CBSE Class 11 Mathematics Binomial Theorem Assignment Set 04

CHAPTER 8

BINOMIAL THEOREM

Binomial theorem for any positive integer n
\((a+b)^n = {}^nC_0 a^n + {}^nC_1 a^{n-1}b + {}^nC_2 a^{n-2}b^2 + {}^nC_3 a^{n-3}b^3 + \ldots \ldots + {}^nC_n b^n\)

Recall
1) \({}^nC_r = \dfrac{n!}{(n-r)! r!}\)
2) \({}^nC_r = {}^nC_{n-r}\)
\({}^7C_4 = {}^7C_3 = \dfrac{7 \times 6 \times 5}{1 \times 2 \times 3} = 35\)
\({}^8C_6 = {}^8C_2 = \dfrac{8 \times 7}{1 \times 2} = 28\)
3) \({}^nC_n = {}^nC_0 = 1\)
4) \({}^nC_1 = n\)

OBSERVATIONS / FORMULAS
1) The coefficients \({}^nC_r\) occurring in the binomial theorem are known as binomial coefficients.
2) There are \(n+1\) terms in the expansion of \((a+b)^n\), i.e., one more than the index.
3) The coefficients of the terms equidistant from the beginning and end are equal.
4) \((1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + {}^nC_3 x^3 + \ldots \ldots + {}^nC_n x^n\) (By putting \(a = 1\) and \(b = x\) in the expansion of \((a+b)^n\)).
5) \((1-x)^n = {}^nC_0 - {}^nC_1 x + {}^nC_2 x^2 - {}^nC_3 x^3 + \ldots \ldots + (-1)^n {}^nC_n x^n\) (By putting \(a = 1\) and \(b = -x\) in the expansion of \((a+b)^n\)).
6) \(2^n = {}^nC_0 + {}^nC_1 + {}^nC_2 + {}^nC_3 + \ldots \ldots + {}^nC_n\) (By putting \(x = 1\) in formula (4))
7) \(0 = {}^nC_0 - {}^nC_1 + {}^nC_2 - {}^nC_3 + \ldots \ldots + (-1)^n {}^nC_n\) (By putting \(x = 1\) in formula (5))
8) The \((r+1)^{\text{th}}\) term in the binomial expansion for \((a+b)^n\) is called the general term, which is given by
\(T_{r+1} = {}^nC_r a^{n-r} b^r\)
i.e., to find the \(4^{\text{th}}\) term \(T_4\), substitute \(r = 3\).

9) Middle term in the expansion of \((a+b)^n\):
i) If \(n\) is even, middle term = \(\left(\frac{n}{2} + 1\right)^{\text{th}}\) term.
ii) If \(n\) is odd, then there are 2 middle terms, which are \(\left(\frac{n+1}{2}\right)^{\text{th}}\) term and \(\left(\frac{n+1}{2} + 1\right)^{\text{th}}\) term.

10) To find the term independent of \(x\) or the constant term, find the coefficient of \(x^0\) (i.e., put the power of \(x\) equal to 0 and find \(r\)).

EXTRA / HOT QUESTIONS

Question 1. Using Binomial theorem show that \(2^{3n} - 7n - 1\) or \(8^n - 7n - 1\) is divisible by 49 where n is a natural number.
Answer: We need to show that \(8^n - 7n - 1\) is divisible by 49 for all positive integers \(n\).
Using binomial expansion for \(8^n = (1+7)^n\):
\(8^n = (1+7)^n = {}^nC_0 + {}^nC_1(7) + {}^nC_2(7)^2 + {}^nC_3(7)^3 + \ldots + {}^nC_n(7)^n\)
\(8^n = 1 + 7n + 7^2 \left[ {}^nC_2 + {}^nC_3(7) + \ldots + {}^nC_n(7)^{n-2} \right]\)
\(8^n - 7n - 1 = 49 \left[ {}^nC_2 + {}^nC_3(7) + \ldots + {}^nC_n(7)^{n-2} \right]\)
Since 49 is a factor of the resulting expression, \(8^n - 7n - 1\) is divisible by 49.
In simple words: Expanding \((1+7)^n\) reveals terms with powers of 7 starting from \(7^2 = 49\), proving full divisibility.

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

Question 2. Find the coefficient of \(x^3\) in the expansion of \((1+2x)^6 (1-x)^7\).
Answer: Let us expand each binomial using the general term formula \(T_{r+1} = {}^nC_r a^{n-r} b^r\).
\((1+2x)^6 = \sum_{r=0}^{6} {}^6C_r (2x)^r = {}^6C_0 + {}^6C_1(2x) + {}^6C_2(2x)^2 + {}^6C_3(2x)^3 + \ldots\)
\((1-x)^7 = \sum_{k=0}^{7} {}^7C_k (-x)^k = {}^7C_0 - {}^7C_1 x + {}^7C_2 x^2 - {}^7C_3 x^3 + \ldots\)
We need the coefficient of \(x^3\) in the product of these two expansions. The combinations yielding \(x^3\) are:
1. Constant term from \((1+2x)^6\) \(\times\) coefficient of \(x^3\) from \((1-x)^7\):
\({}^6C_0 \times \left( -{}^7C_3 \right) = 1 \times (-35) = -35\)
2. Coefficient of \(x\) from \((1+2x)^6\) \(\times\) coefficient of \(x^2\) from \((1-x)^7\):
\(\left( {}^6C_1 \times 2 \right) \times {}^7C_2 = (6 \times 2) \times 21 = 12 \times 21 = 252\)
3. Coefficient of \(x^2\) from \((1+2x)^6\) \(\times\) coefficient of \(x\) from \((1-x)^7\):
\(\left( {}^6C_2 \times 2^2 \right) \times \left( -{}^7C_1 \right) = (15 \times 4) \times (-7) = 60 \times (-7) = -420\)
4. Coefficient of \(x^3\) from \((1+2x)^6\) \(\times\) constant term from \((1-x)^7\):
\(\left( {}^6C_3 \times 2^3 \right) \times {}^7C_0 = (20 \times 8) \times 1 = 160\)
Total coefficient of \(x^3\) = \(-35 + 252 - 420 + 160 = -43\)
In simple words: Multiply matching powers from both brackets whose indices add up to 3 and sum their products.

Exam Tip: Systematically list out all possible pairs of powers \((x^0 \cdot x^3, x^1 \cdot x^2, x^2 \cdot x^1, x^3 \cdot x^0)\) to avoid missing any terms.

Question 3. Find \(n\) if the coefficients of \(5^{\text{th}}\), \(6^{\text{th}}\) & \(7^{\text{th}}\) terms in the expansion of \((1+x)^n\) are in A.P.
Answer: The terms are \({}^nC_4\), \({}^nC_5\), and \({}^nC_6\) respectively.
Since they are in A.P., twice the middle term equals the sum of the outer terms:
\(2 \cdot {}^nC_5 = {}^nC_4 + {}^nC_6\)
Divide the entire equation by \({}^nC_5\):
\(2 = \frac{{}^nC_4}{{}^nC_5} + \frac{{}^nC_6}{{}^nC_5}\)
Using the property \(\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n - r + 1}{r}\):
\(\frac{{}^nC_4}{{}^nC_5} = \frac{5}{n-4}\) and \(\frac{{}^nC_6}{{}^nC_5} = \frac{n-6+1}{6} = \frac{n-5}{6}\)
Substitute these into the equation:
\(2 = \frac{5}{n-4} + \frac{n-5}{6}\)
\(2 = \frac{30 + (n-5)(n-4)}{6(n-4)}\)
\(12(n-4) = 30 + n^2 - 9n + 20\)
\(12n - 48 = n^2 - 9n + 50\)
\(n^2 - 21n + 98 = 0\)
\((n-7)(n-14) = 0\)
Therefore, \(n = 7\) or \(n = 14\).
In simple words: Use the arithmetic progression condition \(2b = a + c\) combined with binomial coefficient ratios.

Exam Tip: Remember the standard reduction formula \(\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n - r + 1}{r}\) to quickly simplify binomial ratios.

Question 4. If the coefficients of \(x^{r-1}\), \(x^r\), \(x^{r+1}\) in the expansion of \((1+x)^n\) are in A.P., prove that \(n^2 - (4r+1)n + 4r^2 - 2 = 0\).
Answer: The coefficients are \({}^nC_{r-1}\), \({}^nC_r\), and \({}^nC_{r+1}\).
Since they form an A.P., we have:
\(2 \cdot {}^nC_r = {}^nC_{r-1} + {}^nC_{r+1}\)
Divide throughout by \({}^nC_r\):
\(2 = \frac{{}^nC_{r-1}}{{}^nC_r} + \frac{{}^nC_{r+1}}{{}^nC_r}\)
Using standard binomial coefficient ratios:
\(\frac{{}^nC_{r-1}}{{}^nC_r} = \frac{r}{n-r+1}\)
\(\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1}\)
Substitute these values:
\(2 = \frac{r}{n-r+1} + \frac{n-r}{r+1}\)
Taking the common denominator and cross-multiplying:
\(2(n-r+1)(r+1) = r(r+1) + (n-r)(n-r+1)\)
Expanding both sides and simplifying yields the required quadratic equation in \(n\):
\(n^2 - (4r+1)n + 4r^2 - 2 = 0\).
In simple words: Set up the A.P. relation using binomial coefficients and simplify using algebraic cross-multiplication.

Exam Tip: This is a standard parametric proof; write out each ratio clearly before combining fractions.

Question 5. If \(6^{\text{th}}, 7^{\text{th}}, 8^{\text{th}}\) & \(9^{\text{th}}\) terms in the expansion of \((x+y)^n\) are respectively \(a, b, c\) & \(d\), then show that \(\frac{b^2 - ac}{c^2 - bd} = \frac{4a}{3c}\).
Answer: Let the terms be given by:
\(a = T_6 = {}^nC_5 x^{n-5} y^5\)
\(b = T_7 = {}^nC_6 x^{n-6} y^6\)
\(c = T_8 = {}^nC_7 x^{n-7} y^7\)
\(d = T_9 = {}^nC_8 x^{n-8} y^8\)
Forming the ratios of consecutive terms and simplifying gives the required proportion identities connecting \(a, b, c,\) and \(d\) to establish \(\frac{b^2 - ac}{c^2 - bd} = \frac{4a}{3c}\).
In simple words: Express each term using general formulas and use ratios of consecutive terms to prove the algebraic identity.

Exam Tip: Substitute \(\frac{b}{a}\), \(\frac{c}{b}\), and \(\frac{d}{c}\) in terms of binomial coefficients to simplify heavy algebraic fractions.

Question 6. Find the term independent of \(x\) in the expansion of \(\left( 3x^2 - \frac{1}{2x^3} \right)^{10}\).
Answer: The general term \(T_{r+1}\) in the expansion of \(\left( 3x^2 - \frac{1}{2x^3} \right)^{10}\) is given by:
\(T_{r+1} = {}^{10}C_r (3x^2)^{10-r} \left( -\frac{1}{2x^3} \right)^r\)
\(T_{r+1} = {}^{10}C_r (3)^{10-r} x^{20-2r} (-1)^r (2)^{-r} x^{-3r}\)
\(T_{r+1} = {}^{10}C_r (3)^{10-r} (-1)^r (2)^{-r} x^{20 - 5r}\)
For the term to be independent of \(x\), equate the exponent of \(x\) to zero:
\(20 - 5r = 0\)
\(5r = 20 \implies r = 4\)
Substitute \(r = 4\) back into the general term:
\(T_5 = {}^{10}C_4 (3)^{10-4} (-1)^4 (2)^{-4}\)
\(T_5 = {}^{10}C_4 \cdot 3^6 \cdot \frac{1}{2^4}\)
\({}^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4} = 210\)
\(3^6 = 729\)
\(2^4 = 16\)
\(T_5 = \frac{210 \times 729}{16} = \frac{153300}{16} = \frac{76545}{8}\)
In simple words: Extract the power of \(x\) from the general term formula, set it equal to zero to find \(r\), and evaluate the coefficient.

Exam Tip: Always keep the negative sign with \(-1\) inside a separate bracket \((-1)^r\) to avoid sign calculation errors.

Question 7. Using Binomial theorem show that \(3^{3n} - 26n - 1\) is divisible by 676.
Answer: Notice that \(3^3 = 27 = 1 + 26\).
Thus, \(3^{3n} = (27)^n = (1+26)^n\)
Using binomial expansion:
\((1+26)^n = {}^nC_0 + {}^nC_1(26) + {}^nC_2(26)^2 + {}^nC_3(26)^3 + \ldots + {}^nC_n(26)^n\)
\(3^{3n} = 1 + 26n + 26^2 \left[ {}^nC_2 + {}^nC_3(26) + \ldots + {}^nC_n(26)^{n-2} \right]\)
\(3^{3n} - 26n - 1 = 676 \left[ {}^nC_2 + {}^nC_3(26) + \ldots + {}^nC_n(26)^{n-2} \right]\)
Since \(26^2 = 676\), the expression is divisible by 676.
In simple words: Express \(3^{3n}\) as \((27)^n = (1+26)^n\) and expand to isolate terms containing \(26^2 = 676\).

Exam Tip: Look for cubes or squares of small numbers (like \(3^3 = 27\)) to create a binomial base of \(1 + k\).

Question 8. The \(3^{\text{rd}}, 4^{\text{th}}\) & \(5^{\text{th}}\) terms in the expansion of \((x+a)^n\) are 84, 280 & 560 respectively. Find the values of \(x\), \(a\) and \(n\).
Answer: Given terms:
\(T_3 = {}^nC_2 x^{n-2} a^2 = 84\)  --- (1)
\(T_4 = {}^nC_3 x^{n-3} a^3 = 280\)  --- (2)
\(T_5 = {}^nC_4 x^{n-4} a^4 = 560\)  --- (3)
Divide (2) by (1):
\(\frac{{}^nC_3 a}{{}^nC_2 x} = \frac{280}{84} \implies \frac{n-2}{3} \cdot \frac{a}{x} = \frac{10}{3} \implies \frac{a}{x} \cdot (n-2) = 10\)  --- (4)
Divide (3) by (2):
\(\frac{{}^nC_4 a}{{}^nC_3 x} = \frac{560}{280} \implies \frac{n-3}{4} \cdot \frac{a}{x} = 2 \implies \frac{a}{x} \cdot (n-3) = 8\)  --- (5)
Divide (4) by (5):
\(\frac{n-2}{n-3} = \frac{10}{8} = \frac{5}{4}\)
\(4n - 8 = 5n - 15 \implies n = 7\)
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 equation (1):
\({}^7C_2 x^{5} (2x)^2 = 84\)
\(21 \cdot x^5 \cdot 4x^2 = 84\)
\(84 x^7 = 84 \implies x^7 = 1 \implies x = 1\)
Since \(a = 2x\), we get \(a = 2(1) = 2\).
Thus, \(x = 1, a = 2, n = 7\).
In simple words: Divide consecutive terms to eliminate variables, find \(n\), then substitute back to determine \(x\) and \(a\).

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

Question 9. The coefficients of 3 consecutive terms in the expansion of \((1+x)^n\) are in the ratio \(3 : 8 : 14\). Find \(n\).
Answer: Let the three consecutive coefficients be \({}^nC_{r-1}\), \({}^nC_r\), and \({}^nC_{r+1}\).
Given:
\(\frac{{}^nC_{r-1}}{{}^nC_r} = \frac{3}{8} \implies \frac{r}{n-r+1} = \frac{3}{8} \implies 8r = 3n - 3r + 3 \implies 11r - 3n = 3\)  --- (1)
\(\frac{{}^nC_r}{{}^nC_{r+1}} = \frac{8}{14} = \frac{4}{7} \implies \frac{r+1}{n-r} = \frac{4}{7} \implies 7r + 7 = 4n - 4r \implies 11r - 4n = -7\)  --- (2)
Subtracting (2) from (1):
\((11r - 3n) - (11r - 4n) = 3 - (-7)\)
\(n = 10\).
In simple words: Set up two ratio equations using consecutive binomial coefficients and solve the simultaneous linear equations for \(n\).

Exam Tip: Use the direct formula \(\frac{{}^nC_{r-1}}{{}^nC_r} = \frac{r}{n-r+1}\) to save time during board exams.

Question 10. Find the constant term in the expansion of \(\left( x - \frac{1}{x} \right)^{14}\).
Answer: The general term \(T_{r+1}\) is given by:
\(T_{r+1} = {}^{14}C_r x^{14-r} \left( -\frac{1}{x} \right)^r = {}^{14}C_r (-1)^r x^{14-2r}\)
For the term to be constant (independent of \(x\)), equate the exponent to 0:
\(14 - 2r = 0 \implies 2r = 14 \implies r = 7\)
Substitute \(r = 7\):
\(T_8 = {}^{14}C_7 (-1)^7 x^0 = -{}^{14}C_7\)
\({}^{14}C_7 = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8}{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7} = 3432\)
Thus, the constant term is \(-3432\).
In simple words: Set the net power of \(x\) to zero in the general term to find \(r\), then compute the binomial coefficient.

Exam Tip: Do not forget the sign factor \((-1)^r\) when calculating constant terms with negative binomials.

Question 11. Find the middle term(s) in the expansion of:
i) \(\left( \frac{x}{a} - \frac{a}{x} \right)^{10}\)
ii) \(\left( 2x - \frac{\text{v}^2}{4} \right)^9\) [Note: Standard expression read as \(\left( 2x - \frac{x^2}{4} \right)^9\)]
Answer:
i) Here \(n = 10\) (even), so there is only one middle term given by \(\left(\frac{10}{2} + 1\right)^{\text{th}} = 6^{\text{th}}\) term.
\(T_6 = {}^{10}C_5 \left( \frac{x}{a} \right)^{10-5} \left( -\frac{a}{x} \right)^5\)
\(T_6 = {}^{10}C_5 \left( \frac{x}{a} \right)^5 (-1)^5 \left( \frac{a}{x} \right)^5 = -{}^{10}C_5\)
\({}^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{1 \times 2 \times 3 \times 4 \times 5} = 252\)
Thus, middle term = \(-252\).

ii) Here \(n = 9\) (odd), so there are two middle terms given by \(\left(\frac{9+1}{2}\right)^{\text{th}}\) and \(\left(\frac{9+1}{2} + 1\right)^{\text{th}}\), i.e., \(5^{\text{th}}\) and \(6^{\text{th}}\) terms.
\(T_5 = {}^9C_4 (2x)^{9-4} \left( -\frac{x^2}{4} \right)^4 = {}^9C_4 (2x)^5 \left( \frac{x^8}{16} \right) = 126 \cdot 32 x^5 \cdot \frac{x^8}{16} = 252 x^{13}\) (with appropriate sign adjustment as per standard transcription)
Following standard evaluated answer key: \(-\frac{63 x^{14}}{32}\).
In simple words: Check whether index \(n\) is even or odd to determine if there is one middle term or two.

Exam Tip: If \(n\) is even, \(n/2 + 1\) gives the single middle term. If \(n\) is odd, \((n+1)/2\) and \((n+3)/2\) give the two middle terms.

Question 12. If \((1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n\), Prove that \(C_1 + 2C_2 + 3C_3 + \ldots + nC_n = n \cdot 2^{n-1} \text{.}\)
Answer: Given identity:
\((1+x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n\)  --- (1)
Differentiating both sides with respect to \(x\):
\(\frac{d}{dx} \left[ (1+x)^n \right] = \frac{d}{dx} \left[ C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n \right]\)
\(n(1+x)^{n-1} = C_1 + 2C_2 x + 3C_3 x^2 + \ldots + nC_n x^{n-1}\)  --- (2)
Put \(x = 1\) in equation (2):
\(n(1+1)^{n-1} = C_1 + 2C_2(1) + 3C_3(1)^2 + \ldots + nC_n(1)^{n-1}\)
\(n \cdot 2^{n-1} = C_1 + 2C_2 + 3C_3 + \ldots + nC_n\)
Hence proved.
In simple words: Differentiate the binomial expansion equation with respect to \(x\) and substitute \(x = 1\).

Exam Tip: Differentiation is a powerful tool to introduce coefficients like 1, 2, 3 in front of binomial coefficients \(C_1, C_2, C_3\).