CBSE Class 11 Mathematics Binomial Theorem Worksheet Set 01

Q1. Using binomial theorem, write down the expansion of the following:-
(i) \(\left(x + 1 - \frac{1}{x}\right)^3\)     (ii) \((\sqrt[3]{x} - \sqrt[3]{a})^6\)     (iii) \((2x + 3y)^4\)
Answer: (i) Let \(a = x + 1\) and \(b = -\frac{1}{x}\). Using \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\):
\(= (x+1)^3 + 3(x+1)^2\left(-\frac{1}{x}\right) + 3(x+1)\left(-\frac{1}{x}\right)^2 + \left(-\frac{1}{x}\right)^3\)
\(= (x^3 + 3x^2 + 3x + 1) - \frac{3(x^2 + 2x + 1)}{x} + \frac{3(x+1)}{x^2} - \frac{1}{x^3}\)
\(= x^3 + 3x^2 + 3x + 1 - 3x - 6 - \frac{3}{x} + \frac{3}{x} + \frac{3}{x^2} - \frac{1}{x^3}\)
\(= x^3 + 3x^2 - 5 - \frac{3}{x} + \frac{3}{x^2} - \frac{1}{x^3}\)
(ii) Using binomial expansion for \((\sqrt[3]{x} - \sqrt[3]{a})^6\):
\(= \sum_{r=0}^{6} \binom{6}{r} (\sqrt[3]{x})^{6-r} (-\sqrt[3]{a})^r\)
\(= \binom{6}{0}x^2 - \binom{6}{1}x^{5/3}a^{1/3} + \binom{6}{2}x^{4/3}a^{2/3} - \binom{6}{3}xa + \binom{6}{4}x^{2/3}a^{4/3} - \binom{6}{5}x^{1/3}a^{5/3} + \binom{6}{6}a^2\)
\(= x^2 - 6x^{5/3}a^{1/3} + 15x^{4/3}a^{2/3} - 20xa + 15x^{2/3}a^{4/3} - 6x^{1/3}a^{5/3} + a^2\)
(iii) Using binomial expansion for \((2x + 3y)^4\):
\(= \binom{4}{0}(2x)^4 + \binom{4}{1}(2x)^3(3y) + \binom{4}{2}(2x)^2(3y)^2 + \binom{4}{3}(2x)(3y)^3 + \binom{4}{4}(3y)^4\)
\(= 1(16x^4) + 4(8x^3)(3y) + 6(4x^2)(9y^2) + 4(2x)(27y^3) + 1(81y^4)\)
\(= 16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4\)
In simple words: We apply the standard binomial expansion formula term by term for each expression.
Exam Tip: Be very careful with alternating signs when negative terms are present inside powers.

Q2. Evaluate the following :-
(i) \((\sqrt{3} + 1)^5 - (\sqrt{3} - 1)^5\)     (ii) \((3 + \sqrt{2})^4 + (3 - \sqrt{2})^4\)     (iii) \((0.99)^3 + (1.01)^3\)
Answer: (i) Let \(a = \sqrt{3}\) and \(b = 1\).
\((\sqrt{3}+1)^5 - (\sqrt{3}-1)^5 = [C_0 a^5 + C_1 a^4 b + C_2 a^3 b^2 + C_3 a^2 b^3 + C_4 a b^4 + C_5 b^5] - [C_0 a^5 - C_1 a^4 b + C_2 a^3 b^2 - C_3 a^2 b^3 + C_4 a b^4 - C_5 b^5]\)
\(= 2[C_1 a^4 b + C_3 a^2 b^3 + C_5 b^5]\)
\(= 2[5(9)(1) + 10(3)(1) + 1(1)] = 2[45 + 30 + 1] = 2(76) = 152\)
(ii) \((3+\sqrt{2})^4 + (3-\sqrt{2})^4 = 2[C_0 3^4 + C_2 3^2 (\sqrt{2})^2 + C_4 (\sqrt{2})^4]\)
\(= 2[1(81) + 6(9)(2) + 1(4)] = 2[81 + 108 + 4] = 2(193) = 386\)
(iii) \((0.99)^3 + (1.01)^3 = (1 - 0.01)^3 + (1 + 0.01)^3\)
Using \((1-x)^3 + (1+x)^3 = 2(1 + 3x^2)\) where \(x = 0.01\):
\(= 2[1 + 3(0.01)^2] = 2[1 + 3(0.0001)] = 2[1.0003] = 2.0006\)
In simple words: Grouping terms with opposite signs helps cancel out identical parts easily.

Exam Tip: Memorize the algebraic shortcuts for the sum and difference of binomial powers.

Q3. Using binomial theorem, prove that \(2^{3n} - 7n - 1\) is divisible by 49, where \(n \in \mathbb{N}\).
Answer: We know that \(2^3 = 8 = 1 + 7\).
So, \(2^{3n} = (2^3)^n = (1 + 7)^n\)
Using binomial theorem:
\((1 + 7)^n = \binom{n}{0} + \binom{n}{1}(7) + \binom{n}{2}(7)^2 + \binom{n}{3}(7)^3 + \dots + \binom{n}{n}(7)^n\)
\(2^{3n} = 1 + 7n + 49\left[\binom{n}{2} + \binom{n}{3}(7) + \dots + \binom{n}{n}(7)^{n-2}\right]\)
\(2^{3n} - 7n - 1 = 49k\)
where \(k = \binom{n}{2} + \binom{n}{3}(7) + \dots + \binom{n}{n}(7)^{n-2}\) is an integer.
Thus, \(2^{3n} - 7n - 1\) is divisible by 49.
In simple words: We express the base in the form of \(1 + 7\) so that higher powers produce a factor of \(49\).

Exam Tip: For divisibility proofs, write the number as \(1 + \text{multiple}\) and factor out the required square term.

Q4. Using binomial theorem determine which number is larger \((1.2)^{4000}\) or \(800\)?
Answer: Express \((1.2)^{4000}\) as \((1 + 0.2)^{4000}\).
Using binomial theorem:
\((1 + 0.2)^{4000} = \binom{4000}{0} + \binom{4000}{1}(0.2) + \text{positive terms}\)
\(= 1 + 4000(0.2) + \text{positive terms}\)
\(= 1 + 800 + \text{positive terms}\)
\(= 801 + \text{positive terms}\)
Since all subsequent terms in the expansion are positive, \((1.2)^{4000} = 801 + \text{positive terms} > 800\).
Therefore, \((1.2)^{4000}\) is larger than 800.
In simple words: Expanding just the first two terms already exceeds 800, proving the number is larger.

Exam Tip: Open the first two or three terms of the binomial expansion to establish inequalities.

Q5. Find the coefficient of \(x^{10}\) in the expansion of \(\left(2x - \frac{1}{x^2}\right)^{25}\).
Answer: The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)
Here, \(n = 25\), \(a = 2x\), and \(b = -\frac{1}{x^2}\).
\(T_{r+1} = \binom{25}{r} (2x)^{25-r} \left(-\frac{1}{x^2}\right)^r\)
\(T_{r+1} = \binom{25}{r} 2^{25-r} (-1)^r x^{25-r} x^{-2r}\)
\(T_{r+1} = \binom{25}{r} 2^{25-r} (-1)^r x^{25-3r}\)
To find the coefficient of \(x^{10}\), equate the power of \(x\) to 10:
\(25 - 3r = 10\)
\(3r = 15 \implies r = 5\)
Substituting \(r = 5\) into the coefficient part:
Coefficient = \(\binom{25}{5} 2^{25-5} (-1)^5 = -\binom{25}{5} 2^{20}\)
In simple words: Find the general term, collect all powers of \(x\), and solve for \(r\) by matching exponents.

Exam Tip: Always isolate \(x\) with its total exponent before equating it to the required power.

Q6. Find the fourth term from the end in the expansion of \(\left(2x - \frac{1}{x^2}\right)^{25}\).
Answer: The total number of terms in the expansion of \((a + b)^n\) is \(n + 1\). Here, \(n = 25\), so there are 26 terms.
The \(k^{\text{th}}\) term from the end is equal to the \((n - k + 2)^{\text{th}}\) term from the beginning.
Fourth term from the end = \(25 - 4 + 2 = 23^{\text{rd}}\) term from the beginning.
We need \(T_{23} = T_{22+1}\). Using the general term formula \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\):
\(T_{23} = \binom{25}{22} (2x)^{25-22} \left(-\frac{1}{x^2}\right)^{22}\)
\(= \binom{25}{3} (2x)^3 \left(\frac{1}{x^{44}}\right)\)
\(= \frac{25 \times 24 \times 23}{3 \times 2 \times 1} \cdot 8x^3 \cdot \frac{1}{x^{44}}\)
\(= 2300 \cdot 8x^{-41} = 18400 x^{-41}\)
In simple words: Convert counting from the end to counting from the start using the formula \((n - k + 2)\).

Exam Tip: The \(r^{\text{th}}\) term from the end in \((a+b)^n\) is the \((n - r + 2)^{\text{th}}\) term from the beginning.

Q7. Find the middle term / terms in the expansion of :-
(i) \(\left(2x - \frac{3}{2x}\right)^{20}\)     (ii) \(\left(x^4 - \frac{1}{x^3}\right)^{11}\)
Answer: (i) Here, \(n = 20\) (even). There is only one middle term, which is given by \(\left(\frac{n}{2} + 1\right)^{\text{th}}\) term = \(11^{\text{th}}\) term.
\(T_{11} = \binom{20}{10} (2x)^{20-10} \left(-\frac{3}{2x}\right)^{10}\)
\(= \binom{20}{10} (2)^{10} x^{10} \frac{3^{10}}{2^{10} x^{10}} = \binom{20}{10} 3^{10}\)
(ii) Here, \(n = 11\) (odd). There are two middle terms: \(\left(\frac{n+1}{2}\right)^{\text{th}}\) and \(\left(\frac{n+3}{2}\right)^{\text{th}}\) terms, i.e., \(6^{\text{th}}\) and \(7^{\text{th}}\) terms.
\(T_6 = \binom{11}{5} (x^4)^{11-5} \left(-\frac{1}{x^3}\right)^5 = \binom{11}{5} x^{24} \left(-\frac{1}{x^{15}}\right) = -\binom{11}{5} x^9\)
\(T_7 = \binom{11}{6} (x^4)^{11-6} \left(-\frac{1}{x^3}\right)^6 = \binom{11}{6} x^{20} \left(\frac{1}{x^{18}}\right) = \binom{11}{6} x^2\)
In simple words: If \(n\) is even, there is 1 middle term. If \(n\) is odd, there are 2 middle terms.

Exam Tip: Check whether \(n\) is even or odd first to determine the number of middle terms.

Q8. Find the term independent of x in the expansion of :-
(i) \(\left(2x - \frac{1}{x^2}\right)^{10}\)     (ii) \(\left(3x^2 - \frac{1}{2x}\right)^9\)
Answer: (i) General term \(T_{r+1} = \binom{10}{r} (2x)^{10-r} \left(-\frac{1}{x^2}\right)^r = \binom{10}{r} 2^{10-r} (-1)^r x^{10-3r}\)
For the term independent of \(x\), set power of \(x\) to 0:
\(10 - 3r = 0 \implies 3r = 10\) (Not an integer). Wait, let us re-check the expression.
Correction for (i): \(T_{r+1} = \binom{10}{r} (2x)^{10-r}(-1)^r x^{-2r} = \binom{10}{r} 2^{10-r} (-1)^r x^{10-3r}\). If the exponent is 12: Let's solve with standard matching. Since 10 is not divisible by 3, let's assume the power was 12 or evaluate as \(10 - 3r = 0 \implies r = 10/3\) (No term independent of \(x\)).
Let us solve (ii):
\(T_{r+1} = \binom{9}{r} (3x^2)^{9-r} \left(-\frac{1}{2x}\right)^r = \binom{9}{r} 3^{9-r} (-1)^r \left(\frac{1}{2^r}\right) x^{18-2r-r}\)
\(T_{r+1} = \binom{9}{r} 3^{9-r} (-1)^r 2^{-r} x^{18-3r}\)
Set power of \(x\) to 0:
\(18 - 3r = 0 \implies r = 6\)
Substituting \(r = 6\):
\(T_7 = \binom{9}{6} 3^{9-6} (-1)^6 2^{-6} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \cdot 27 \cdot \frac{1}{64} = 84 \times \frac{27}{64} = \frac{567}{16}\)
In simple words: A term independent of \(x\) means the exponent of \(x\) equals 0.

Exam Tip: Combine all indices of \(x\) into a single exponent and equate to 0 to find \(r\).

Q9. If the fourth term in the expansion of \(\left(ax + \frac{1}{x}\right)^n\) is \(\frac{5}{2}\), then find the value of 'a' and 'n'.
Answer: Fourth term \(T_4 = T_{3+1} = \binom{n}{3} (ax)^{n-3} \left(\frac{1}{x}\right)^3 = \binom{n}{3} a^{n-3} x^{n-6}\)
We are given \(T_4 = \frac{5}{2}\) and typically for such problems \(x^{n-6} = 1 \implies n = 6\).
Substituting \(n = 6\):
\(\binom{6}{3} a^{3} = \frac{5}{2}\)
\(20 a^3 = \frac{5}{2} \implies a^3 = \frac{5}{40} = \frac{1}{8} \implies a = \frac{1}{2}\)
Thus, \(a = \frac{1}{2}\) and \(n = 6\).
In simple words: Match the given term with the general formula by equating coefficients and powers of \(x\).

Exam Tip: Equate the exponent of \(x\) to zero first to find \(n\), then solve for \(a\).

Q10. Find the value of a so that the term independent of x in \(\left(\sqrt{x} + \frac{a}{x^2}\right)^{10}\) is 405.
Answer: General term \(T_{r+1} = \binom{10}{r} (\sqrt{x})^{10-r} \left(\frac{a}{x^2}\right)^r = \binom{10}{r} x^{\frac{10-r}{2}} a^r x^{-2r}\)
\(T_{r+1} = \binom{10}{r} a^r x^{\frac{10-r}{2} - 2r} = \binom{10}{r} a^r x^{\frac{10 - 5r}{2}}\)
For the term independent of \(x\):
\(\frac{10 - 5r}{2} = 0 \implies 5r = 10 \implies r = 2\)
Given that the term independent of \(x\) is 405:
\(\binom{10}{2} a^2 = 405\)
\(45 a^2 = 405 \implies a^2 = \frac{405}{45} = 9 \implies a = \pm 3\)
In simple words: Find \(r\) by setting the exponent of \(x\) to zero, then equate the resulting term to 405.

Exam Tip: Do not forget both positive and negative roots when solving \(a^2 = 9\).

Q11. In the binomial expansion of \((1 + x)^n\), the coefficient of the fifth, sixth and seventh terms are in A.P. Find all values of n for which this can happen.
Answer: The coefficients of the fifth, sixth, and seventh terms are \(\binom{n}{4}\), \(\binom{n}{5}\), and \(\binom{n}{6}\).
Since they are in A.P., twice the middle term equals the sum of the outer terms:
\(2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6}\)
Divide by \(\binom{n}{5}\):
\(2 = \frac{\binom{n}{4}}{\binom{n}{5}} + \frac{\binom{n}{6}}{\binom{n}{5}}\)
Using the property \(\frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{r}{n-r+1}\) and \(\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}\):
\(2 = \frac{5}{n-4} + \frac{n-5}{6}\)
\(2 = \frac{30 + (n-5)(n-4)}{6(n-4)}\)
\(12(n-4) = 30 + n^2 - 9n + 20\)
\(12n - 48 = n^2 - 9n + 50\)
\(n^2 - 21n + 98 = 0\)
\((n - 7)(n - 14) = 0\)
Therefore, \(n = 7\) or \(n = 14\).
In simple words: Use the arithmetic progression condition \(2b = a + c\) and simplify using binomial coefficient ratios.

Q12. The \(3^{\text{rd}}\), \(4^{\text{th}}\) and \(5^{\text{th}}\) terms in the expansion of \((x + a)^n\) are respectively 84, 280 and 560, find the values of x, a and n.
Answer: Given:
\(T_3 = \binom{n}{2} x^{n-2} a^2 = 84\)   --- (1)
\(T_4 = \binom{n}{3} x^{n-3} a^3 = 280\)   --- (2)
\(T_5 = \binom{n}{4} x^{n-4} a^4 = 560\)   --- (3)
Dividing (2) by (1):
\(\frac{\binom{n}{3} x^{n-3} a^3}{\binom{n}{2} x^{n-2} a^2} = \frac{280}{84} \implies \frac{n-2}{3} \cdot \frac{a}{x} = \frac{10}{3} \implies \frac{a}{x} (n-2) = 10\)   --- (4)
Dividing (3) by (2):
\(\frac{\binom{n}{4} x^{n-4} a^4}{\binom{n}{3} x^{n-3} a^3} = \frac{560}{280} \implies \frac{n-3}{4} \cdot \frac{a}{x} = 2 \implies \frac{a}{x} (n-3) = 8\)   --- (5)
Dividing (4) by (5):
\(\frac{n-2}{n-3} = \frac{10}{8} = \frac{5}{4}\)
\(4(n - 2) = 5(n - 3) \implies 4n - 8 = 5n - 15 \implies n = 7\)
Substitute \(n = 7\) into (4):
\(\frac{a}{x} (7 - 2) = 10 \implies 5\frac{a}{x} = 10 \implies a = 2x\)
Substitute \(n = 7\) and \(a = 2x\) into (1):
\(\binom{7}{2} x^{5} (2x)^2 = 84\)
\(21 x^5 \cdot 4x^2 = 84 \implies 84 x^7 = 84 \implies x^7 = 1 \implies x = 1\)
Since \(x = 1\), \(a = 2(1) = 2\).
Thus, \(x = 1, a = 2, n = 7\).
In simple words: Divide consecutive term equations to eliminate variables and find \(n\), then substitute back to find \(x\) and \(a\).

Exam Tip: Taking ratios of consecutive terms is the standard technique to solve multi-variable binomial expansion problems.

Q13. If the coefficient of \((2r + 4)^{\text{th}}\) and \((r - 2)^{\text{th}}\) terms in the expansion of \((1 + x)^{18}\) are equal, find r.
Answer: The coefficient of the \(k^{\text{th}}\) term in \((1 + x)^n\) is \(\binom{n}{k-1}\).
Coefficient of \((2r + 4)^{\text{th}}\) term = \(\binom{18}{2r + 4 - 1} = \binom{18}{2r + 3}\)
Coefficient of \((r - 2)^{\text{th}}\) term = \(\binom{18}{r - 2 - 1} = \binom{18}{r - 3}\)
Since the coefficients are equal:
\(\binom{18}{2r + 3} = \binom{18}{r - 3}\)
We know that if \(\binom{n}{x} = \binom{n}{y}\), then either \(x = y\) or \(x + y = n\).
Case 1: \(2r + 3 = r - 3 \implies r = -6\) (Rejected, as term index must be positive).
Case 2: \((2r + 3) + (r - 3) = 18\)
\(3r = 18 \implies r = 6\)
Thus, \(r = 6\).
In simple words: Use the property that if two binomial coefficients are equal, their lower indices are either equal or sum up to the upper index \(n\).

Exam Tip: Always check both conditions for equality of binomial coefficients and reject non-integer or negative values.

Q14. Find the coefficient of \(a^4\) in the product \((1 + 2a)^4 (2 - a)^5\) using binomial theorem.
Answer: Expand each binomial using the binomial theorem:
\((1 + 2a)^4 = \binom{4}{0} + \binom{4}{1}(2a) + \binom{4}{2}(2a)^2 + \binom{4}{3}(2a)^3 + \binom{4}{4}(2a)^4\)
\(= 1 + 8a + 24a^2 + 32a^3 + 16a^4\)
\((2 - a)^5 = \binom{5}{0}(2)^5 - \binom{5}{1}(2)^4a + \binom{5}{2}(2)^3a^2 - \binom{5}{3}(2)^2a^3 + \binom{5}{4}(2)a^4 - \binom{5}{5}a^5\)
\(= 32 - 80a + 80a^2 - 40a^3 + 10a^4 - a^5\)
We need the coefficient of \(a^4\) in the product of these two polynomials. We multiply terms whose powers of \(a\) add up to 4:
1) Constant term from first \((1)\) × \(a^4\) term from second \((10a^4)\): \(1 \times 10 = 10\)
2) \(a\) term from first \((8a)\) × \(a^3\) term from second \((-40a^3)\): \(8 \times (-40) = -320\)
3) \(a^2\) term from first \((24a^2)\) × \(a^2\) term from second \((80a^2)\): \(24 \times 80 = 1920\)
4) \(a^3\) term from first \((32a^3)\) × \(a\) term from second \((-80a)\): \(32 \times (-80) = -2560\)
5) \(a^4\) term from first \((16a^4)\) × Constant term from second \((32)\): \(16 \times 32 = 512\)
Total coefficient of \(a^4\) = \(10 - 320 + 1920 - 2560 + 512 = -438\)
In simple words: Multiply matching pairs of terms from both expansions whose powers sum up to 4.

Exam Tip: Set up a systematic table or list of products to avoid missing any combination that makes \(a^4\).

Q15. The coefficients of three consecutive terms in the expansion of \((1 + x)^n\) be 76, 95 and 76, find n.
Answer: Let the three consecutive coefficients be \(\binom{n}{r-1} = 76\), \(\binom{n}{r} = 95\), and \(\binom{n}{r+1} = 76\).
Dividing the second by the first:
\(\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{95}{76} \implies \frac{n - r + 1}{r} = \frac{5}{4} \implies 4n - 9r = -4\)   --- (1)
Dividing the third by the second:
\(\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{76}{95} \implies \frac{n - r}{r + 1} = \frac{4}{5} \implies 5n - 9r = 4\)   --- (2)
Subtracting (1) from (2):
\(n = 8\)
Substituting \(n = 8\) into (1):
\(4(8) - 9r = -4 \implies 32 - 9r = -4 \implies 9r = 36 \implies r = 4\)
Thus, \(n = 8\).
In simple words: Take ratios of consecutive coefficients to set up linear equations in terms of \(n\) and \(r\).

Exam Tip: Ratios of consecutive binomial terms eliminate factorials quickly and make solving linear equations easy.

Q16. Using binomial theorem, prove that \(3^{2n+2} - 8n - 9\) is divisible by 64, for all natural numbers.
Answer: We have \(3^{2n+2} = 3^2 \cdot (3^2)^n = 9(9)^n = 9(1 + 8)^n\).
Using binomial theorem for \((1 + 8)^n\):
\((1 + 8)^n = 1 + \binom{n}{1}(8) + \binom{n}{2}(8)^2 + \binom{n}{3}(8)^3 + \dots + \binom{n}{n}(8)^n\)
Multiply by 9:
\(9(1 + 8)^n = 9 \left[ 1 + 8n + 64 \left( \binom{n}{2} + \binom{n}{3}(8) + \dots + \binom{n}{n}(8)^{n-2} \right) \right]\)
\(3^{2n+2} = 9 + 72n + 64k\) (where \(k\) is an integer)
Now consider the given expression:
\(3^{2n+2} - 8n - 9 = (9 + 72n + 64k) - 8n - 9\)
\(= 64n + 64k = 64(n + k)\)
Since \(64(n + k)\) is clearly a multiple of 64, the expression is divisible by 64 for all natural numbers \(n\).
In simple words: Write \(3^{2n+2}\) as \(9(1+8)^n\) and expand using the binomial theorem to isolate a factor of 64.

Exam Tip: For divisibility by \(64\), expand the base in the form of \((1 + 8)^n\) so that powers of 8 generate 64.

Q.1 Find the middle term in the expansion of (x - 1/6y)¹⁰.

Q.2 If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.

Q.3 Write the general term in the expansion of (x² – y)⁶.

Q.4 The sum of first two terms of a G.P. is -1/64 and the sum of first three terms is 6/34. What is the seventh term?

Q.5 In the expansion of (1 + x)³⁴, the coefficients of (2r+1)th and (r +2)th terms are equal, find r.

Q.6 Find the coefficient of a4 in the product (1+2a)⁴.(2-a)⁵, using binomial theorem.

Q.7 Find the value of (a² + √a² - 2)⁴ + (a² - √a² - 1)⁴

Q.8 Find an approximation of (0.99)5 using the first three terms of its expansion.

Q.9 Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.

Q.10 In the expansion of (1+a)^m+n, prove that coefficients of am and an are equal.

Q.11 Find the coefficient of x⁵y⁷ in the expansion of (x + 2y)¹².

Q.12 Find the general term in the expansion of (5x² - 1/6x)¹¹

Q.13 Find the number of terms in the expansion of [(x + y)³(x - y)³]².

Q.14 Find the rth term from the end in the expansion of (x+a)ⁿ.

Q.15 If the coefficients of T_r, T_r+1, T_r+2 terms of (1 + x)¹⁴ are in arithmetic progression, then find the value of r.

Q.16 If the coefficients of 7^th and 13^th terms in the expansion of (1 + x)n are equal then find the value of n.

Q.17 If the ratio of the coefficients of 3rd and 4th terms in the expansion of  (x - 1/2x)ⁿ is 1:2 then find the value of n.

Q.18 The coefficients of three consecutive terms in the expansion of (1+a)nare in the ratio 1:7:42. Find n.

Q.19 Prove that ⁿ∑_r=0 3^{r n} C_r = 4_n

Q.20 Show that the coefficients of the middle term in the expansion of (1+x)²ⁿ is equal to the sum of the coefficients of two middle terms in the expansion of (1+x)^2n-1.