 + ^5C_4 x y^4 + ^5C_5 x^0 y^5\)
\( = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5\) ...(1)
\((x - y)^5 = ^5C_0 x^5 (-y)^0 + ^5C_1 x^4 (-y)^1 + ^5C_2 x^3 (-y)^2 + ^5C_3 x^2 (-y)^3 + ^5C_4 x (-y)^4 + ^5C_5 x^0 (-y)^5\)
\( = x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5\) ...(2)
Now, we add equations (1) and (2) for the first part of the question:
\((x + y)^5 + (x - y)^5 = (x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5) + (x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5)\)
\( = 2x^5 + 20x^3y^2 + 10xy^4\) ...(3)

Next, we subtract equation (2) from equation (1) for the second part of the question:
\((x + y)^5 – (x - y)^5 = (x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5) - (x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5)\)
\( = 10x^4y + 20x^2y^3 + 2y^5\) ...(4)

To find the value of \([(\sqrt{3} + 1)^5 – (\sqrt{3} – 1)^5]\), we substitute \(x = \sqrt{3}\) and \(y = 1\) into equation (4).
\((\sqrt{3} + 1)^5 – (\sqrt{3} – 1)^5 = 10(\sqrt{3})^4 (1) + 20(\sqrt{3})^2 (1)^3 + 2(1)^5\)
\( = 10 \times 9 \times 1 + 20 \times 3 \times 1 + 2 \times 1\)
\( = 90 + 60 + 2\)
\( = 152\)
In simple words: First, we expand \((x+y)^5\) and \((x-y)^5\) using the binomial theorem. When we subtract these two expansions, many terms cancel out, leaving a simpler expression. Then, we put \(x = \sqrt{3}\) and \(y = 1\) into this simplified expression to find the final numerical answer.

🎯 Exam Tip: When dealing with sums or differences of binomial expansions, identify and cancel out the terms that have opposite signs. This simplifies the expression and reduces the chances of calculation errors. Ensure you substitute the values of x and y correctly into the simplified form.

Question 5. Evaluate the following :
(i) \((2 + \sqrt{5})^5 + (2 – \sqrt{5})^5\)
(ii) \(( \sqrt{3} + 1)^5 – ( \sqrt{3} – 1)^5\)
Hence, show in (ii), without using tables, that the value of \(( \sqrt{3} + 1)^5\) lies between 152 and 153.
Answer: We will use the expanded form of \((x+y)^5 + (x-y)^5 = 2x^5 + 20x^3y^2 + 10xy^4\) from Question 4, equation (3), and \((x+y)^5 - (x-y)^5 = 10x^4y + 20x^2y^3 + 2y^5\) from Question 4, equation (4).
(i) For \((2 + \sqrt{5})^5 + (2 – \sqrt{5})^5\), we put \(x = 2\) and \(y = \sqrt{5}\) into equation (3):
\((2 + \sqrt{5})^5 + (2 – \sqrt{5})^5 = 2(2)^5 + 20(2)^3(\sqrt{5})^2 + 10(2)(\sqrt{5})^4\)
\( = 2 \times 32 + 20 \times 8 \times 5 + 20 \times 25\)
\( = 64 + 800 + 500\)
\( = 1364\)
(ii) For \(( \sqrt{3} + 1)^5 – ( \sqrt{3} – 1)^5\), we put \(x = \sqrt{3}\) and \(y = 1\) into equation (4):
\(( \sqrt{3} + 1)^5 – ( \sqrt{3} – 1)^5 = 10(\sqrt{3})^4 (1) + 20(\sqrt{3})^2 (1)^3 + 2(1)^5\)
\( = 10 \times 9 \times 1 + 20 \times 3 \times 1 + 2 \times 1\)
\( = 90 + 60 + 2\)
\( = 152\)
Now, we need to show that \(( \sqrt{3} + 1)^5\) lies between 152 and 153.
From our calculation, we know that \(( \sqrt{3} + 1)^5 – ( \sqrt{3} – 1)^5 = 152\).
This means \(( \sqrt{3} + 1)^5 = 152 + ( \sqrt{3} – 1)^5\).
We also know that \(\sqrt{3}\) is approximately 1.732.
So, \( \sqrt{3} – 1 \approx 1.732 – 1 = 0.732\).
Since \(0 < 0.732 < 1\), it means \(0 < ( \sqrt{3} – 1) < 1\).
If a number is between 0 and 1, any positive power of that number will also be between 0 and 1.
So, \(0 < ( \sqrt{3} – 1)^5 < 1\).
Therefore, \(( \sqrt{3} + 1)^5 = 152 + (\text{a number between 0 and 1})\).
This shows that \(( \sqrt{3} + 1)^5\) is greater than 152 but less than \(152+1\), which is 153.
Hence, \(( \sqrt{3} + 1)^5\) lies between 152 and 153.
In simple words: We use simplified formulas for sums and differences of binomial expansions. For part (i), we substitute the values into the sum formula. For part (ii), we use the difference formula. To show the range, we use the result from part (ii) and understand that a number between 0 and 1, when raised to a power, stays between 0 and 1. This helps us see that the value is just a little more than 152.

🎯 Exam Tip: When evaluating complex expressions, look for opportunities to use previously derived formulas or simplified forms. For showing bounds, estimating the value of terms like \( (\sqrt{3} – 1)^5 \) by comparing them to 0 and 1 can save time and prevent the need for calculators.

Question 6. If the first three terms in the expansion of \((1 + ax)^n\) in ascending powers of x are \(1 + 12x + 64x^2\), find n and a.
Answer: We use the binomial theorem to expand \((1 + ax)^n\).
\((1 + ax)^n = ^nC_0 (1)^n (ax)^0 + ^nC_1 (1)^{n-1} (ax)^1 + ^nC_2 (1)^{n-2} (ax)^2 + \dots\)
\( = 1 \times 1 \times 1 + n \times 1 \times ax + \frac{n(n-1)}{2 \times 1} \times 1 \times a^2x^2 + \dots\)
\( = 1 + nax + \frac{n(n-1)}{2} a^2x^2 + \dots\)
We are given that the first three terms are \(1 + 12x + 64x^2\).
By comparing the coefficients of x:
\(na = 12\) ...(1)
By comparing the coefficients of \(x^2\):
\(\frac{n(n-1)}{2} a^2 = 64\) ...(2)
From equation (1), we can find \(a = \frac{12}{n}\).
Substitute this value of 'a' into equation (2):
\(\frac{n(n-1)}{2} \left(\frac{12}{n}\right)^2 = 64\)
\( \implies \frac{n(n-1)}{2} \times \frac{144}{n^2} = 64\)
\( \implies \frac{144(n-1)}{2n} = 64\)
\( \implies \frac{72(n-1)}{n} = 64\)
\( \implies 72(n-1) = 64n\)
\( \implies 72n - 72 = 64n\)
\( \implies 72n - 64n = 72\)
\( \implies 8n = 72\)
\( \implies n = 9\)
Now, substitute \(n=9\) back into equation (1) to find 'a':
\(9a = 12\)
\( \implies a = \frac{12}{9}\)
\( \implies a = \frac{4}{3}\)
So, \(n=9\) and \(a=\frac{4}{3}\).
In simple words: We first write out the beginning part of the binomial expansion for \((1 + ax)^n\). Then, we match the numbers in front of \(x\) and \(x^2\) with the given values. This gives us two equations with 'n' and 'a'. We solve these equations together to find the values of 'n' and 'a'.

🎯 Exam Tip: When comparing coefficients, ensure you correctly expand the binomial up to the required power of x. Remember that \(^nC_0 = 1\), \(^nC_1 = n\), and \(^nC_2 = \frac{n(n-1)}{2}\). Carefully solve the system of equations for 'n' and 'a'.

Question 7. Find the first three terms in the expansion of \([2 + x(3 + 4x)]^5\) in ascending powers of x.
Answer: We need to find the first three terms in the expansion of \([2 + x(3 + 4x)]^5\).
First, let's simplify the expression inside the bracket:
\(2 + x(3 + 4x) = 2 + 3x + 4x^2\)
So, we need to expand \((2 + 3x + 4x^2)^5\).
We can group terms and use the binomial theorem. Let \(y = (2 + 3x)\).
Then the expression becomes \((y + 4x^2)^5\).
Using binomial theorem: \((y + 4x^2)^5 = ^5C_0 y^5 (4x^2)^0 + ^5C_1 y^4 (4x^2)^1 + ^5C_2 y^3 (4x^2)^2 + \dots\)
\( = 1 \times y^5 \times 1 + 5 \times y^4 \times (4x^2) + 10 \times y^3 \times (16x^4) + \dots\)
\( = y^5 + 20x^2y^4 + 160x^4y^3 + \dots\)
Now, substitute \(y = (2 + 3x)\) back into the terms and expand them to find powers of x up to \(x^2\).
The first term is \(y^5 = (2 + 3x)^5\).
\((2 + 3x)^5 = ^5C_0 2^5 (3x)^0 + ^5C_1 2^4 (3x)^1 + ^5C_2 2^3 (3x)^2 + \dots\)
\( = 1 \times 32 \times 1 + 5 \times 16 \times 3x + 10 \times 8 \times 9x^2 + \dots\)
\( = 32 + 240x + 720x^2 + \dots\)
The second term is \(20x^2y^4 = 20x^2 (2 + 3x)^4\).
We only need terms up to \(x^2\). So, from \((2 + 3x)^4\), we only need the constant term. The constant term is \(^4C_0 2^4 (3x)^0 = 1 \times 16 \times 1 = 16\).
So, \(20x^2y^4 = 20x^2 (16 + \text{terms with x, x^2, etc.})\)
\( = 320x^2 + \dots\)
The third term \(160x^4y^3\) will have powers of \(x\) starting from \(x^4\), which is higher than \(x^2\). So, we don't need to consider it for the first three terms up to \(x^2\).
Combining the terms we found:
\((2 + 3x + 4x^2)^5 = (32 + 240x + 720x^2 + \dots) + (320x^2 + \dots) + (\text{terms with higher powers of x})\)
\( = 32 + 240x + (720x^2 + 320x^2) + \dots\)
\( = 32 + 240x + 1040x^2 + \dots\)
The first three terms in the expansion are \(32 + 240x + 1040x^2\).
In simple words: First, simplify the expression inside the bracket. Then, use the binomial theorem by treating the expression as two parts. Expand the first part fully and for the second part, take only the constant term to get terms up to \(x^2\). Add up the similar powers of \(x\) to get the final answer.

🎯 Exam Tip: When expanding expressions with more than two terms raised to a power, group them into two terms and apply the binomial theorem. If asked for terms up to a certain power, only expand each sub-binomial (like \((2+3x)^5\)) to the necessary power to avoid unnecessary calculations.

Question 8. Expand \((1 + 2x + 3x^2)^n\) in a series of ascending powers of x up to and including the term in \(x^2\).
Answer: We need to expand \((1 + 2x + 3x^2)^n\) up to the term in \(x^2\).
We can group the terms inside the bracket. Let \(y = (1 + 2x)\).
Then the expression becomes \((y + 3x^2)^n\).
Using the binomial theorem: \((y + 3x^2)^n = ^nC_0 y^n (3x^2)^0 + ^nC_1 y^{n-1} (3x^2)^1 + ^nC_2 y^{n-2} (3x^2)^2 + \dots\)
\( = 1 \times y^n \times 1 + n \times y^{n-1} \times 3x^2 + \frac{n(n-1)}{2} \times y^{n-2} \times (3x^2)^2 + \dots\)
\( = y^n + 3nx^2 y^{n-1} + \dots\) (Terms like \((3x^2)^2\) will have \(x^4\) or higher, so we don't need them.)
Now, substitute \(y = (1 + 2x)\) back into the relevant terms:
\(y^n = (1 + 2x)^n\). Expand this up to \(x^2\):
\((1 + 2x)^n = ^nC_0 1^n (2x)^0 + ^nC_1 1^{n-1} (2x)^1 + ^nC_2 1^{n-2} (2x)^2 + \dots\)
\( = 1 + n(2x) + \frac{n(n-1)}{2} (4x^2) + \dots\)
\( = 1 + 2nx + 2n(n-1)x^2 + \dots\)
Next, consider the term \(3nx^2 y^{n-1} = 3nx^2 (1 + 2x)^{n-1}\).
To find terms up to \(x^2\), we only need the constant term from \((1 + 2x)^{n-1}\), which is \(^ {n-1}C_0 1^{n-1} (2x)^0 = 1\).
So, \(3nx^2 (1 + 2x)^{n-1} = 3nx^2 (1 + \text{terms with x, x^2, etc.})\)
\( = 3nx^2 + \dots\)
Now, combine these two parts:
\((1 + 2x + 3x^2)^n = (1 + 2nx + 2n(n-1)x^2 + \dots) + (3nx^2 + \dots)\)
\( = 1 + 2nx + [2n(n-1) + 3n]x^2 + \dots\)
\( = 1 + 2nx + [2n^2 - 2n + 3n]x^2 + \dots\)
\( = 1 + 2nx + (2n^2 + n)x^2 + \dots\)
The expansion in ascending powers of x up to and including the term in \(x^2\) is \(1 + 2nx + (2n^2 + n)x^2\).
In simple words: To expand an expression with three terms raised to a power, treat the first two terms as one group and the third term as another. Then use the binomial theorem. Expand each part only as far as needed to get terms up to \(x^2\). Finally, combine the coefficients of similar powers of \(x\).

🎯 Exam Tip: When dealing with trinomial expansions and seeking terms up to a specific power, it is crucial to strategically group terms. Only expand inner binomials (like \((1+2x)^{n-1}\)) to the extent that they contribute to the desired power of x in the final expression, ignoring higher-order terms from sub-expansions.

Question 9. Write down the expansion by the binomial theorem of \(\left(3 x-\frac{y}{2}\right)^4\). By giving x and y suitable values, deduce the value of \((29.5)^4\) correct to four significant figures.
Answer: First, we expand \(\left(3 x-\frac{y}{2}\right)^4\) using the binomial theorem.
\(\left(3 x-\frac{y}{2}\right)^4 = ^4C_0 (3x)^4 \left(-\frac{y}{2}\right)^0 + ^4C_1 (3x)^3 \left(-\frac{y}{2}\right)^1 + ^4C_2 (3x)^2 \left(-\frac{y}{2}\right)^2 + ^4C_3 (3x)^1 \left(-\frac{y}{2}\right)^3 + ^4C_4 (3x)^0 \left(-\frac{y}{2}\right)^4\)
\( = 1 \times 81x^4 \times 1 + 4 \times 27x^3 \left(-\frac{y}{2}\right) + 6 \times 9x^2 \left(\frac{y^2}{4}\right) + 4 \times 3x \left(-\frac{y^3}{8}\right) + 1 \times 1 \times \frac{y^4}{16}\)
\( = 81x^4 - 54x^3y + \frac{27}{2}x^2y^2 - \frac{3}{2}xy^3 + \frac{1}{16}y^4\) ...(1)

Next, to deduce the value of \((29.5)^4\), we need to choose values for x and y such that \(\left(3 x-\frac{y}{2}\right)\) becomes 29.5.
Let \(3x - \frac{y}{2} = 29.5\). A simple choice is to let \(y=1\) so that \(\frac{y}{2} = 0.5\).
Then, \(3x - 0.5 = 29.5\)
\(3x = 29.5 + 0.5\)
\(3x = 30\)
\(x = 10\)
So, we substitute \(x = 10\) and \(y = 1\) into the expanded expression (1):
\((29.5)^4 = 81(10)^4 - 54(10)^3(1) + \frac{27}{2}(10)^2(1)^2 - \frac{3}{2}(10)(1)^3 + \frac{1}{16}(1)^4\)
\( = 81 \times 10000 - 54 \times 1000 \times 1 + \frac{27}{2} \times 100 \times 1 - \frac{3}{2} \times 10 \times 1 + \frac{1}{16}\)
\( = 810000 - 54000 + (27 \times 50) - (3 \times 5) + 0.0625\)
\( = 810000 - 54000 + 1350 - 15 + 0.0625\)
\( = 757335.0625\)
Rounding to four significant figures, we get 757300. (The original problem statement implies rounding to 4 significant figures, and 757335.0625 rounded to 4 sig figs is 757300).
In simple words: First, we write out the full expansion using the binomial theorem. Then, we pick simple numbers for 'x' and 'y' that make the expression inside the bracket equal to 29.5. We use \(x=10\) and \(y=1\). Finally, we put these numbers into our expanded form and calculate the answer, then round it to four significant figures.

🎯 Exam Tip: When deducing values, choose 'x' and 'y' carefully to simplify calculations. Often, whole numbers or easy fractions (like 1 or 2) make the arithmetic much easier. Always remember to round your final answer to the specified number of significant figures.

Question 10. Using binomial theorem, evaluate : \((999)^3\).
Answer: We need to evaluate \((999)^3\) using the binomial theorem.
We can write 999 as \((1000 - 1)\).
So, \((999)^3 = (1000 - 1)^3\).
Using the binomial theorem for \((a - b)^3 = a^3 - ^3C_1 a^2 b^1 + ^3C_2 a^1 b^2 - ^3C_3 a^0 b^3\):
\((1000 - 1)^3 = ^3C_0 (1000)^3 (-1)^0 + ^3C_1 (1000)^2 (-1)^1 + ^3C_2 (1000)^1 (-1)^2 + ^3C_3 (1000)^0 (-1)^3\)
\( = 1 \times (1000)^3 \times 1 + 3 \times (1000)^2 \times (-1) + 3 \times (1000)^1 \times (1) + 1 \times 1 \times (-1)\)
\( = 1,000,000,000 - 3 \times 1,000,000 + 3 \times 1000 - 1\)
\( = 1,000,000,000 - 3,000,000 + 3,000 - 1\)
\( = 997,003,000 - 1\)
\( = 997,002,999\)
In simple words: To calculate \((999)^3\) using the binomial theorem, we rewrite 999 as \((1000 - 1)\). Then, we use the binomial expansion formula for \((a-b)^3\). This helps us avoid direct multiplication of large numbers by breaking it down into simpler additions and subtractions of powers of 1000.

🎯 Exam Tip: For numbers close to 10, 100, 1000, etc., rewrite them as \((10^k \pm 1)\). This makes binomial expansion calculations much easier as powers of 10 are simple to handle. Be careful with the alternating signs in a negative binomial expansion.

Question 11. Write down in terms of x and n, the term containing \(x^3\) in the expansion of \(\left(1-\frac{x}{n}\right)^n\) by the binomial theorem. If this term equals \(\frac{7}{8}\) when \(x = -2\), and n is a positive integer, calculate the value of n.
Answer: We need to find the term containing \(x^3\) in the expansion of \(\left(1-\frac{x}{n}\right)^n\).
The general term \(T_{r+1}\) in the expansion of \((a+b)^n\) is \(^nC_r a^{n-r} b^r\).
Here, \(a = 1\), \(b = -\frac{x}{n}\). We want the term with \(x^3\), so we need \(r=3\).
The term containing \(x^3\) is \(T_{3+1} = T_4\).
\(T_4 = ^nC_3 (1)^{n-3} \left(-\frac{x}{n}\right)^3\)
\( = \frac{n(n-1)(n-2)}{3 \times 2 \times 1} \times 1 \times \left(-\frac{x^3}{n^3}\right)\)
\( = \frac{n(n-1)(n-2)}{6} \times -\frac{x^3}{n^3}\)
\( = -\frac{(n-1)(n-2)}{6n^2} x^3\)

We are given that this term equals \(\frac{7}{8}\) when \(x = -2\).
So, \(-\frac{(n-1)(n-2)}{6n^2} (-2)^3 = \frac{7}{8}\)
\(-\frac{(n-1)(n-2)}{6n^2} (-8) = \frac{7}{8}\)
\( \implies \frac{8(n-1)(n-2)}{6n^2} = \frac{7}{8}\)
\( \implies \frac{4(n-1)(n-2)}{3n^2} = \frac{7}{8}\)
\( \implies 32(n-1)(n-2) = 21n^2\)
\( \implies 32(n^2 - 3n + 2) = 21n^2\)
\( \implies 32n^2 - 96n + 64 = 21n^2\)
\( \implies 32n^2 - 21n^2 - 96n + 64 = 0\)
\( \implies 11n^2 - 96n + 64 = 0\)
This is a quadratic equation. We can solve it using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Here, \(a=11\), \(b=-96\), \(c=64\).
\(n = \frac{-(-96) \pm \sqrt{(-96)^2 - 4(11)(64)}}{2(11)}\)
\(n = \frac{96 \pm \sqrt{9216 - 2816}}{22}\)
\(n = \frac{96 \pm \sqrt{6400}}{22}\)
\(n = \frac{96 \pm 80}{22}\)
Two possible values for n:
\(n_1 = \frac{96 + 80}{22} = \frac{176}{22} = 8\)
\(n_2 = \frac{96 - 80}{22} = \frac{16}{22} = \frac{8}{11}\)
Since n must be a positive integer, we choose \(n=8\).
In simple words: First, we use the binomial theorem's general term formula to find the expression for the \(x^3\) term. Then, we are given that this term has a specific value when \(x = -2\). We substitute \(x = -2\) into our \(x^3\) term expression and set it equal to the given value. This creates a quadratic equation for 'n', which we solve. We pick the integer solution for 'n' because it's stated that 'n' is a positive integer.

🎯 Exam Tip: Always remember the general term formula \(T_{r+1} = ^nC_r a^{n-r} b^r\) for binomial expansions. When solving for 'n' in a quadratic equation, pay attention to any conditions on 'n' (like being a positive integer) to select the correct solution.

Question 12.
(i) Obtain the binomial expansion of \((2 – \sqrt{3})^6\) in the form \(a + b\sqrt{3}\), where a and b are integers. State the corresponding result for the expansion \((2 + \sqrt{3})^6\).
(ii) show that \((2 – \sqrt{3})^6\) is the reciprocal of \((2 + \sqrt{3})^6\).
Answer:
(i) We use the binomial theorem to expand \((2 – \sqrt{3})^6\).
\((2 – \sqrt{3})^6 = ^6C_0 2^6 (\sqrt{3})^0 - ^6C_1 2^5 (\sqrt{3})^1 + ^6C_2 2^4 (\sqrt{3})^2 - ^6C_3 2^3 (\sqrt{3})^3 + ^6C_4 2^2 (\sqrt{3})^4 - ^6C_5 2^1 (\sqrt{3})^5 + ^6C_6 2^0 (\sqrt{3})^6\)
\( = 1 \times 64 \times 1 - 6 \times 32 \times \sqrt{3} + 15 \times 16 \times 3 - 20 \times 8 \times 3\sqrt{3} + 15 \times 4 \times 9 - 6 \times 2 \times 9\sqrt{3} + 1 \times 1 \times 27\)
\( = 64 - 192\sqrt{3} + 720 - 480\sqrt{3} + 540 - 108\sqrt{3} + 27\)
Group the rational and irrational terms:
\( = (64 + 720 + 540 + 27) - (192\sqrt{3} + 480\sqrt{3} + 108\sqrt{3})\)
\( = 1351 - 780\sqrt{3}\)
So, \((2 – \sqrt{3})^6 = 1351 - 780\sqrt{3}\). Here, \(a = 1351\) and \(b = -780\).

For the expansion of \((2 + \sqrt{3})^6\), the terms involving odd powers of \(\sqrt{3}\) will be positive instead of negative.
\((2 + \sqrt{3})^6 = ^6C_0 2^6 (\sqrt{3})^0 + ^6C_1 2^5 (\sqrt{3})^1 + ^6C_2 2^4 (\sqrt{3})^2 + ^6C_3 2^3 (\sqrt{3})^3 + ^6C_4 2^2 (\sqrt{3})^4 + ^6C_5 2^1 (\sqrt{3})^5 + ^6C_6 2^0 (\sqrt{3})^6\)
\( = 64 + 192\sqrt{3} + 720 + 480\sqrt{3} + 540 + 108\sqrt{3} + 27\)
\( = (64 + 720 + 540 + 27) + (192\sqrt{3} + 480\sqrt{3} + 108\sqrt{3})\)
\( = 1351 + 780\sqrt{3}\)

(ii) To show that \((2 – \sqrt{3})^6\) is the reciprocal of \((2 + \sqrt{3})^6\), we need to show that their product is 1.
\((2 – \sqrt{3})^6 (2 + \sqrt{3})^6 = [(2 – \sqrt{3})(2 + \sqrt{3})]^6\)
We know that \((A-B)(A+B) = A^2 - B^2\).
So, \([(2 – \sqrt{3})(2 + \sqrt{3})]^6 = [(2)^2 - (\sqrt{3})^2]^6\)
\( = [4 - 3]^6\)
\( = [1]^6\)
\( = 1\)
Since the product of \((2 – \sqrt{3})^6\) and \((2 + \sqrt{3})^6\) is 1, it means they are reciprocals of each other.
Therefore, \((2 – \sqrt{3})^6 = \frac{1}{(2 + \sqrt{3})^6}\).
In simple words: First, we expand \((2 – \sqrt{3})^6\) using the binomial theorem, carefully combining the whole numbers and the terms with \(\sqrt{3}\). We write it as \(a + b\sqrt{3}\). The expansion for \((2 + \sqrt{3})^6\) will have the same whole number part but a positive \(\sqrt{3}\) part. For the second part, we multiply \((2 – \sqrt{3})^6\) by \((2 + \sqrt{3})^6\). We can group them as \([(2 – \sqrt{3})(2 + \sqrt{3})]^6\). Since \((2 – \sqrt{3})(2 + \sqrt{3})\) equals \(2^2 - (\sqrt{3})^2 = 4-3=1\), the whole expression becomes \(1^6\), which is 1. This shows they are reciprocals.

🎯 Exam Tip: When dealing with binomials involving square roots, remember the difference of squares identity \((a-b)(a+b) = a^2 - b^2\). This can significantly simplify calculations, especially when products of conjugates are involved. Pay attention to signs when calculating expansions with negative terms.

Question 13. Find the coefficient of \(x^5\) in the expansion of \((1 + 2x)^6 (1 – x)^7\).
Answer: We need to find the coefficient of \(x^5\) in the product of two expansions: \((1 + 2x)^6\) and \((1 – x)^7\).
First, let's find the expansion of \((1 + 2x)^6\):
\((1 + 2x)^6 = ^6C_0 (2x)^0 + ^6C_1 (2x)^1 + ^6C_2 (2x)^2 + ^6C_3 (2x)^3 + ^6C_4 (2x)^4 + ^6C_5 (2x)^5 + ^6C_6 (2x)^6\)
\( = 1 + 6(2x) + 15(4x^2) + 20(8x^3) + 15(16x^4) + 6(32x^5) + 1(64x^6)\)
\( = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + 64x^6\)
Next, let's find the expansion of \((1 – x)^7\):
\((1 – x)^7 = ^7C_0 (-x)^0 + ^7C_1 (-x)^1 + ^7C_2 (-x)^2 + ^7C_3 (-x)^3 + ^7C_4 (-x)^4 + ^7C_5 (-x)^5 + ^7C_6 (-x)^6 + ^7C_7 (-x)^7\)
\( = 1 - 7x + 21x^2 - 35x^3 + 35x^4 - 21x^5 + 7x^6 - x^7\)
Now, we multiply these two expansions and collect terms that result in \(x^5\).
A term in \(x^5\) can be formed by multiplying:
(constant from first expansion) \(\times\) (\(x^5\) term from second expansion)
(\(x\) term from first expansion) \(\times\) (\(x^4\) term from second expansion)
(\(x^2\) term from first expansion) \(\times\) (\(x^3\) term from second expansion)
(\(x^3\) term from first expansion) \(\times\) (\(x^2\) term from second expansion)
(\(x^4\) term from first expansion) \(\times\) (\(x\) term from second expansion)
(\(x^5\) term from first expansion) \(\times\) (constant from second expansion)

Coefficient of \(x^5\):
\((1 \times -21) + (12 \times 35) + (60 \times -35) + (160 \times 21) + (240 \times -7) + (192 \times 1)\)
\( = -21 + 420 - 2100 + 3360 - 1680 + 192\)
\( = 399 - 2100 + 3360 - 1680 + 192\)
\( = -1701 + 3360 - 1680 + 192\)
\( = 1659 - 1680 + 192\)
\( = -21 + 192\)
\( = 171\)
The coefficient of \(x^5\) in the expansion of \((1 + 2x)^6 (1 – x)^7\) is 171.
In simple words: First, we expand both binomial expressions, \((1 + 2x)^6\) and \((1 – x)^7\), up to the \(x^5\) term. Then, to find the coefficient of \(x^5\) in their product, we look for all the ways we can multiply a term from the first expansion by a term from the second expansion such that their powers of \(x\) add up to 5. We multiply these coefficients and sum them all up.

🎯 Exam Tip: To find the coefficient of a specific power of x in the product of two expansions, list the relevant terms from each expansion. For \((x^m \text{ from first})(x^n \text{ from second})\), ensure \(m+n\) equals the target power (e.g., 5). Be careful with negative signs in the coefficients.

Question 14. If the coefficients of second, third and fourth terms in the expansion of \((1 + x)^{2n}\) are in A.P., show that \(2n^2 – 9n + 7 = 0\).
Answer: Let the coefficients of the second, third, and fourth terms in the expansion of \((1 + x)^{2n}\) be \(C_2\), \(C_3\), and \(C_4\) respectively.
Using the binomial theorem, the general term for \((1+x)^N\) is \(^NC_r x^r\). The coefficient is \(^NC_r\).
Here, \(N = 2n\).
Coefficient of the 2nd term (when \(r=1\)) is \(C_2 = ^{2n}C_1\).
Coefficient of the 3rd term (when \(r=2\)) is \(C_3 = ^{2n}C_2\).
Coefficient of the 4th term (when \(r=3\)) is \(C_4 = ^{2n}C_3\).
If these coefficients are in Arithmetic Progression (A.P.), then the middle term is the average of the first and third terms. So, \(2C_3 = C_2 + C_4\).
\( \implies 2 \times ^{2n}C_2 = ^{2n}C_1 + ^{2n}C_3\)
Now, we write out the combination formulas:
\(^NC_r = \frac{N!}{r!(N-r)!}\)
\( \implies 2 \times \frac{(2n)!}{2!(2n-2)!} = \frac{(2n)!}{1!(2n-1)!} + \frac{(2n)!}{3!(2n-3)!}\)
We can divide all terms by \((2n)!\) and simplify the denominators:
\( \implies \frac{2}{2(2n-2)!} = \frac{1}{(2n-1)!} + \frac{1}{6(2n-3)!}\)
Rewrite factorials to the smallest one, which is \((2n-3)!\):
\((2n-2)! = (2n-2)(2n-3)!\)
\((2n-1)! = (2n-1)(2n-2)(2n-3)!\)
\( \implies \frac{1}{(2n-2)(2n-3)!} = \frac{1}{(2n-1)(2n-2)(2n-3)!} + \frac{1}{6(2n-3)!}\)
Now, multiply the entire equation by \(6(2n-1)(2n-2)(2n-3)!\) to clear denominators:
\( \implies 6(2n-1) = 6 + (2n-1)(2n-2)\)
\( \implies 12n - 6 = 6 + (4n^2 - 4n - 2n + 2)\)
\( \implies 12n - 6 = 6 + 4n^2 - 6n + 2\)
\( \implies 12n - 6 = 4n^2 - 6n + 8\)
Move all terms to one side to form a quadratic equation:
\( \implies 4n^2 - 6n - 12n + 8 + 6 = 0\)
\( \implies 4n^2 - 18n + 14 = 0\)
Divide the entire equation by 2:
\( \implies 2n^2 - 9n + 7 = 0\)
This proves the required statement.
In simple words: First, we write down the coefficients of the second, third, and fourth terms using the combination formula. Since these terms are in A.P., we know that twice the middle term equals the sum of the first and third terms. We set up this equation and then simplify the factorial expressions. After clearing denominators and rearranging the terms, we arrive at the desired quadratic equation.

🎯 Exam Tip: For problems involving terms in A.P., remember the property: if a, b, c are in A.P., then \(2b = a+c\). When simplifying factorial expressions like \(\frac{N!}{r!(N-r)!}\), expand the larger factorials to cancel out the smaller ones, especially the \((N-r)!\) terms.

Question 15. Let n be a positive integer. If the coefficients of 2nd, 3rd, 4th terms in the expansion of \((1 + x)^n\) are in A.P., then find the value of n.
Answer: Let the coefficients of the 2nd, 3rd, and 4th terms in the expansion of \((1 + x)^n\) be \(C_2\), \(C_3\), and \(C_4\) respectively.
Coefficient of the 2nd term is \(C_2 = ^nC_1\).
Coefficient of the 3rd term is \(C_3 = ^nC_2\).
Coefficient of the 4th term is \(C_4 = ^nC_3\).
Since these coefficients are in Arithmetic Progression (A.P.), we have \(2C_3 = C_2 + C_4\).
\( \implies 2 \times ^nC_2 = ^nC_1 + ^nC_3\)
Now, we write out the combination formulas:
\( \implies 2 \times \frac{n!}{2!(n-2)!} = \frac{n!}{1!(n-1)!} + \frac{n!}{3!(n-3)!}\)
We can divide all terms by \(n!\) and simplify the denominators:
\( \implies \frac{2}{2(n-2)!} = \frac{1}{(n-1)!} + \frac{1}{6(n-3)!}\)
Rewrite factorials to the smallest one, which is \((n-3)!\):
\((n-2)! = (n-2)(n-3)!\)
\((n-1)! = (n-1)(n-2)(n-3)!\)
\( \implies \frac{1}{(n-2)(n-3)!} = \frac{1}{(n-1)(n-2)(n-3)!} + \frac{1}{6(n-3)!}\)
Now, multiply the entire equation by \(6(n-1)(n-2)(n-3)!\) to clear denominators:
\( \implies 6(n-1) = 6 + (n-1)(n-2)\)
\( \implies 6n - 6 = 6 + (n^2 - 3n + 2)\)
\( \implies 6n - 6 = n^2 - 3n + 8\)
Move all terms to one side to form a quadratic equation:
\( \implies n^2 - 3n - 6n + 8 + 6 = 0\)
\( \implies n^2 - 9n + 14 = 0\)
Factorize the quadratic equation:
\( \implies n^2 - 2n - 7n + 14 = 0\)
\( \implies n(n-2) - 7(n-2) = 0\)
\( \implies (n-2)(n-7) = 0\)
This gives two possible values for n:
\(n = 2\) or \(n = 7\)
If \(n=2\), the expansion \((1+x)^2\) only has three terms: \(^2C_0, ^2C_1, ^2C_2\). The second, third, and fourth terms would be \(^2C_1, ^2C_2\), and the fourth term would not exist. For the fourth term to exist, 'n' must be at least 3.
However, the phrasing "coefficients of 2nd, 3rd, 4th terms" implies these terms *exist*. If \(n=2\), there is no 4th term. So, \(n=2\) is not a valid solution in this context.
Therefore, the value of \(n\) is 7.
In simple words: We list the coefficients for the second, third, and fourth terms of \((1+x)^n\) using the combination formula. Since they are in A.P., we set up the equation where twice the middle coefficient equals the sum of the other two. We simplify this equation by cancelling factorials and solve for 'n'. After getting two possible values for 'n', we check if they are reasonable. For example, 'n' must be at least 3 for a 4th term to exist, which helps us pick the correct answer.

🎯 Exam Tip: When solving for 'n' and you get multiple solutions, always check if they are valid within the context of the binomial expansion. For a term \(T_{r+1}\) (i.e., with index r), you must have \(n \ge r\). So, for the 4th term (r=3) to exist, \(n\) must be at least 3.

Question 16. In the binomial expansion of \(\left(\sqrt[3]{3}+\sqrt{2}\right)^5\) find the term which does not contain Irrational expression.
Answer: We need to find the term in the expansion of \(\left(\sqrt[3]{3}+\sqrt{2}\right)^5\) that does not contain an irrational expression (i.e., a rational term).
The general term \(T_{r+1}\) in the expansion of \((a+b)^n\) is \(^nC_r a^{n-r} b^r\).
Here, \(a = \sqrt[3]{3} = 3^{1/3}\), \(b = \sqrt{2} = 2^{1/2}\), and \(n=5\).
So, the general term is \(T_{r+1} = ^5C_r (3^{1/3})^{5-r} (2^{1/2})^r\)
\( = ^5C_r 3^{\frac{5-r}{3}} 2^{\frac{r}{2}}\)
For the term to be rational, the powers of 3 and 2 must both be integers. This means \(\frac{5-r}{3}\) must be an integer, and \(\frac{r}{2}\) must be an integer.
From \(\frac{r}{2}\) being an integer, 'r' must be an even number. Possible values for 'r' (since \(0 \le r \le 5\)) are \(0, 2, 4\).
Let's check these values for \(\frac{5-r}{3}\):
If \(r=0\): \(\frac{5-0}{3} = \frac{5}{3}\) (not an integer, so the term is irrational).
If \(r=2\): \(\frac{5-2}{3} = \frac{3}{3} = 1\) (an integer, so this is a rational term).
If \(r=4\): \(\frac{5-4}{3} = \frac{1}{3}\) (not an integer, so the term is irrational).
Thus, the only value of 'r' that yields a rational term is \(r=2\).
This corresponds to the \(T_{2+1} = T_3\) term.
The term is \(^5C_2 (3^{1/3})^{5-2} (2^{1/2})^2\)
\( = ^5C_2 (3^{1/3})^3 (2^{1/2})^2\)
\( = ^5C_2 \times 3^1 \times 2^1\)
\( = \frac{5 \times 4}{2 \times 1} \times 3 \times 2\)
\( = 10 \times 3 \times 2\)
\( = 60\)
The term that does not contain an irrational expression is 60.
In simple words: We write down the general formula for any term in the expansion. For the term to be a whole number (rational), the powers of the numbers with roots must become whole numbers. This means the fractions in the exponents must simplify to integers. We check different possible values of 'r' to find the one that makes both exponents integer. In this case, when \(r=2\), both \(\frac{5-r}{3}\) and \(\frac{r}{2}\) become integers, leading to a rational term. Then we calculate this term.

🎯 Exam Tip: For binomial expansions involving roots, express the bases as fractional powers (e.g., \(\sqrt[3]{3} = 3^{1/3}\)). The key to finding rational terms is to ensure that the exponents of all prime factors in the general term become integers, which typically requires finding common multiples for the denominators of the fractional powers.