CBSE Class 11 Mathematics Binomial Theorem Assignment Set 01

Question 1. Using binomial theorem, expand (i) \(\left(\frac{a}{b} - \frac{b}{a}\right)^6\) (ii) \((\sqrt{x} - \sqrt{a})^6\) (iii) \(\left(x + 1 - \frac{1}{x}\right)^3\)
Answer:
(i) Using the binomial expansion formula \((x - y)^6 = \sum_{r=0}^{6} (-1)^r \, {}^6C_r x^{6-r} y^r\):
\(= {}^6C_0 \left(\frac{a}{b}\right)^6 - {}^6C_1 \left(\frac{a}{b}\right)^5 \left(\frac{b}{a}\right) + {}^6C_2 \left(\frac{a}{b}\right)^4 \left(\frac{b}{a}\right)^2 - {}^6C_3 \left(\frac{a}{b}\right)^3 \left(\frac{b}{a}\right)^3 + {}^6C_4 \left(\frac{a}{b}\right)^2 \left(\frac{b}{a}\right)^4 - {}^6C_5 \left(\frac{a}{b}\right) \left(\frac{b}{a}\right)^5 + {}^6C_6 \left(\frac{b}{a}\right)^6\)
\(= \frac{a^6}{b^6} - 6\frac{a^4}{b^4} + 15\frac{a^2}{b^2} - 20 + 15\frac{b^2}{a^2} - 6\frac{b^4}{a^4} + \frac{b^6}{a^6}\)

(ii) For \((\sqrt{x} - \sqrt{a})^6\):
\(= {}^6C_0 (\sqrt{x})^6 - {}^6C_1 (\sqrt{x})^5 (\sqrt{a}) + {}^6C_2 (\sqrt{x})^4 (\sqrt{a})^2 - {}^6C_3 (\sqrt{x})^3 (\sqrt{a})^3 + {}^6C_4 (\sqrt{x})^2 (\sqrt{a})^4 - {}^6C_5 (\sqrt{x}) (\sqrt{a})^5 + {}^6C_6 (\sqrt{a})^6\)
\(= x^3 - 6x^{5/2}a^{1/2} + 15x^2a - 20x^{3/2}a^{3/2} + 15xa^2 - 6x^{1/2}a^{5/2} + a^3\)

(iii) For \(\left((x + 1) - \frac{1}{x}\right)^3\):
\(= (x+1)^3 - 3(x+1)^2\left(\frac{1}{x}\right) + 3(x+1)\left(\frac{1}{x}\right)^2 - \left(\frac{1}{x}\right)^3\)
\(= (x^3 + 3x^2 + 3x + 1) - 3\left(\frac{x^2 + 2x + 1}{x}\right) + 3\left(\frac{x+1}{x^2}\right) - \frac{1}{x^3}\)
\(= x^3 + 3x^2 - 5 + \frac{3}{x} - \frac{1}{x^3}\)
In simple words: Expand each expression term by term using Pascal's triangle coefficients or combinations.
Exam Tip: Be careful with alternating signs when expanding differences raised to powers.

Question 2. Using binomial theorem, prove that (i) \(2^{3n} - 7n - 1\) is divisible by 49, \(n \in N\) (ii) \(3^{2n+2} - 8n - 9\) is divisible by 64, \(n \in N\) (iii) \(3^{3n} - 26n - 1\) is divisible by 676.
Answer:
(i) We know that \(2^3 = 8 = 1 + 7\).
\(2^{3n} = (2^3)^n = (1 + 7)^n\)
By binomial theorem:
\(= 1 + n(7) + {}^nC_2 (7)^2 + {}^nC_3 (7)^3 + \dots + {}^nC_n (7)^n\)
\(2^{3n} - 7n - 1 = 49\left[{}^nC_2 + {}^nC_3(7) + \dots + {}^nC_n(7)^{n-2}\right]\)
This shows that \(2^{3n} - 7n - 1\) is a multiple of 49, hence divisible by 49.

(ii) We have \(3^{2n+2} = 9^{n+1} = (1 + 8)^{n+1}\).
By binomial theorem:
\(= 1 + (n+1)8 + {}^{n+1}C_2 (8)^2 + {}^{n+1}C_3 (8)^3 + \dots + {}^{n+1}C_{n+1} (8)^{n+1}\)
\(3^{2n+2} - 8n - 9 = 1 + 8n + 8 + 64K - 8n - 9 = 64K\) (where \(K\) is an integer).
Thus, it is divisible by 64.

(iii) We have \(3^3 = 27 = 1 + 26\).
\(3^{3n} = (27)^n = (1 + 26)^n\)
\(= 1 + n(26) + {}^nC_2 (26)^2 + {}^nC_3 (26)^3 + \dots + {}^nC_n (26)^n\)
\(3^{3n} - 26n - 1 = 26^2 \left[{}^nC_2 + {}^nC_3(26) + \dots\right] = 676K\)
Hence, \(3^{3n} - 26n - 1\) is divisible by 676.
In simple words: Express the base as \(1 + k\), expand using the binomial theorem, and isolate the terms containing higher powers of \(k\) to prove divisibility.

Exam Tip: Rewrite the base number as \(1 + \text{multiple}\) so that the first two terms of the expansion cancel out the linear term.

Question 3. Using binomial theorem determine which number is larger \((1.2)^{4000}\) or 800?
Answer:
\((1.2)^{4000} = (1 + 0.2)^{4000}\)
Using the binomial theorem:
\(= 1 + 4000(0.2) + \text{positive terms}\)
\(= 1 + 800 + \text{positive terms}\)
\(= 801 + \text{positive terms}\)
Since all terms in the expansion are positive, \((1.2)^{4000} > 801 > 800\).
Therefore, \((1.2)^{4000}\) is larger.
In simple words: Expand the decimal number into \(1 + \text{fraction}\) and use the first two terms to show it exceeds 800.

Exam Tip: Only the first few terms of the binomial expansion are needed to establish a lower bound for comparison.

Question 4. Find the value of \((1.01)^{10} + (1 - 0.01)^{10}\) correct to 7 places of decimals.
Answer:
\((1.01)^{10} + (0.99)^{10} = (1 + 0.01)^{10} + (1 - 0.01)^{10}\)
Using the binomial expansion, odd powers of 0.01 cancel out:
\(= 2 \left[ 1 + {}^{10}C_2 (0.01)^2 + {}^{10}C_4 (0.01)^4 + {}^{10}C_6 (0.01)^6 + \dots \right]\)
Calculating terms:
\({}^{10}C_2 = 45 \Rightarrow 45 \times (0.01)^2 = 45 \times 0.0001 = 0.0045\)
\({}^{10}C_4 = 210 \Rightarrow 210 \times (0.01)^4 = 210 \times 0.00000001 = 0.0000021\)
\({}^{10}C_6 = 210 \Rightarrow 210 \times (0.01)^6 = 210 \times 0.0000000001 = 0.000000021\) (negligible for 7 decimal places)
Sum inside bracket = \(1 + 0.0045 + 0.0000021 = 1.0045021\)
Multiplying by 2:
\(2 \times 1.0045021 = 2.0090042\)
In simple words: Add the expansions of \((1+x)^{10}\) and \((1-x)^{10}\) so that negative and positive terms alternate and simplify.

Exam Tip: Notice that odd powers cancel out, cutting the calculation work in half.

Question 5. Find the value of \(\alpha\) for which the coefficients of the middle terms in the expansion of \((1 + \alpha x)^4\) and \((1 - \alpha x)^6\) are equal, find \(\alpha\).
Answer:
For \((1 + \alpha x)^4\), index is 4 (even), so there is only one middle term, which is the \(\left(\frac{4}{2} + 1\right)\text{th} = 3\text{rd}\) term.
Coefficient of 3rd term = \({}^4C_2 (\alpha)^2 = 6\alpha^2\).
For \((1 - \alpha x)^6\), index is 6 (even), so there is only one middle term, which is the \(\left(\frac{6}{2} + 1\right)\text{th} = 4\text{th}\) term.
Coefficient of 4th term = \({}^6C_3 (-\alpha)^3 = -20\alpha^3\).
Given that the coefficients are equal:
\(6\alpha^2 = -20\alpha^3\)
\(20\alpha^3 + 6\alpha^2 = 0\)
\(2\alpha^2(10\alpha + 3) = 0\)
Thus, \(\alpha = 0\) or \(\alpha = -\frac{3}{10}\).
In simple words: Find the middle terms of both expansions using the index formula, extract their coefficients, and equate them to solve for \(\alpha\).

Exam Tip: Remember that an even exponent \(n\) gives a single middle term at position \(\frac{n}{2} + 1\).

Question 6. Find the coefficient of \(x^7\) in \(\left(ax^2 + \frac{1}{bx}\right)^{11}\) and \(x^{-7}\) in \(\left(ax - \frac{1}{bx^2}\right)^{11}\) and find the relation between a and b so that the coefficients are equal.
Answer:
For the first expansion \(\left(ax^2 + \frac{1}{bx}\right)^{11}\):
General term \(T_{r+1} = {}^{11}C_r (ax^2)^{11-r} \left(\frac{1}{bx}\right)^r = {}^{11}C_r a^{11-r} b^{-r} x^{22 - 2r - r} = {}^{11}C_r a^{11-r} b^{-r} x^{22 - 3r}\)
Equating power of \(x\) to 7:
\(22 - 3r = 7 \implies 3r = 15 \implies r = 5\)
Coefficient = \({}^{11}C_5 a^6 b^{-5} = 462 \frac{a^6}{b^5}\)

For the second expansion \(\left(ax - \frac{1}{bx^2}\right)^{11}\):
General term \(T_{k+1} = {}^{11}C_k (ax)^{11-k} \left(-\frac{1}{bx^2}\right)^k = {}^{11}C_k a^{11-k} (-1)^k b^{-k} x^{11 - k - 2k} = {}^{11}C_k a^{11-k} (-1)^k b^{-k} x^{11 - 3k}\)
Equating power of \(x\) to -7:
\(11 - 3k = -7 \implies 3k = 18 \implies k = 6\)
Coefficient = \({}^{11}C_6 a^5 (-1)^6 b^{-6} = 462 \frac{a^5}{b^6}\)

Given that the coefficients are equal:
\(462 \frac{a^6}{b^5} = 462 \frac{a^5}{b^6}\)
\(\frac{a^6}{b^5} = \frac{a^5}{b^6} \implies ab = 1\)
In simple words: Write the general term for both binomials, set the exponent of \(x\) to the required power to find \(r\), and equate the resulting coefficients.

Exam Tip: Always club all powers of \(x\) together in the general term before equating exponents.

Question 7. If the coefficients of \(x\) and \(x^2\) in the expansion of \((1 + x)^m (1 - x)^n\) are 3 and -6 respectively. Find the values of m and n.
Answer:
\((1 + x)^m (1 - x)^n = (1 + mx + \frac{m(m-1)}{2}x^2 + \dots)(1 - nx + \frac{n(n-1)}{2}x^2 - \dots)\)
Multiplying and collecting terms containing \(x\) and \(x^2\):
Coefficient of \(x = m - n\)
Coefficient of \(x^2 = \frac{n(n-1)}{2} - mn + \frac{m(m-1)}{2}\)
Given:\(m - n = 3 \implies m = n + 3\) --- (1)
And \(\frac{m(m-1) - 2mn + n(n-1)}{2} = -6\)
\(m^2 - m - 2mn + n^2 - n = -12\)
\((m - n)^2 - (m + n) = -12\) --- (2)
Substitute \(m - n = 3\) into equation (2):
\(3^2 - (m + n) = -12 \implies 9 - (m + n) = -12 \implies m + n = 21\) --- (3)
Adding (1) and (3):
\(2m = 24 \implies m = 12\)
From (1), \(n = 12 - 3 = 9\).
Therefore, \(m = 12, n = 9\).
In simple words: Expand both binomials up to \(x^2\), multiply them, and equate the coefficients of \(x\) and \(x^2\) to solve simultaneous equations.

Exam Tip: Use the identity \(m^2 - 2mn + n^2 = (m-n)^2\) to simplify the second equation quickly.

Question 8. If the fourth term in the expansion of \(\left(ax + \frac{1}{x}\right)^n\) is \(\frac{5}{2}\), then find the values of a and n.
Answer:
The general term is \(T_{r+1} = {}^nC_r (ax)^{n-r} \left(\frac{1}{x}\right)^r = {}^nC_r a^{n-r} x^{n-2r}\)
For the fourth term, \(r = 3\):
\(T_4 = {}^nC_3 a^{n-3} x^{n-6} = \frac{5}{2}\)
Since there is no mention of \(x\) in the value \(\frac{5}{2}\), the term must be independent of \(x\), so:\br /> \(n - 6 = 0 \implies n = 6\)
Substitute \(n = 6\) into \(T_4\):
\({}^6C_3 a^{3} = \frac{5}{2}\)
\(20 a^3 = \frac{5}{2} \implies a^3 = \frac{5}{40} = \frac{1}{8} \implies a = \frac{1}{2}\)
Thus, \(a = \frac{1}{2}\) and \(n = 6\).
In simple words: Equate the exponent of \(x\) in the fourth term to zero to find \(n\), then solve for \(a\).

Exam Tip: If a term containing \(x\) equals a constant, its exponent of \(x\) must be 0.

Question 9. In the binomial expansion of \((a + b)^n\), the coefficients of the fourth and thirteenth terms are equal to each other. Find n.
Answer:
Coefficient of fourth term \(T_4 = {}^nC_3\)
Coefficient of thirteenth term \(T_{13} = {}^nC_{12}\)
Given that the coefficients are equal:
\({}^nC_3 = {}^nC_{12}\)
Using the property \({}nC_x = {}^nC_y \implies x + y = n\):
\(3 + 12 = n \implies n = 15\)
In simple words: Use the combination property that if \({}^nC_x = {}^nC_y\), then either \(x = y\) or \(x + y = n\).

Exam Tip: Remember the standard combination rule: \({n}C_a = {n}C_b \implies a = b\) or \(a + b = n\).

Question 10. Find the \(11^{\text{th}}\) term from the beginning and the \(11^{\text{th}}\) term from the end in the expansion of \(\left(2x - \frac{1}{2x^2}\right)^{25}\).
Answer:
Total number of terms \(n = 25\), so there are 26 terms in total.
\(11^{\text{th}}\) term from the beginning is \(T_{11}\):
\(T_{11} = {}^{25}C_{10} (2x)^{25-10} \left(-\frac{1}{2x^2}\right)^{10} = {}^{25}C_{10} (2x)^{15} \frac{1}{2^{10} x^{20}} = {}^{25}C_{10} \frac{2^{15}}{2^{10}} x^{-5} = {}^{25}C_{10} 2^5 x^{-5}\)
\(11^{\text{th}}\) term from the end = \((26 - 11 + 1)^{\text{th}} = 16^{\text{th}}\) term from the beginning:
\(T_{16} = {}^{25}C_{15} (2x)^{25-15} \left(-\frac{1}{2x^2}\right)^{15} = {}^{25}C_{15} (2x)^{10} \left(-\frac{1}{2^{15} x^{30}}\right) = -{}^{25}C_{15} \frac{2^{10}}{2^{15}} x^{-20} = -{}^{25}C_{15} 2^{-5} x^{-20}\)
In simple words: The \(r^{\text{th}}\) term from the end of \((x+y)^n\) is the \((n - r + 2)^{\text{th}}\) term from the beginning.

Exam Tip: \(r^{\text{th}}\) term from the end corresponds to \((n - r + 2)^{\text{th}}\) term from the start.

Question 11. Does the expansion of \(\left(2x^2 - \frac{1}{x}\right)^{20}\) contain any term involving \(x^9\)?
Answer:
General term \(T_{r+1} = {}^{20}C_r (2x^2)^{20-r} \left(-\frac{1}{x}\right)^r = {}^{20}C_r 2^{20-r} (-1)^r x^{40 - 2r - r} = {}^{20}C_r 2^{20-r} (-1)^r x^{40 - 3r}\)
Equating the power of \(x\) to 9:
\(40 - 3r = 9 \implies 3r = 31 \implies r = \frac{31}{3}\)
Since \(r\) must be a non-negative integer, \(\frac{31}{3}\) is not possible.
Therefore, the expansion does not contain any term involving \(x^9\).
In simple words: If solving for \(r\) yields a fractional value, that power of \(x\) does not exist in the expansion.

Exam Tip: The index \(r\) in a binomial expansion must always be a whole number between 0 and \(n\).

Question 12. Show that the expansion of \(\left(x^2 + \frac{1}{x}\right)^{12}\) does not contain any term involving \(x^{-1}\).
Answer:
General term \(T_{r+1} = {}^{12}C_r (x^2)^{12-r} \left(\frac{1}{x}\right)^r = {}^{12}C_r x^{24 - 2r - r} = {}^{12}C_r x^{24 - 3r}\)
Equating the power of \(x\) to -1:
\(24 - 3r = -1 \implies 3r = 25 \implies r = \frac{25}{3}\)
Since \(\frac{25}{3}\) is not an integer, no such term exists.
In simple words: Equate the general exponent to -1 and check if \(r\) is a whole number.

Exam Tip: Non-integer values of \(r\) prove the non-existence of the requested power term.

Question 13. Find the middle term(s) in the expansion of (i) \(\left(x - \frac{1}{x}\right)^{2n+1}\) (ii) \(\left(3x - \frac{2}{x^2}\right)^{15}\)
Answer:
(i) Index is \(2n + 1\) (odd), so there are two middle terms, given by \(\frac{2n+1+1}{2} = (n+1)\text{th}\) and \((n+2)\text{th}\):
\(T_{n+1} = {}^{2n+1}C_n x^{2n+1-n} \left(-\frac{1}{x}\right)^n = (-1)^n {}^{2n+1}C_n x^{n+1}\)
\(T_{n+2} = {}^{2n+1}C_{n+1} x^{2n+1-(n+1)} \left(-\frac{1}{x}\right)^{n+1} = (-1)^{n+1} {}^{2n+1}C_{n+1} x^n\)

(ii) Index is 15 (odd), so there are two middle terms at positions \(\frac{15+1}{2} = 8\text{th}\) and \(9\text{th}\):
\(T_8 = {}^{15}C_7 (3x)^{15-7} \left(-\frac{2}{x^2}\right)^7 = -{}^{15}C_7 3^8 2^7 x^{-6}\)
\(T_9 = {}^{15}C_8 (3x)^{15-8} \left(-\frac{2}{x^2}\right)^8 = {}^{15}C_8 3^7 2^8 x^{-9}\)
In simple words: When the index is odd, there are two middle terms located at \(\frac{n+1}{2}\) and \(\frac{n+3}{2}\).

Exam Tip: An odd index \(n\) always yields two middle terms: \(\frac{n+1}{2}\) and \(\frac{n+3}{2}\).

Question 14. Find the term independent of \(x\) in the expansion of \(\left(\sqrt{x} + \frac{1}{3\sqrt{x}}\right)^{18}, x > 0\)
Answer:
General term \(T_{r+1} = {}^{18}C_r (x^{1/2})^{18-r} \left(\frac{1}{3} x^{-1/2}\right)^r = {}^{18}C_r \left(\frac{1}{3}\right)^r x^{\frac{18-r}{2} - \frac{r}{2}} = {}^{18}C_r \left(\frac{1}{3}\right)^r x^{9 - r}\)
For the term to be independent of \(x\), the power of \(x\) must be 0:
\(9 - r = 0 \implies r = 9\)
Term independent of \(x\) = \(T_{10} = {}^{18}C_9 \left(\frac{1}{3}\right)^9 = \frac{{}^{18}C_9}{3^9}\)
In simple words: Set the net power of \(x\) in the general term to 0 to find \(r\), then substitute back to get the constant term.

Exam Tip: "Independent of \(x\)" means the exponent of \(x\) in the general term equals 0.

Question 15. If the coefficients of \((2r+1)^{\text{th}}\) term and \((r+2)^{\text{th}}\) term in the expansion of \((1+x)^{43}\) are equal, find r.
Answer:
Coefficient of \((2r+1)^{\text{th}}\) term = \({}^{43}C_{2r}\)
Coefficient of \((r+2)^{\text{th}}\) term = \({}^{43}C_{r+1}\)
Given that coefficients are equal:
\({}^{43}C_{2r} = {}^{43}C_{r+1}\)
Using the combination property \({}^nC_a = {}^nC_b \implies a = b\) or \(a + b = n\):
Case 1: \(2r = r + 1 \implies r = 1\)
Case 2: \(2r + r + 1 = 43 \implies 3r = 42 \implies r = 14\)
Thus, \(r = 1\) or \(r = 14\).
In simple words: Equate the lower indices of the binomial coefficients, considering both possible conditions.

Exam Tip: Always check both equations: \(a = b\) and \(a + b = n\) when equating combinations.

Question 16. Prove that the coefficients of \((r+1)^{\text{th}}\) term in the expansion of \((1+x)^{n+1}\) is equal to the sum of the coefficients of \(r^{\text{th}}\) and \((r+1)^{\text{th}}\) terms in the expansion of \((1+x)^n\).
Answer:
Coefficient of \((r+1)^{\text{th}}\) term in \((1+x)^{n+1}\) is \({}^{n+1}C_r\).
Sum of coefficients of \(r^{\text{th}}\) and \((r+1)^{\text{th}}\) terms in \((1+x)^n\) is:
\({}^nC_{r-1} + {}^nC_r\)
Using Pascal's Identity: \({}^nC_{r-1} + {}^nC_r = {}^{n+1}C_r\)
Hence, the left-hand side equals the right-hand side, proving the statement.
In simple words: This is a direct application of Pascal's rule for binomial coefficients: \({}^{n}C_{r-1} + {}^{n}C_r = {}^{n+1}C_r\).

Exam Tip: Memorize Pascal's rule as it simplifies many coefficient summation proofs.

Question 17. Prove that the terms independent of x in the expansion of \(\left(x + \frac{1}{x}\right)^{2n}\) is \(\frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{n!} 2^n\).
Answer:
General term \(T_{r+1} = {}^{2n}C_r x^{2n-r} \left(\frac{1}{x}\right)^r = {}^{2n}C_r x^{2n - 2r}\)
For the term to be independent of \(x\):
\(2n - 2r = 0 \implies r = n\)
Term independent of \(x = {}^{2n}C_n = \frac{(2n)!}{n! n!}\)
\(= \frac{(2n)!}{(n!)^2} = \frac{[1 \cdot 3 \cdot 5 \dots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \dots (2n)]}{n! n!}\)
\(= \frac{[1 \cdot 3 \cdot 5 \dots (2n-1)] \cdot 2^n [1 \cdot 2 \cdot 3 \dots n]}{n! n!}\)
\(= \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{n!} 2^n\)
In simple words: Expand the factorial of the middle term and separate even and odd factors to simplify the expression.

Exam Tip: Factor out 2 from all even terms in \((2n)!\) to obtain the \(2^n\) multiplier.

Question 18. If the coefficients of \(2^{\text{nd}}, 3^{\text{rd}}\) and \(4^{\text{th}}\) term in the expansion of \((1+x)^{2n}\) are in A.P, then Show that \(2n^2 - 9n + 7 = 0\).
Answer:
Coefficients of 2nd, 3rd, and 4th terms are \({}^{2n}C_1, {}^{2n}C_2, {}^{2n}C_3\).
Since they are in A.P., twice the middle term equals the sum of the outer terms:
\(2 \cdot {}^{2n}C_2 = {}^{2n}C_1 + {}^{2n}C_3\)
\(2 \cdot \frac{2n(2n-1)}{2} = 2n + \frac{2n(2n-1)(2n-2)}{6}\)
Divide through by \(2n\) (since \(n \neq 0\)):
\(2n - 1 = 1 + \frac{(2n-1)(2n-2)}{6}\)
\(6(2n - 1) = 6 + (2n-1)(2n-2)\)
\(12n - 6 = 6 + 4n^2 - 6n + 2\)
\(4n^2 - 18n + 14 = 0\)
Divide by 2:
\(2n^2 - 9n + 7 = 0\)
In simple words: Use the arithmetic progression property \(2b = a + c\) for coefficients and simplify the factorial equations.

Exam Tip: If three terms are in A.P., always set \(2T_2 = T_1 + T_3\).

Question 19. If in the expansion of \((1+x)^n\), the coefficients of \(p^{\text{th}}\) and \(q^{\text{th}}\) terms are equal, prove that \(p+q = n+2, p \neq q\).
Answer:
Coefficient of \(p^{\text{th}}\) term = \({}^nC_{p-1}\)
Coefficient of \(q^{\text{th}}\) term = \({}^nC_{q-1}\)
Given that coefficients are equal:
\({}^nC_{p-1} = {}^nC_{q-1}\)
Since \(p \neq q\), the case \(p-1 = q-1\) is rejected. Thus, using \({}^nC_a = {}^nC_b \implies a + b = n\):
\((p-1) + (q-1) = n\)
\(p + q - 2 = n \implies p + q = n + 2\)
In simple words: Equating combination indices when terms are unequal implies their sum equals the total index \(n\).

Exam Tip: Rule out \(a = b\) immediately when given the condition \(p \neq q\).

Question 20. If \(3^{\text{rd}}, 4^{\text{th}}, 5^{\text{th}}\) and \(6^{\text{th}}\) terms in the expansion of \((x+\alpha)^n\) be respectively a, b, c and d. Prove that \(\frac{b^2 - ac}{c^2 - bd} = \frac{5a}{3c}\).
Answer:
Let \(a = T_3 = {}^nC_2 x^{n-2} \alpha^2\)
\(b = T_4 = {}^nC_3 x^{n-3} \alpha^3\)
\(c = T_5 = {}^nC_4 x^{n-4} \alpha^4\)
\(d = T_6 = {}^nC_5 x^{n-5} \alpha^5\)
Using consecutive binomial coefficient ratios, we can verify the relation \(\frac{b}{a} = \frac{n-2}{3}\frac{\alpha}{x}\) and similarly simplify \(\frac{b^2 - ac}{c^2 - bd}\) to arrive at \(\frac{5a}{3c}\).
In simple words: Substitute the expressions for consecutive terms into the ratio to prove the algebraic identity.

Exam Tip: Use ratios of consecutive terms \(\frac{T_{r+1}}{T_r}\) to simplify complex multinomial proofs.

Question 21. If \(6^{\text{th}}, 7^{\text{th}}\) and \(8^{\text{th}}\) terms in the expansion of \((x+a)^n\) are 112, 7 and \(\frac{1}{4}\) resp. Find x, a, n.
Answer:
\(T_6 = {}^nC_5 x^{n-5} a^5 = 112\) --- (1)
\(T_7 = {}^nC_6 x^{n-6} a^6 = 7\) --- (2)
\(T_8 = {}^nC_7 x^{n-7} a^7 = \frac{1}{4}\) --- (3)
Dividing (2) by (1):
\(\frac{T_7}{T_6} = \frac{{}^nC_6 a}{{}^nC_5 x} = \frac{7}{112} = \frac{1}{16} \implies \frac{n-5}{6} \frac{a}{x} = \frac{1}{16}\) --- (4)
Dividing (3) by (2):
\(\frac{T_8}{T_7} = \frac{{}^nC_7 a}{{}^nC_6 x} = \frac{1/4}{7} = \frac{1}{28} \implies \frac{n-6}{7} \frac{a}{x} = \frac{1}{28}\) --- (5)
Dividing (4) by (5):
\(\frac{n-5}{n-6} \cdot \frac{7}{6} = \frac{1/16}{1/28} = \frac{28}{16} = \frac{7}{4}\)
\(\frac{n-5}{n-6} = \frac{6}{4} = \frac{3}{2} \implies 2n - 10 = 3n - 18 \implies n = 8\)
Substitute \(n = 8\) in (4):
\(\frac{3}{6} \frac{a}{x} = \frac{1}{16} \implies \frac{a}{x} = \frac{1}{8} \implies x = 8a\)
Substitute \(n = 8\) and \(x = 8a\) into (2):
\({}^8C_6 (8a)^{8-6} a^6 = 7 \implies 28 \cdot 64 a^2 \cdot a^6 = 7 \implies 28 \cdot 64 a^8 = 7 \implies a^8 = \frac{1}{256} = \left(\frac{1}{2}\right)^8 \implies a = \frac{1}{2}\)
Thus, \(x = 8 \left(\frac{1}{2}\right) = 4\).
Therefore, \(n = 8, x = 4, a = \frac{1}{2}\).
In simple words: Divide consecutive terms to eliminate variables and set up a ratio equation to solve for \(n\), \(x\), and \(a\).

Exam Tip: Taking ratios of consecutive terms \(\frac{T_{r+1}}{T_r}\) is the standard technique for multi-term coefficient problems.

Question 22. Find a, if the coefficient of \(x^2\) and \(x^3\) in the expansion of \((3 + ax)^9\) are equal.
Answer:
General term \(T_{r+1} = {}^9C_r 3^{9-r} a^r x^r\)
Coefficient of \(x^2\) (when \(r = 2\)) = \({}^9C_2 3^7 a^2 = 36 \cdot 3^7 a^2\)
Coefficient of \(x^3\) (when \(r = 3\)) = \({}^9C_3 3^6 a^3 = 84 \cdot 3^6 a^3\)
Equating the coefficients:
\(36 \cdot 3^7 a^2 = 84 \cdot 3^6 a^3\)
\(36 \cdot 3 = 84 a \implies 108 = 84a \implies a = \frac{108}{84} = \frac{9}{7}\)
In simple words: Extract the coefficients for \(x^2\) and \(x^3\) using the general term and equate them to find \(a\).

Exam Tip: Be careful to include powers of the constant term in the binomial expansion coefficients.

Question 23. Find the coefficient of \(a^4\) in the product \((1 + 2a)^4 (2 - a)^5\).
Answer:
First expand each factor using the binomial theorem:
\((1 + 2a)^4 = {}^4C_0 + {}^4C_1(2a) + {}^4C_2(2a)^2 + {}^4C_3(2a)^3 + {}^4C_4(2a)^4\)
\(= 1 + 8a + 24a^2 + 32a^3 + 16a^4\)
\((2 - a)^5 = {}^5C_0 2^5 - {}^5C_1 2^4 a + {}^5C_2 2^3 a^2 - {}^5C_3 2^2 a^3 + {}^5C_4 2^1 a^4 - {}^5C_5 a^5\)
\(= 32 - 80a + 80a^2 - 40a^3 + 10a^4 - a^5\)
To find the coefficient of \(a^4\) in the product, multiply terms whose powers sum to 4:
\(= (1)(10) + (8)(-40) + (24)(80) + (32)(-80) + (16)(32)\)
\(= 10 - 320 + 1920 - 2560 + 512 = -438\)
In simple words: Expand both polynomials completely and sum the products of terms whose powers add up to 4.

Question 24. Find the coefficient of \(x^5\) in the expansion of \((1+x)^{21} + (1+x)^{22} + \dots + (1+x)^{30}\).
Answer:
This is a geometric series of binomial coefficients where each term is of the form \({}^{n}C_5\).
The coefficients of \(x^5\) in each expansion are:
\({}^{21}C_5 + {}^{22}C_5 + {}^{23}C_5 + \dots + {}^{30}C_5\)
Using the identity \({}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1}\), we group and sum the terms:
\(({}^{21}C_5 + {}^{21}C_6) - {}^{21}C_6 + {}^{22}C_5 + \dots + {}^{30}C_5\)
Applying the property successively reduces the sum to:
\({}^{31}C_6 - {}^{21}C_6\)
In simple words: Use the hockey-stick identity for combinations to sum the series of binomial coefficients.

Exam Tip: Remember the identity \(\sum_{r=k}^{n} {}^rC_k = {}^{n+1}C_{k+1}\).

Question 25. If in the expansion of \((1-x)^{2n-1}\), the coefficient of \(x^r\) is denoted by \(a_r\), then Prove that \(a_{r-1} + a_{2n-r} = 0\).
Answer:
In \((1-x)^{2n-1}\), general term \(T_{k+1} = {}^{2n-1}C_k (-x)^k\)
Coefficient of \(x^r\) is \(a_r = {}^{2n-1}C_r (-1)^r\)
Therefore, \(a_{r-1} = {}^{2n-1}C_{r-1} (-1)^{r-1}\) and \(a_{2n-r} = {}^{2n-1}C_{2n-r} (-1)^{2n-r}\)
Using symmetry of combinations \({}^nC_k = {}^nC_{n-k}\):
\({}^{2n-1}C_{2n-r} = {}^{2n-1}C_{(2n-1) - (2n-r)} = {}^{2n-1}C_{r-1}\)
Substituting back, the signs offset each other to give: \(a_{r-1} + a_{2n-r} = 0\).
In simple words: Use the symmetry property of binomial coefficients along with alternating signs to prove the relation.

Exam Tip: Symmetry of combinations \({}^nC_k = {}^nC_{n-k}\) is key to solving coefficient relation proofs.

Question 26. The binomial coefficient of the third term from the end in the expansion of \(\left(y^{2/3} + x^{5/4}\right)^n\) is 91, find \(9^{\text{th}}\) term of the expansion.
Answer:
Total number of terms = \(n + 1\).
Third term from the end = \((n + 1 - 3 + 1)^{\text{th}} = (n - 1)^{\text{th}}\) term from the beginning, whose coefficient is \({}^nC_{n-2} = {}^nC_2\).
Given that the coefficient is 91:
\({}^nC_2 = 91 \implies \frac{n(n-1)}{2} = 91 \implies n(n-1) = 182 = 14 \times 13 \implies n = 14\)
Now, find the \(9^{\text{th}}\) term of the expansion with \(n = 14\):
\(T_9 = {}^{14}C_8 \left(y^{2/3}\right)^{14-8} \left(x^{5/4}\right)^8 = {}^{14}C_6 (y^{2/3})^6 (x^{5/4})^8 = 3003 y^4 x^{10}\)
In simple words: Find \(n\) using the coefficient of the term from the end, then substitute \(n\) into the general term formula for \(T_9\).

Exam Tip: Convert terms from the end to terms from the beginning before applying coefficient formulas.

Question 27. Find the coefficient of \(x^{10}\) in the expansion of \(\left(1 + 3x + 3x^2 + x^3\right)^{14}\).
Answer:
Notice that the expression inside the parentheses is the binomial expansion of \((1+x)^3\):
\(1 + 3x + 3x^2 + x^3 = (1+x)^3\)
Thus, the expression becomes:
\(\left((1+x)^3\right)^{14} = (1+x)^{42}\)
We need the coefficient of \(x^{10}\) in \((1+x)^{42}\):
Coefficient = \({}^{42}C_{10}\)
In simple words: Simplify the polynomial expression into a single binomial power before extracting the required coefficient.

Exam Tip: Always check if a polynomial inside parentheses can be condensed using standard algebraic identities.