CBSE Class 11 Mathematics Binomial Theorem Notes Set 01

Important Notes and Formulas

Binomial Theorem

If a, b ∈ R and n ∈ N, then
\( (a + b)^n = {}^{n}C_{0}\,a^{n}b^{0} + {}^{n}C_{1}\,a^{n-1}b^{1} + {}^{n}C_{2}\,a^{n-2}b^{2} + \ldots + {}^{n}C_{n}\,a^{0}b^{n} \)

Remarks:
1. If the index of the binomial is n then the expansion contains n + 1 terms.
2. In each term, the sum of indices of a and b is always n.
3. Coefficients of the terms in binomial expansion equidistant from both the ends are equal.
4. \( (a - b)^n = {}^{n}C_{0}\,a^{n}b^{0} - {}^{n}C_{1}\,a^{n-1}b^{1} + {}^{n}C_{2}\,a^{n-2}b^{2} - \ldots + (-1)^{n}\,{}^{n}C_{n}\,a^{0}b^{n} \)

Binomial Coefficients

The coefficients \( {}^{n}C_{0}, {}^{n}C_{1}, {}^{n}C_{2}, \ldots, {}^{n}C_{n} \) in the expansion of \( (a + b)^n \) are called the binomial coefficients and are denoted by \( C_0, C_1, C_2, \ldots, C_n \) respectively.
Now \( (1 + x)^n = {}^{n}C_{0}x^{0} + {}^{n}C_{1}x^{1} + {}^{n}C_{2}x^{2} + \ldots + {}^{n}C_{n}x^{n} \) ..... (i)
Put x = 1: \( (1 + 1)^n = {}^{n}C_{0} + {}^{n}C_{1} + {}^{n}C_{2} + \ldots + {}^{n}C_{n} \)
∴ \( C_0 + C_1 + C_2 + \ldots + C_n = 2^n \)
∴ The sum of all binomial coefficients is \( 2^n \).
Put x = −1 in equation (i): \( (1 - 1)^n = {}^{n}C_{0} - {}^{n}C_{1} + {}^{n}C_{2} - \ldots + (-1)^{n}\,{}^{n}C_{n} \)
∴ \( {}^{n}C_{0} - {}^{n}C_{1} + {}^{n}C_{2} - {}^{n}C_{3} + \ldots + (-1)^{n}\,{}^{n}C_{n} = 0 \)
∴ \( C_0 + C_2 + C_4 + \ldots = C_1 + C_3 + C_5 + \ldots \)
\( C_0, C_2, C_4, \ldots \) are called even coefficients and \( C_1, C_3, C_5, \ldots \) are called odd coefficients.
Let \( C_0 + C_2 + C_4 + \ldots = C_1 + C_3 + C_5 + \ldots = k \)
Now \( C_0 + C_1 + C_2 + C_3 + \ldots + C_n = 2^n \)
∴ \( (C_0 + C_2 + C_4 + \ldots) + (C_1 + C_3 + C_5 + \ldots) = 2^n \)
∴ k + k = \( 2^n \), i.e. 2k = \( 2^n \)
∴ \( k = \frac{2^n}{2} = 2^{n-1} \)
∴ \( C_0 + C_2 + C_4 + \ldots = C_1 + C_3 + C_5 + \ldots = 2^{n-1} \)
∴ The sum of even coefficients = the sum of odd coefficients = \( 2^{n-1} \).

Properties of Binomial Coefficients

For the sake of convenience the coefficients \( {}^{n}C_{0}, {}^{n}C_{1}, \ldots, {}^{n}C_{r}, \ldots, {}^{n}C_{n} \) are usually denoted by \( C_0, C_1, \ldots, C_r, \ldots, C_n \) respectively.
(i) \( C_0 + C_1 + C_2 + \ldots + C_n = 2^n \)
(ii) \( C_0 - C_1 + C_2 - \ldots + (-1)^{n} C_n = 0 \)
(iii) \( C_0 + C_2 + C_4 + \ldots = C_1 + C_3 + C_5 + \ldots = 2^{n-1} \)
(iv) \( {}^{n}C_{r_1} = {}^{n}C_{r_2} \Rightarrow r_1 = r_2 \) or \( r_1 + r_2 = n \)
(v) \( {}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r} \)
(vi) \( r\,{}^{n}C_{r} = n\,{}^{n-1}C_{r-1} \)

Solved Examples

Example 1. If \( {}^{n}C_{8} = {}^{n}C_{6} \), determine n and hence \( {}^{n}C_{2} \).
Answer:
Given, \( {}^{n}C_{8} = {}^{n}C_{6} \)
⇒ n = 8 + 6 ⇒ n = 14
Now \( {}^{n}C_{2} = {}^{14}C_{2} = \frac{14 \times 13}{2!} = 91 \)

Example 2. If \( {}^{15}C_{3r} = {}^{15}C_{r+3} \), find r.
Answer:
We know that if \( {}^{n}C_{x} = {}^{n}C_{y} \), then x = y or x + y = n.
Here \( {}^{15}C_{3r} = {}^{15}C_{r+3} \)
∴ Either 3r = r + 3 ⇒ \( r = \frac{3}{2} \), which is not possible, since r is an integer,
or 3r + r + 3 = 15 ⇒ r = 3.
Hence r = 3.

Example 3. Prove that \( \sum_{r=1}^{5} {}^{5}C_{r} = 31 \).
Answer:
\( \sum_{r=1}^{5} {}^{5}C_{r} = {}^{5}C_{1} + {}^{5}C_{2} + {}^{5}C_{3} + {}^{5}C_{4} + {}^{5}C_{5} \)
\( = \frac{5!}{1!\,4!} + \frac{5!}{2!\,3!} + \frac{5!}{3!\,2!} + \frac{5!}{4!\,1!} + \frac{5!}{5!\,0!} \)
\( = \frac{5}{1} + \frac{5 \cdot 4}{2} + \frac{5 \cdot 4}{2} + \frac{5}{1} + 1 \)
= 5 + 10 + 10 + 5 + 1 = 31

Example 4. Find n, if \( {}^{n}C_{6} : {}^{n-3}C_{3} = 33 : 4 \).
Answer:
Given, \( {}^{n}C_{6} : {}^{n-3}C_{3} = 33 : 4 \)
∴ \( \frac{n!}{6!\,(n-6)!} \times \frac{3!\,(n-3-3)!}{(n-3)!} = \frac{33}{4} \)
or \( \frac{n!}{(n-3)!} \cdot \frac{3!}{6!} = \frac{33}{4} \) or \( \frac{n(n-1)(n-2)}{6 \cdot 5 \cdot 4} = \frac{33}{4} \)
or n(n − 1)(n − 2) = 6 · 5 · 33 = 11 · 3 · 3 · 2 · 5
or n(n − 1)(n − 2) = 11 · (3 · 3) · (2 · 5) = 11 · 10 · 9  ∴ n = 11

Example 5. Find the value of the expression \( {}^{47}C_{4} + \sum_{j=1}^{5} {}^{52-j}C_{3} \).
Answer:
Given expression \( = {}^{47}C_{4} + \sum_{j=1}^{5} {}^{52-j}C_{3} \)
\( = {}^{47}C_{4} + ({}^{51}C_{3} + {}^{50}C_{3} + {}^{49}C_{3} + {}^{48}C_{3} + {}^{47}C_{3}) \)
\( = ({}^{47}C_{4} + {}^{47}C_{3}) + {}^{48}C_{3} + {}^{49}C_{3} + {}^{50}C_{3} + {}^{51}C_{3} \)
\( = {}^{48}C_{4} + {}^{48}C_{3} + {}^{49}C_{3} + {}^{50}C_{3} + {}^{51}C_{3} \)  [∵ \( {}^{47}C_{4} + {}^{47}C_{3} = {}^{48}C_{4} \)]
\( = ({}^{49}C_{4} + {}^{49}C_{3}) + {}^{50}C_{3} + {}^{51}C_{3} = ({}^{50}C_{4} + {}^{50}C_{3}) + {}^{51}C_{3} \)
\( = {}^{51}C_{4} + {}^{51}C_{3} = {}^{52}C_{4} \)

Example 6. Expand (i) \( (2x^2 + 3)^4 \)   (ii) \( \left(2x^2 - \frac{1}{x}\right)^6 \)
Answer:
(i) \( (2x^2 + 3)^4 = {}^{4}C_{0}(2x^2)^4(3)^0 + {}^{4}C_{1}(2x^2)^3(3)^1 + {}^{4}C_{2}(2x^2)^2(3)^2 + {}^{4}C_{3}(2x^2)^1(3)^3 + {}^{4}C_{4}(2x^2)^0(3)^4 \)
∵ \( {}^{4}C_{0} = {}^{4}C_{4} = 1,\ {}^{4}C_{1} = {}^{4}C_{3} = 4,\ {}^{4}C_{2} = \frac{4!}{2!\,2!} = 6 \)
= (1)(16x⁸)(1) + 4(8x⁶)(3) + 6(4x⁴)(9) + 4(2x²)(27) + (1)(1)(81)
= 16x⁸ + 96x⁶ + 216x⁴ + 216x² + 81
(ii) \( \left(2x^2 - \frac{1}{x}\right)^6 = {}^{6}C_{0}(2x^2)^6 - {}^{6}C_{1}(2x^2)^5\left(\frac{1}{x}\right) + {}^{6}C_{2}(2x^2)^4\left(\frac{1}{x}\right)^2 - {}^{6}C_{3}(2x^2)^3\left(\frac{1}{x}\right)^3 + {}^{6}C_{4}(2x^2)^2\left(\frac{1}{x}\right)^4 - {}^{6}C_{5}(2x^2)\left(\frac{1}{x}\right)^5 + {}^{6}C_{6}\left(\frac{1}{x}\right)^6 \)
∵ \( {}^{6}C_{0} = {}^{6}C_{6} = 1,\ {}^{6}C_{1} = {}^{6}C_{5} = 6,\ {}^{6}C_{2} = {}^{6}C_{4} = \frac{6!}{2!\,4!} = 15,\ {}^{6}C_{3} = \frac{6!}{3!\,3!} = 20 \)
\( = 64x^{12} - 6(32x^{10})\frac{1}{x} + 15(16x^8)\frac{1}{x^2} - 20(8x^6)\frac{1}{x^3} + 15(4x^4)\frac{1}{x^4} - 6(2x^2)\frac{1}{x^5} + \frac{1}{x^6} \)
\( = 64x^{12} - 192x^9 + 240x^6 - 160x^3 + 60 - \frac{12}{x^3} + \frac{1}{x^6} \)

Example 7. Expand \( (1 + x + x^2)^3 \).
Answer:
Let y = x + x². Then,
\( (1 + x + x^2)^3 = (1 + y)^3 = {}^{3}C_{0} + {}^{3}C_{1}y + {}^{3}C_{2}y^2 + {}^{3}C_{3}y^3 \)
\( = 1 + 3y + 3y^2 + y^3 = 1 + 3(x + x^2) + 3(x + x^2)^2 + (x + x^2)^3 \)
\( = 1 + 3(x + x^2) + 3(x^2 + 2x^3 + x^4) + \{{}^{3}C_{0}x^3(x^2)^0 + {}^{3}C_{1}x^2(x^2)^1 + {}^{3}C_{2}x(x^2)^2 + {}^{3}C_{3}x^0(x^2)^3\} \)
\( = 1 + 3(x + x^2) + 3(x^2 + 2x^3 + x^4) + (x^3 + 3x^4 + 3x^5 + x^6) \)
\( = x^6 + 3x^5 + 6x^4 + 7x^3 + 6x^2 + 3x + 1 \)

Example 8. Prove that \( (\sqrt{5} + 1)^5 - (\sqrt{5} - 1)^5 = 352 \).
Answer:
\( (\sqrt{5} + 1)^5 = {}^{5}C_{0}(\sqrt{5})^5 + {}^{5}C_{1}(\sqrt{5})^4(1) + {}^{5}C_{2}(\sqrt{5})^3(1)^2 + {}^{5}C_{3}(\sqrt{5})^2(1)^3 + {}^{5}C_{4}(\sqrt{5})(1)^4 + {}^{5}C_{5}(1)^5 \)
\( (\sqrt{5} - 1)^5 = {}^{5}C_{0}(\sqrt{5})^5 - {}^{5}C_{1}(\sqrt{5})^4(1) + {}^{5}C_{2}(\sqrt{5})^3(1)^2 - {}^{5}C_{3}(\sqrt{5})^2(1)^3 + {}^{5}C_{4}(\sqrt{5})(1)^4 - {}^{5}C_{5}(1)^5 \)
∵ \( {}^{5}C_{0} = {}^{5}C_{5} = 1;\ {}^{5}C_{1} = {}^{5}C_{4} = 5;\ {}^{5}C_{2} = {}^{5}C_{3} = \frac{5 \cdot 4}{2 \cdot 1} = 10 \)
\( (\sqrt{5} + 1)^5 = 25\sqrt{5} + 5(25) + 10(5\sqrt{5}) + 10(5) + 5\sqrt{5} + 1 = 25\sqrt{5} + 125 + 50\sqrt{5} + 50 + 5\sqrt{5} + 1 \)
\( (\sqrt{5} - 1)^5 = 25\sqrt{5} - 125 + 50\sqrt{5} - 50 + 5\sqrt{5} - 1 \)
Subtracting, the terms containing \( \sqrt{5} \) cancel:
\( (\sqrt{5} + 1)^5 - (\sqrt{5} - 1)^5 = 2(125 + 50 + 1) = 2 \times 176 = 352 \)

Example 9. Using binomial theorem compute \( (99)^5 \).
Answer:
\( (99)^5 = (100 - 1)^5 = {}^{5}C_{0}(100)^5 - {}^{5}C_{1}(100)^4 + {}^{5}C_{2}(100)^3 - {}^{5}C_{3}(100)^2 + {}^{5}C_{4}(100)^1 - {}^{5}C_{5}(100)^0 \)
\( = (100)^5 - 5 \times (100)^4 + 10 \times (100)^3 - 10 \times (100)^2 + 5 \times 100 - 1 \)
\( = 10^{10} - 5 \times 10^8 + 10^7 - 10^5 + 5 \times 10^2 - 1 \)
\( = (10^{10} + 10^7 + 5 \times 10^2) - (5 \times 10^8 + 10^5 + 1) \)
= 10010000500 − 500100001 = 9509900499

Example 10. Use the binomial theorem to find the exact value of \( (10.1)^5 \).
Answer:
\( (10.1)^5 = (10 + 0.1)^5 \)
\( = 10^5 + {}^{5}C_{1}10^4(0.1) + {}^{5}C_{2}10^3(0.1)^2 + {}^{5}C_{3}10^2(0.1)^3 + {}^{5}C_{4}10(0.1)^4 + {}^{5}C_{5}(0.1)^5 \)
\( = 100000 + 5 \times 10^4(0.1) + 10 \times 10^3(0.01) + 10 \times 10^2(0.001) + 5 \times 10(0.0001) + 0.00001 \)
= 100000 + 5000 + 100 + 1 + 0.005 + 0.00001 = 105101.00501

Example 11. Which number is larger, \( (1.2)^{4000} \) or 800?
Answer:
\( (1.2)^{4000} = (1 + 0.2)^{4000} \)
\( = {}^{4000}C_{0} + {}^{4000}C_{1}(0.2) + \) sum of positive terms
= 1 + 4000(0.2) + a positive number
= 1 + 800 + a positive number > 800
Hence \( (1.2)^{4000} > 800 \).

Example 12. Prove that \( \sum_{r=0}^{n} 3^r \cdot {}^{n}C_{r} = 4^n \).
Answer:
\( (1 + x)^n = {}^{n}C_{0}x^0 + {}^{n}C_{1}x + {}^{n}C_{2}x^2 + \ldots + {}^{n}C_{n}x^n \)
∴ \( {}^{n}C_{0}x^0 + {}^{n}C_{1}x + {}^{n}C_{2}x^2 + \ldots + {}^{n}C_{n}x^n = (1 + x)^n \) ....(1)
Now \( \sum_{r=0}^{n} 3^r \cdot {}^{n}C_{r} = {}^{n}C_{0}3^0 + {}^{n}C_{1}3^1 + {}^{n}C_{2}3^2 + \ldots + {}^{n}C_{n}3^n \)
\( = (1 + 3)^n \) [putting x = 3 in (1)]
\( = 4^n \)

Example 13. Find the coefficient of \( x^9 \) in \( (1 + 3x + 3x^2 + x^3)^{15} \).
Answer:
\( (1 + 3x + 3x^2 + x^3)^{15} = [(1 + x)^3]^{15} = (1 + x)^{45} \)
∴ Coefficient of \( x^9 \) in \( (1 + 3x + 3x^2 + x^3)^{15} \) = coefficient of \( x^9 \) in \( (1 + x)^{45} \)
\( = {}^{45}C_{9} \)  [since in the expansion of \( (1 + x)^n \), coefficient of \( x^r = {}^{n}C_{r} \)]
\( = \frac{45!}{9!\,36!} \)

Example 14. Find the coefficient of \( x^5 \) in the expansion of the product \( (1 + 2x)^6 (1 - x)^7 \).
Answer:
\( (1 + 2x)^6 = 1 + {}^{6}C_{1}(2x) + {}^{6}C_{2}(2x)^2 + {}^{6}C_{3}(2x)^3 + {}^{6}C_{4}(2x)^4 + {}^{6}C_{5}(2x)^5 + {}^{6}C_{6}(2x)^6 \) .....(1)
Again, \( (1 - x)^7 = 1 - {}^{7}C_{1}x + {}^{7}C_{2}x^2 - {}^{7}C_{3}x^3 + {}^{7}C_{4}x^4 - {}^{7}C_{5}x^5 + {}^{7}C_{6}x^6 - {}^{7}C_{7}x^7 \)
\( = 1 - 7x + 21x^2 - 35x^3 + 35x^4 - 21x^5 + 7x^6 - x^7 \) .....(2)
Now \( (1 + 2x)^6 (1 - x)^7 = (1 + 12x + 60x^2 + 160x^3 + 240x^4 + 192x^5 + \ldots) \times (1 - 7x + 21x^2 - 35x^3 + 35x^4 - 21x^5 + \ldots) \)
∴ Required coefficient of \( x^5 \) in the product
= 1 × (−21) + 12 × 35 + 60 × (−35) + 160 × 21 + 240 × (−7) + 192 × 1
= −21 + 420 − 2100 + 3360 − 1680 + 192 = 171