Binomial Theorem PDF

Binomial Theorem Factorial notation The notation 𝑛! or ⌊𝑛 represents the product of first 𝑛 natural numbers. We read this symbol as ‘𝑛 factorial’. Thus, 𝑛! = 1 × 2 × 3 × 4 ×...× (𝑛 – 1) × 𝑛 = 𝑛 × (𝑛 – 1) ×...× 3 × 2 × 1. Example : 1! = 1 2! = 1 × 2 = 2 3! = 1 × 2 × 3 = 6 4! = 1 × 2 × 3 × 4 = 24 5! =...

Binomial Theorem Factorial notation The notation 𝑛! or ⌊𝑛 represents the product of first 𝑛 natural numbers. We read this symbol as ‘𝑛 factorial’. Thus, 𝑛! = 1 × 2 × 3 × 4 ×...× (𝑛 – 1) × 𝑛 = 𝑛 × (𝑛 – 1) ×...× 3 × 2 × 1. Example : 1! = 1 2! = 1 × 2 = 2 3! = 1 × 2 × 3 = 6 4! = 1 × 2 × 3 × 4 = 24 5! = 1 × 2 × 3 × 4 × 5 = 120 We define 0! = 1. Note that, 𝑛! = 𝑛 × (𝑛 – 1) ×...× 3 × 2 × 1 = 𝑛(𝑛 − 1)! 𝑛! = 𝑛(𝑛 − 1)! = 𝑛(𝑛 − 1)(𝑛 − 2)! [Provided (𝑛 ≥ 2)] 𝑛! = 𝑛(𝑛 − 1)! = 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)! [Provided (𝑛 ≥ 3)] We can write, 6! = 6 × 5! = 6 × 120 = 720 12! 12×11×10! 10!2! = 10!×2 = 6 × 11 = 66 Permutations A permutation is an arrangement in a definite order of a number of objects taken some or all at a time. The number of permutations of 𝑛 different things taken 𝑟 at a time, where repetition is not allowed, is denoted by 𝑛𝑃𝑟 and is given by 𝑛! 𝑛𝑃𝑟 = (𝑛−𝑟)! , where 0 ≤ 𝑟 ≤ 𝑛. Combinations The number of combinations of 𝑛 different things taken 𝑟 at a time, denoted by 𝑛𝐶𝑟 , is given by 𝑛! 𝑛𝐶𝑟 = 𝑟!(𝑛−𝑟)! , where 0 ≤ 𝑟 ≤ 𝑛. Example : 10! 10! 10 × 9 × 8 × 7 × 6! 10𝐶4 = = = = 210 4! (10 − 4)! 4! 6! 4 × 3 × 2 × 1 × 6! 10! 10! 10 × 9 × 8 × 7 × 6! 10𝐶6 = = = = 210 6! (10 − 6)! 6! 4! 6! × 4 × 3 × 2 × 1 6! 6! 6 × 5 × 4 × 3! 6𝑃3 = = = = 120 (6 − 3)! 3! 3! 𝑛! 𝑛! 𝑛𝐶0 = = =1 (𝑛 0! − 0)! 1 × 𝑛! 𝑛! 𝑛 × (𝑛 − 1)! 𝑛𝐶1 = = =𝑛 1! (𝑛 − 1)! 1 × (𝑛 − 1)! 𝑛! 𝑛(𝑛 − 1)(𝑛 − 2)! 𝑛(𝑛 − 1) 𝑛𝐶2 = = = 2! (𝑛 − 2)! 2! (𝑛 − 2)! 2! 𝑛! 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)! 𝑛(𝑛 − 1)(𝑛 − 2) 𝑛𝐶3 = = = 3! (𝑛 − 3)! 3! (𝑛 − 3)! 3!  𝑛! 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 4)! 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) 𝑛𝐶4 = = = 4! (𝑛 − 4)! 4! (𝑛 − 4)! 4! ……………………………………………………………………………………… 𝑛! 𝑛(𝑛 − 1)(𝑛 − 2) … [𝑛 − (𝑟 − 1)] 𝑛(𝑛 − 1)(𝑛 − 2) … (𝑛 − 𝑟 + 1) 𝑛𝐶𝑟 = = = 𝑟! (𝑛 − 𝑟)! 𝑟! 𝑟! ………………………………………………………………………………………………. 𝑛! 𝑛! 𝑛! 𝑛𝐶𝑛 = = = =1 𝑛! (𝑛 − 𝑛)! 𝑛! 0! 𝑛! × 1 Some Properties 1) 𝑛𝑃𝑟 = 𝑟! 𝑛𝐶𝑟 2) 𝑛𝐶𝑟 = 𝑛𝐶𝑛−𝑟 3) 𝑛𝐶𝑟 + 𝑛𝐶𝑟−1 = 𝑛 + 1𝐶𝑟 4) 𝑛𝐶𝑟 = 𝑛𝐶𝑘 ⇒ 𝑟 = 𝑘 or 𝑟 = 𝑛 − 𝑘 𝑛𝐶𝑟 𝑛−𝑟+1 5) 𝑛𝐶𝑟−1 = 𝑟 Binomial Theorem for Positive Integral Indices Let us have a look at the following identities done earlier: (𝑎 + 𝑏)0 = 1 , 𝑎 + 𝑏 ≠ 0 (𝑎 + 𝑏)1 = 𝑎 + 𝑏 (𝑎 + 𝑏)2 = 𝑎 2 + 2𝑎𝑏 + 𝑏2 (𝑎 + 𝑏)3 = 𝑎 3 + 3𝑎 2 𝑏 + 3𝑎𝑏2 + 𝑏3 (𝑎 + 𝑏)4 = (𝑎 + 𝑏)3 (𝑎 + 𝑏) = 𝑎 4 + 4𝑎 3 𝑏 + 6𝑎 2 𝑏2 + 4𝑎𝑏3 + 𝑏4 In these expansions, we observe that (i) The total number of terms in the expansion is one more than the index. For example, in the expansion of (𝑎 + 𝑏)2 , number of terms is 3 whereas the index of (𝑎 + 𝑏)2 is 2. (ii) Powers of the first quantity ‘𝑎 ’ go on decreasing by 1 whereas the powers of the second quantity ‘𝑏’ increase by 1, in the successive terms. (iii) In each term of the expansion, the sum of the indices of 𝑎 and 𝑏 is the same and is equal to the index of (𝑎 + 𝑏). Binomial theorem for any positive integer 𝒏 (𝑎 + 𝑏)𝑛 = 𝑛𝐶0 𝑎 𝑛 + 𝑛𝐶1 𝑎 𝑛−1 𝑏 + 𝑛𝐶2 𝑎 𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎 𝑛−3 𝑏 3 + ⋯ + 𝑛𝐶𝑛−1 𝑎𝑏𝑛−1 + 𝑛𝐶𝑛 𝑏𝑛 = 𝑎 𝑛 + 𝑛𝐶1 𝑎 𝑛−1 𝑏 + 𝑛𝐶2 𝑎 𝑛−2 𝑏2 + 𝑛𝐶3 𝑎 𝑛−3 𝑏3 + ⋯ + 𝑛𝐶𝑛−1 𝑎𝑏 𝑛−1 + 𝑏𝑛 Here, 𝑛𝐶0 , 𝑛𝐶1 , 𝑛𝐶2 , 𝑛𝐶3 , … , 𝑛𝐶𝑛 are called binomial co-efficients in the expansion of (𝑎 + 𝑏)𝑛. Observations (i) There are (𝑛 + 1) terms in the expansion of (𝑎 + 𝑏)𝑛, i.e., one more than the index. (ii) In the successive terms of the expansion the index of ‘𝑎’ goes on decreasing by unity. It is 𝑛 in the first term, (𝑛– 1) in the second term, and so on ending with zero in the last term. At the same time  the index of ‘𝑏’ increases by unity, starting with zero in the first term, 1 in the second and so on ending with 𝑛 in the last term. (iii) In the expansion of (𝑎 + 𝑏)𝑛 , the sum of the indices of ‘𝑎’ and ‘𝑏’ is 𝑛 + 0 = 𝑛 in the first term, (𝑛 – 1) + 1 = 𝑛 in the second term and so on 0 + 𝑛 = 𝑛 in the last term. Thus, it can be seen that the sum of the indices of ‘𝑎’ and ‘𝑏’ is n in every term of the expansion. Some Special Cases In the expansion of (𝑎 + 𝑏)𝑛, (i) Taking 𝑎 = 𝑥 and 𝑏 = −𝑦 , we obtain (𝑥 − 𝑦)𝑛 = [𝑥 + (−𝑦)]𝑛 = 𝑛𝐶0 𝑥 𝑛 + 𝑛𝐶1 𝑥 𝑛−1 (−𝑦) + 𝑛𝐶2 𝑥 𝑛−2 (−𝑦)2 + 𝑛𝐶3 𝑥 𝑛−3 (−𝑦)3 + ⋯ + 𝑛𝐶𝑛 (−𝑦)𝑛 = 𝑛𝐶0 𝑥 𝑛 − 𝑛𝐶1 𝑥 𝑛−1 𝑦 + 𝑛𝐶2 𝑥 𝑛−2 𝑦 2 − 𝑛𝐶3 𝑥 𝑛−3 𝑦 3 + ⋯ + (−1)𝑛 𝑛𝐶𝑛 𝑦 𝑛 Thus, (𝑥 − 𝑦)𝑛 = 𝑛𝐶0 𝑥 𝑛 − 𝑛𝐶1 𝑥 𝑛−1 𝑦 + 𝑛𝐶2 𝑥 𝑛−2 𝑦 2 − 𝑛𝐶3 𝑥 𝑛−3 𝑦 3 + ⋯ + (−1)𝑛 𝑛𝐶𝑛 𝑦 𝑛 (ii) Taking 𝑎 = 1 and 𝑏 = 𝑥 , we obtain (1 + 𝑥)𝑛 = 𝑛𝐶0 1𝑛 + 𝑛𝐶1 1𝑛−1 𝑥 + 𝑛𝐶2 1𝑛−2 𝑥 2 + 𝑛𝐶3 1𝑛−3 𝑥 3 + ⋯ + +𝑛𝐶𝑛 𝑥 𝑛 Thus, (1 + 𝑥)𝑛 = 𝑛𝐶0 + 𝑛𝐶1 𝑥 + 𝑛𝐶2 𝑥 2 + 𝑛𝐶3 𝑥 3 + ⋯ + 𝑛𝐶𝑛 𝑥 𝑛 In particular, for 𝑥 = 1 ,we have 2𝑛 = 𝑛𝐶0 + 𝑛𝐶1 + 𝑛𝐶2 + 𝑛𝐶3 + ⋯ + +𝑛𝐶𝑛 Thus, the sum of the binomial co-efficients in the expansion of (𝑎 + 𝑏)𝑛 is 2𝑛. (iii) Taking 𝑎 = 1 and 𝑏 = −𝑥 , we obtain (1 − 𝑥)𝑛 = 𝑛𝐶0 1𝑛 + 𝑛𝐶1 1𝑛−1 (−𝑥) + 𝑛𝐶2 1𝑛−2 (−𝑥)2 + 𝑛𝐶3 1𝑛−3 (−𝑥)3 + ⋯ + 𝑛𝐶𝑛 (−𝑥)𝑛 Thus, (1 − 𝑥)𝑛 = 𝑛𝐶0 − 𝑛𝐶1 𝑥 + 𝑛𝐶2 𝑥 2 − 𝑛𝐶3 𝑥 3 + ⋯ + (−1)𝑛 𝑛𝐶𝑛 𝑥 𝑛 In particular, for 𝑥 = 1 ,we have 0 = 𝑛𝐶0 − 𝑛𝐶1 + 𝑛𝐶2 − 𝑛𝐶3 + ⋯ + (−1)𝑛 𝑛𝐶𝑛 𝟑 𝟒 Problem : Expand (𝒙𝟐 + ) , 𝒙 ≠ 𝟎. 𝒙 Solution : By using binomial theorem, we have 3 4 3 3 2 3 3 3 4 (𝑥 2 + ) = 4𝐶0 (𝑥 2 )4 + 4𝐶1 (𝑥 2 )3 ( ) + 4𝐶2 (𝑥 2 )2 ( ) + 4𝐶3 (𝑥 2 ) ( ) + 4𝐶4 ( ) 𝑥 𝑥 𝑥 𝑥 𝑥 8 6 3 4 9 2 27 81 = 𝑥 + 4. 𝑥. 𝑥 + 6. 𝑥. 𝑥 2 + 4. 𝑥. 𝑥 3 + 𝑥 4 108 81 = 𝑥 8 + 12𝑥 5 + 54𝑥 2 + 𝑥 + 𝑥4 Problem : Expand (𝒙 − 𝟐𝒚)𝟓. Solution : By using binomial theorem, we have (𝑥 − 2𝑦)5 = 5𝐶0 𝑥 5 − 5𝐶1 𝑥 4 (2𝑦) + 5𝐶2 𝑥 3 (2𝑦)2 − 5𝐶3 𝑥 2 (2𝑦)3 + 5𝐶4 𝑥(2𝑦)4 − 5𝐶5 (2𝑦)5 = 𝑥 5 − 10𝑥 4 𝑦 + 40𝑥 3 𝑦 2 − 80𝑥 2 𝑦 3 + 80𝑥𝑦 4 − 32𝑦 5 General Term In Binomial Expansion (𝑎 + 𝑏)𝑛 = 𝑛𝐶0 𝑎 𝑛 + 𝑛𝐶1 𝑎 𝑛−1 𝑏 + 𝑛𝐶2 𝑎 𝑛−2 𝑏 2 + 𝑛𝐶3 𝑎 𝑛−3 𝑏 3 + ⋯ + 𝑛𝐶𝑛−1 𝑎𝑏𝑛−1 + 𝑛𝐶𝑛 𝑏𝑛 We observe that, the first term= 𝑡1 = 𝑛𝐶0 𝑎 𝑛 = 𝑛𝐶0 𝑎 𝑛 𝑏0 the second term= 𝑡2 = 𝑛𝐶1 𝑎 𝑛−1 𝑏 the third term= 𝑡3 = 𝑛𝐶2 𝑎 𝑛−2 𝑏 2 the fourth term= 𝑡4 = 𝑛𝐶3 𝑎 𝑛−3 𝑏3 From the above, we can generalise that the (𝑟 + 1)𝑡ℎ term= 𝑡𝑟+1 = 𝑛𝐶𝑟 𝑎 𝑛−𝑟 𝑏𝑟 Thus, we have 𝑡𝑟+1 = 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟 This is called the general term in the binomial expansion of (𝑎 + 𝑏)𝑛. Problem : Find the number of terms in the expansion of (𝒂 + 𝒃)𝒏 , where 𝒏 is a positive integer. Solution Since the index of (𝑎 + 𝑏) is 𝑛 , so the number of terms in the expansion of (𝑎 + 𝑏)𝑛 is (𝑛 + 1). 𝟕 Problem : Find the number of terms in the expansion of (𝟏 + 𝟑𝒙 + 𝟑𝒙𝟐 + 𝒙𝟑 ). Solution Note that,(1 + 3𝑥 + 3𝑥 2 + 𝑥 3 )7 = [(1 + 𝑥)3 ]7 = (1 + 𝑥)21. Therefore, the number of terms in the expansion of (1 + 3𝑥 + 3𝑥 2 + 𝑥 3 )7 is (21 + 1) = 22. 𝟏 𝟔 Problem : Find the number of terms in the expansion of (𝒙𝟐 − 𝟐 + 𝒙𝟐 ). Solution 6 1 6 1 2 1 12 Note that, (𝑥 2 − 2 + ) = [(𝑥 − ) ] = (𝑥 − ). 𝑥2 𝑥 𝑥 1 6 Therefore, the number of terms in the expansion of (𝑥 2 − 2 + 𝑥 2) is (12 + 1) = 13. Problem : Find the number of terms in the expansion of (𝒙 + 𝒚)𝟕 (𝒙 − 𝒚)𝟕. Solution Note that, (𝑥 + 𝑦)7 (𝑥 − 𝑦)7 = [(𝑥 + 𝑦)(𝑥 − 𝑦)]7 = (𝑥 2 − 𝑦 2 )7 Therefore, the number of terms in the expansion of (𝑥 + 𝑦)7 (𝑥 − 𝑦)7 is (7 + 1) = 8. Problem : Find the second term in the expansion of (𝟐𝒙 + 𝟑𝒚)𝟓. Solution Second term= 𝑡2 = 𝑡1+1 = 5𝐶1 (2𝑥)5−1 (3𝑦)1 = 5. 24. 𝑥 4. 3𝑦 = 240𝑥 4 𝑦. Problem : Find the fifth term in the expansion of (𝟏 + 𝒙)𝟓. Solution 5! 5×4! Fifth term= 𝑡5 = 𝑡4+1 = 5𝐶4. 16−4. 𝑥 4 = 4!1! 𝑥 4 = 4!×1 𝑥 4 = 5𝑥 4.  𝟏 𝟗 Problem : Find the sixth term in the expansion of (𝒙 − 𝒙). Solution 1 5 9! 1 9×8×7×6×5! 1 126 Sixth term= 𝑡6 = 𝑡5+1 = 9𝐶5 𝑥 9−5 (− ) = −. 𝑥4. =−. =−. 𝑥 5!4! 𝑥5 5!×4×3×2×1 𝑥 𝑥 𝟏 𝟏𝟎 Problem : Find the term independent of 𝒙 in the expansion of (𝟏 − 𝒙). Solution : By using binomial theorem, we have 1 10 1 1 2 1 3 1 4 10 1 10 (1 − ) = 10𝐶0 − 10𝐶1. + 10𝐶2. ( ) − 10𝐶3. ( ) + 10𝐶4. ( ) − ⋯ + (−1). 10𝐶10. ( ) 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 st st From the above expansion, it is clear that 1 term is independent of 𝑥 and the 1 term is 10𝐶0 = 1. 𝟏 𝟏𝟎 Problem : Find the term independent of 𝒙 in the expansion of (𝒙 − 𝒙). Solution : 1 10 Let (𝑟 + 1)𝑡ℎ term be independent of 𝑥 in the expansion of (𝑥 − 𝑥). 1 𝑟 Now, 𝑡𝑟+1 = 10𝐶𝑟 (𝑥)10−𝑟 (− 𝑥) 1 = (−1)𝑟 10𝐶𝑟 𝑥10−𝑟. 𝑥 𝑟 = (−1)𝑟 10𝐶𝑟 𝑥10−2𝑟 Since 𝑡𝑟+1 is independent of 𝑥, so index of 𝑥 is zero. ∴ 10 − 2𝑟 = 0 ⇒ 2𝑟 = 10 ⇒ 𝑟 = 5. Hence (5 + 1)𝑡ℎ 𝑖. 𝑒. , 6th term is independent of 𝑥 and is given by 𝑡6 = 𝑡5+1 = (−1)5 10𝐶5 10! =− 5! 5! 10.9.8.7.6.5! =− 5.4.3.2.1.5! = −252 𝟏 𝟏𝟎 Problem : Find the term independent of 𝒙 in the expansion of (𝒙 + 𝒙). Solution : 1 10 Let (𝑟 + 1)𝑡ℎ term be independent of 𝑥 in the expansion of (𝑥 + 𝑥). 1 𝑟 Now, 𝑡𝑟+1 = 10𝐶𝑟 (𝑥)10−𝑟 (𝑥) 1 = 10𝐶𝑟 𝑥10−𝑟. 𝑥 𝑟 = 10𝐶𝑟 𝑥10−2𝑟 Since 𝑡𝑟+1 is independent of 𝑥, so index of 𝑥 is zero. ∴ 10 − 2𝑟 = 0 ⇒ 2𝑟 = 10 ⇒ 𝑟 = 5. Hence (5 + 1)𝑡ℎ , 𝑖. 𝑒. , 6th term is independent of 𝑥 and is given by  𝑡6 = 𝑡5+1 = 10𝐶5 10! = 5! 5! 10.9.8.7.6.5! = 5.4.3.2.1.5! = 252 𝟏 𝟏𝟐 Problem : Find the term independent of 𝒙 in the expansion of (𝟗𝒙𝟐 − 𝟑𝒙). Solution : 1 12 Let (𝑟 + 1)𝑡ℎ term be independent of 𝑥 in the expansion of (9𝑥 2 − 3𝑥). 1 𝑟 Now, 𝑡𝑟+1 = 12𝐶𝑟 (9𝑥 2 )12−𝑟 (− 3𝑥) 1 = (−1)𝑟 12𝐶𝑟 912−𝑟. 𝑥 24−2𝑟. 3𝑟 𝑥 𝑟 1 = (−1)𝑟 12𝐶𝑟 912−𝑟. 𝑥 24−3𝑟. 𝑟 3 Since 𝑡𝑟+1 is independent of 𝑥, so index of 𝑥 is zero. ∴ 24 − 3𝑟 = 0 ⇒ 3𝑟 = 24 ⇒ 𝑟 = 8. Hence (8 + 1)𝑡ℎ , 𝑖. 𝑒. , 9th term is independent of 𝑥 and is given by 1 𝑡9 = 𝑡8+1 = (−1)8 12𝐶8 912−8. 38 12! 912−8 = × 8 8! 4! 3 12 × 11 × 10 × 9 × 8! 94 = × 8 8! × 4 × 3 × 2 × 1 3 38 = 11 × 5 × 9 × 8 3 = 495 𝟏𝟎 √𝒎 Problem : If the 𝒙 independent term in the expansion of (√𝒙 − 𝒙𝟐 ) 𝐢𝐬 𝟒𝟓𝟎 , then find the value of 𝒎. Solution : 10 √𝑚 Let (𝑟 + 1)𝑡ℎ term be independent of 𝑥 in the expansion of (√𝑥 − 𝑥 2 ). 10−𝑟 𝑟 √𝑚 Now, 𝑡𝑟+1 = 10𝐶𝑟 (√𝑥) (− 𝑥2 ) 10−𝑟 𝑟 (√𝑚) = (−1)𝑟 10𝐶𝑟. 𝑥 2. 𝑥 2𝑟 𝑟 10−𝑟 −2𝑟 = (−1)𝑟 10𝐶𝑟. 𝑚2. 𝑥 2 𝑟 10−5𝑟 = (−1)𝑟 10𝐶𝑟. 𝑚2. 𝑥 2 Since 𝑡𝑟+1 is independent of 𝑥,so index of 𝑥 is zero. 10−5𝑟 ∴ 2 = 0 ⇒ 10 − 5𝑟 = 0 ⇒ 5𝑟 = 10 ⇒ 𝑟 = 2. Hence (2 + 1)rd , 𝑖. 𝑒. , 3rd term is independent of 𝑥 and is given by  𝑟 𝑡3 = 𝑡2+1 = (−1)𝑟 10𝐶𝑟. 𝑚2 2 = (−1)2 10𝐶2. 𝑚2 10! = 2!8! 𝑚 10.9.8! = 2.8! 𝑚 = 45𝑚 By given condition, 45𝑚 = 450 ⇒ 𝑚 = 10 Problem : Find the co-efficient of 𝒙𝟐 in the expansion of (𝟏 + 𝒑𝒙)𝟏𝟎. Solution By using binomial theorem, we have (1 + 𝑝𝑥)10 = 10𝐶0 + 10𝐶1 (𝑝𝑥) + 10𝐶2 (𝑝𝑥)2 + 10𝐶3 (𝑝𝑥)3 + ⋯ + 10𝐶10 (𝑝𝑥)10 = 1 + 10𝐶1 (𝑝𝑥) + 10𝐶2 𝑝 2 𝑥 2 + 10𝐶3 𝑝 3 𝑥 3 + ⋯ + 𝑝10 𝑥10 So, the co-efficient of 𝑥 2 in the expansion of (1 + 𝑝𝑥)10 is 10𝐶2 𝑝 2 10! = 2!8! 𝑝 2 10.9.8! 2 = 𝑝 2.8! = 45𝑝 2. 𝟏 𝟏 𝟏𝟎 Problem : Find the co-efficient of 𝒙𝟐 in the expansion of (𝟏 − 𝒙). Solution : By using binomial theorem, we have 1 10 1 1 2 1 3 1 10 (1 − ) = 10𝐶0 − 10𝐶1 ( ) + 10𝐶2 ( ) − 10𝐶3 ( ) + ⋯ + 10𝐶10 ( ) 𝑥 𝑥 𝑥 𝑥 𝑥 1 1 10 10! 10.9.8! So, the co-efficient of 𝑥 2 in the expansion of (1 − 𝑥) is 10𝐶2 = 2!8! = 2.8! = 45. 𝟏 𝟏𝟎 Problem : Find the co-efficient of 𝒙−𝟑 in the expansion of (𝟏 − 𝒙 ). Solution : By using binomial theorem, we have 1 10 1 1 2 1 3 1 10 (1 − ) = 10𝐶0 − 10𝐶1 ( ) + 10𝐶2 ( ) − 10𝐶3 ( ) + ⋯ + 10𝐶10 ( ) 𝑥 𝑥 𝑥 𝑥 𝑥 1 10 10! 10.9.8.7! So, the co-efficient of 𝑥 −3 in the expansion of (1 − 𝑥) is −10𝐶3 = − 3!7! = − 6.7! = −120. 𝟏𝟎 Problem : Find the co-efficient of 𝒙𝟑 in the expansion of (𝟏 + 𝟑𝒙 + 𝟑𝒙𝟐 + 𝒙𝟑 ). Solution : Note that, (1 + 3𝑥 + 3𝑥 2 + 𝑥 3 )10 = [(1 + 𝑥)3 ]10 = (1 + 𝑥)30 = 30𝐶0 + 30𝐶1 𝑥 + 30𝐶2 𝑥 2 + 30𝐶3 𝑥 3 + 30𝐶4 𝑥 4 + ⋯ + 30𝐶30 𝑥 30 So, the co-efficient of 𝑥 3 in the expansion of (1 + 3𝑥 + 3𝑥 2 + 𝑥 3 )10 is 30𝐶3. 𝟏 𝟏 𝟏𝟎 Problem : Find the co-efficient of 𝒙𝟔 in the expansion of (𝟏 − 𝒙𝟐 ). Solution : By using binomial theorem, we have 1 10 1 1 2 1 3 1 10 (1 − 2 ) = 10𝐶0 − 10𝐶1 ( 2 ) + 10𝐶2 ( 2 ) − 10𝐶3 ( 2 ) + ⋯ + 10𝐶10 ( 2 ) 𝑥 𝑥 𝑥 𝑥 𝑥 1 1 10 10! 10.9.8.7! So, the co-efficient of 𝑥6 in the expansion of (1 − 𝑥 2 ) is −10𝐶3 = − 3!7! = − 6.7! = −120. 𝟏 𝟏𝟎 Problem : Find the co-efficient of sixth term in the expansion of (𝒂 − 𝟐𝒂). Solution : 1 5 Sixth term= 𝑡6 = 𝑡5+1 = 10𝐶5 𝑎10−5 (− 2𝑎) 1 = −10𝐶5. 𝑎 5. 25𝑎 5 1 = −10𝐶5. 25 1 10 Thus, the co-efficient of sixth term in the expansion of (𝑎 − 2𝑎) is 1 −10𝐶5. 25 10! = − 5!5! 10.9.8.7.6.5! 1 =− 5!5! × 32 10.9.8.7.6 1 = − 120 × 32 63 = −8 𝟕 𝟖 Problem : Find the co-efficient of 𝒙−𝟐 in the expansion of (𝟑𝒙 − 𝒙). Solution : 7 8 Let (𝑟 + 1)𝑡ℎ term contain 𝑥 −2 in the expansion of (3𝑥 − 𝑥). 7 𝑟 Now, 𝑡𝑟+1 = 8𝐶𝑟 (3𝑥)8−𝑟 (− 𝑥) 7𝑟 = (−1)𝑟 8𝐶𝑟 38−𝑟. 𝑥 8−𝑟. 𝑥 𝑟 = (−1)𝑟 8𝐶𝑟 38−𝑟. 7𝑟. 𝑥 8−2𝑟 Since 𝑡𝑟+1 contains 𝑥 −2 , so index of 𝑥 is −2. ∴ 8 − 2𝑟 = −2 ⇒ 2𝑟 = 10 ⇒ 𝑟 = 5. Hence (5 + 1)𝑡ℎ 𝑖. 𝑒. 6th term contains 𝑥 −2. Therefore, the co-efficient of 𝑥 −2  = (−1)𝑟 8𝐶𝑟 38−𝑟. 7𝑟 = (−1)5 8𝐶5 38−5. 75 = −8𝐶5. 33. 75 = −25412184 𝟓 𝒂𝟐 Problem : Find the co-efficient of 𝒙 in the expansion of (𝒙𝟐 + 𝒙 ). Solution : 5 𝑎2 Let (𝑟 + 1)𝑡ℎ term contain 𝑥 in the expansion of (𝑥 2 + 𝑥 ). 𝑟 𝑎2 Now, 𝑡𝑟+1 = 5𝐶𝑟 (𝑥 2 )5−𝑟 ( 𝑥 ) 𝑎 2𝑟 = 5𝐶𝑟. 𝑥10−2𝑟. 𝑥𝑟 2𝑟 10−3𝑟 = 5𝐶𝑟 𝑎 𝑥 Since 𝑡𝑟+1 contains 𝑥, so index of 𝑥 is 1. ∴ 10 − 3𝑟 = 1 ⇒ 3𝑟 = 9 ⇒ 𝑟 = 3. Hence (3 + 1)𝑡ℎ 𝑖. 𝑒. 4th term contains 𝑥. Therefore, the co-efficient of 𝑥 = 5𝐶𝑟 𝑎 2𝑟 = 5𝐶3 𝑎 6 5! 6 = 𝑎 3! 2! 5.4.3! 6 = 𝑎 3! × 2 = 10𝑎 6 𝟏 𝟏𝟎 Problem : Find the co-efficient of 𝒙𝟏𝟎 in the expansion of (𝒙𝟐 − 𝒙𝟑 ). Solution : 1 10 Let (𝑟 + 1)𝑡ℎ term contain 𝑥10 in the expansion of (𝑥 2 − 𝑥 3). 1 𝑟 Now, 𝑡𝑟+1 = 10𝐶𝑟 (𝑥 2 )10−𝑟 (− 𝑥 3) 1 = (−1)𝑟 10𝐶𝑟. 𝑥 20−2𝑟. 𝑥 3𝑟 = (−1)𝑟 10𝐶𝑟 𝑥 20−5𝑟 Since 𝑡𝑟+1 contains 𝑥10 , so index of 𝑥 is 10. ∴ 20 − 5𝑟 = 10 ⇒ 5𝑟 = 10 ⇒ 𝑟 = 2. Hence (2 + 1)𝑟𝑑 , 𝑖. 𝑒. , 3rd term contains 𝑥10. Therefore, the co-efficient of 𝑥10 = (−1)𝑟 10𝐶𝑟 = (−1)2 10𝐶2 10! = 2!8!  10.9.8! = 2.8! = 45 𝟏 𝟏𝟎 Problem : Find the co-efficient of 𝒙 in the expansion of (𝟏 − 𝟐𝒙𝟑 + 𝟑𝒙𝟓 ) (𝟏 + 𝒙 ). Solution : 1 10 (1 − 2𝑥 3 + 3𝑥 5 ) (1 + ) 𝑥 1 1 1 1 1 1 = (1 − 2𝑥 3 + 3𝑥 5 ) (1 + 10𝐶1. + 10𝐶2. 2 + 10𝐶3. 3 + 10𝐶4. 4 + 10𝐶5. 5 + ⋯ + 10 ) 𝑥 𝑥 𝑥 𝑥 𝑥 𝑥 From the above multiplication, it is clear that the co-efficient of 𝑥 in the expansion of 1 10 (1 − 2𝑥 3 + 3𝑥 5 ) (1 + ) is 𝑥 −2 × 10𝐶2 + 3 × 10𝐶4 10! 10! = −2 × +3× 2! 8! 4! 6! 10 × 9 × 8! 10 × 9 × 8 × 7 × 6! = −2 × +3× 2 × 8! 4 × 3 × 2 × 1 × 6! = −90 + 630 = 540 𝟏 𝒏 Problem : If the 4th term in the expansion of (𝒑𝒙 + 𝒙) is independent of 𝒙, find the value of 𝒏. 𝟓 Calculate 𝒑 if the 4th term be 𝟐. Solution : 1 3 Here, 𝑡4 = 𝑡3+1 = 𝑛𝐶3 (𝑝𝑥)𝑛−3 (𝑥) 1 = 𝑛𝐶3. 𝑝 𝑛−3. 𝑥 𝑛−3. 𝑥 3 = 𝑛𝐶3. 𝑝 𝑛−3. 𝑥 𝑛−6 Since 𝑡4 is independent of 𝑥, so index of 𝑥 is zero. ∴ 𝑛 − 6 = 0 ⇒ 𝑛 = 6. 5 If the 4th term be 2, then 5 𝑡4 = 2 5 ⇒ 𝑛𝐶3. 𝑝 𝑛−3. 𝑥 𝑛−6 = 2 5 ⇒ 6𝐶3. 𝑝 6−3. 𝑥 6−6 = 2 6! 5 ⇒ 𝑝3 = 3! (6 − 3)! 2 6 × 5 × 4 × 3! 3 5 ⇒ 𝑝 = 6 × 3! 2 6 × 5 × 4 × 3! 3 5 ⇒ 𝑝 = 6 × 3! 2  1 ⇒ 𝑝3 = 8 1 ⇒𝑝= 2 𝟏 𝒏 Problem : If the fifth term in the expansion of (𝒙𝟐 − 𝒙) is independent of 𝒙, find the value of 𝒏. Solution : 1 4 Here, 𝑡5 = 𝑡4+1 = 𝑛𝐶4 (𝑥 2 )𝑛−4 (− ) 𝑥 1 = 𝑛𝐶4. 𝑥 2𝑛−8. 𝑥 4 = 𝑛𝐶4 𝑥 2𝑛−12 Since 𝑡5 is independent of 𝑥, so index of 𝑥 is zero. ∴ 2𝑛 − 12 = 0 ⇒ 2𝑛 = 12 ⇒ 𝑛 = 6. 𝟏 𝟏𝟎 Problem : If the third and fourth terms in the expansion of (𝟐𝒙 + 𝟖) are equal , then find the value of 𝒙. Solution : 1 2 1 Here , 𝑡3 = 𝑡2+1 = 10𝐶2 (2𝑥)10−2 ( ) = 10𝐶2 (2𝑥) 8. 8 82 1 3 1 and 𝑡4 = 𝑡3+1 = 10𝐶3 (2𝑥)10−3 (8) = 10𝐶3 (2𝑥) 7. 83 By the given condition, 𝑡3 = 𝑡4 1 1 ⇒ 10𝐶2 (2𝑥)8. = 10𝐶3 (2𝑥)7. 82 83 10! 10! 1 ⇒. 2𝑥 =. 2!8! 3!7! 8 10! 10! 1 ⇒. 2𝑥 =. 2! × 8 × 7! 3 × 2! × 7! 8 1 ⇒ 2𝑥 = 3 1 ⇒𝑥= 6 Problem : If in the expansion of (𝟏 + 𝒙)𝟐𝟎 the co-efficients of 𝒓𝒕𝒉 term and (𝒓 + 𝟏)𝒕𝒉 term is in the ratio 𝟏 ∶ 𝟐 , then find the value of 𝒓. Solution : In the expansion of (1 + 𝑥)20 𝑡𝑟 = 𝑡(𝑟−1)+1 = 20𝐶𝑟−1 𝑥 𝑟−1 𝑡𝑟+1 = 20𝐶𝑟 𝑥 𝑟 By the given condition, 𝑡𝑟 1 𝑡𝑟+1 =2 20𝐶𝑟−1 1 ⇒ = 20𝐶𝑟 2  20𝐶𝑟 ⇒ 20 =2 𝐶𝑟−1 20−𝑟+1 𝑛𝐶𝑟 𝑛−𝑟+1 ⇒ =2 [∵ = ] 𝑟 𝑛𝐶𝑟−1 𝑟 ⇒ 20 − 𝑟 + 1 = 2𝑟 ⇒ 21 = 3𝑟 ⇒𝑟=7 𝟏 𝟏𝟏 Problem : If the co-efficient of 𝒙𝟕 in the expansion of (𝒑𝒙𝟐 + 𝒒𝒙) be equal to the co-efficient of 𝟏 𝟏𝟏 𝒙−𝟕 in the expansion of (𝒑𝒙 − ) , then prove that 𝒑𝒒 = 𝟏. 𝒒𝒙𝟐 Solution : 1 11 Let (𝑟 + 1)𝑡ℎ term contain 𝑥 7 in the expansion of (𝑝𝑥 2 + 𝑞𝑥). 1 𝑟 1 1 Now, 𝑡𝑟+1 = 11𝐶𝑟 (𝑝𝑥 2 )11−𝑟 (𝑞𝑥) = 11𝐶𝑟. 𝑝11−𝑟. 𝑥 22−2𝑟. 𝑞𝑟𝑥 𝑟 = 11𝐶𝑟. 𝑝11−𝑟. 𝑞𝑟. 𝑥 22−3𝑟 ∴ 22 − 3𝑟 = 7 ⇒ 3𝑟 = 15 ⇒ 𝑟 = 5 1 ∴ 𝑡6 = 11𝐶5. 𝑝 6. 5. 𝑥 7 𝑞 1 11 Again, let (𝑘 + 1)𝑡ℎ term contain 𝑥 −7 in the expansion of (𝑝𝑥 − 𝑞𝑥 2). 1 𝑘 Now, 𝑡𝑘+1 = 11𝐶𝑘 (𝑝𝑥)11−𝑘 (− 𝑞𝑥 2 ) 1 = (−1)𝑘 11𝐶𝑘. 𝑝11−𝑘. 𝑥11−𝑘. 𝑞𝑘 𝑥 2𝑘 1 11−3𝑘 = (−1)𝑘 11𝐶𝑘. 𝑝11−𝑘..𝑥 𝑞𝑘 ∴ 11 − 3𝑘 = −7 ⇒ 3𝑘 = 18 ⇒ 𝑘 = 6 1 1 1 ∴ 𝑡7 = (−1)6 11𝐶6. 𝑝 5. 𝑞6. 𝑥 −7 = 11𝐶6. 𝑝 5. 𝑞6. 𝑥 −7 = 11𝐶5. 𝑝 5. 𝑞6. 𝑥 −7 [∵ 𝑛𝐶𝑟 = 𝑛𝐶𝑛−𝑟 ] By given condition, 1 1 11𝐶5. 𝑝 6. 𝑞5 = 11𝐶5. 𝑝 5. 𝑞6 1 ⇒𝑝= 𝑞 ⇒ 𝑝𝑞 = 1 Middle terms in the expansion of (𝒂 + 𝒃)𝒏 Regarding the middle term in the expansion of (𝑎 + 𝑏)𝑛 , we have (i) If 𝑛 is even, then the number of terms in the expansion will be 𝑛 + 1. Since 𝑛 is even, so 𝑛 + 1 𝑛+1+1 𝑡ℎ 𝑛 𝑡ℎ is odd. Therefore, the middle term is ( 2 ) ,i.e., ( 2 + 1) term. 8 𝑡ℎ For example, in the expansion of (𝑥 + 2𝑦)8 , the middle term is ( + 1) ,i.e., 5th term. 2 (ii) If 𝑛 is odd, then 𝑛 + 1 is even. So, there will be two middle terms in the expansion, namely,  𝑛+1 𝑡ℎ 𝑛+1 𝑡ℎ ( ) term and ( + 1) term. 2 2 7+1 𝑡ℎ For example, in the expansion of (2𝑥 − 𝑦)7 , the middle terms are ( ) ,i.e., 4th term and 2 7+1 𝑡ℎ ( + 1) term, i.e., 5th term. 2 𝟏.𝟑.𝟓…(𝟐𝒏−𝟏) Problem : Prove that the middle term in the expansion of (𝟏 + 𝒙)𝟐𝒏 is 𝒏!. 𝟐𝒏. 𝒙𝒏 , where 𝒏 is a positive integer. Solution : 2𝑛 𝑡ℎ Since 2𝑛 is even, so the middle term in the expansion of (1 + 𝑥)2𝑛 is ( 2 + 1) , i.e., (𝑛 + 1)𝑡ℎ term which is given by 𝑡𝑛+1 = 2𝑛𝐶𝑛 (1)2𝑛−𝑛 𝑥 𝑛 (2𝑛)! 𝑛 = 𝑛!𝑛! 𝑥 1.2.3.4….(2𝑛−2)(2𝑛−1)2𝑛 𝑛 = 𝑛!𝑛! 𝑥 [1.3.5…(2𝑛−1)][2.4.6….(2𝑛−2)2𝑛] 𝑛 = 𝑛!𝑛! 𝑥 𝑛 [1.3.5…(2𝑛−1)]2 [1.2.3…(𝑛−1)𝑛] 𝑛 = 𝑛!𝑛! 𝑥 [1.3.5…(2𝑛−1)]𝑛! 𝑛 𝑛 = 𝑛!𝑛! 2 𝑥 1.3.5…(2𝑛−1) 𝑛 𝑛 = 𝑛! 2 𝑥 𝟏 𝟖 Problem : Find the middle term in the expansion of (𝟐𝒙𝟐 − ). 𝒙 Solution : 1 8 Here 𝑛 = 8 which is even. So, there is only one middle term in the expansion of (2𝑥 2 − 𝑥) and it is 8 𝑡ℎ (2 + 1) , i.e., 5𝑡ℎ term which is given by 1 4 𝑡5 = 𝑡4+1 = 8𝐶4 (2𝑥 2 )8−4 (− 𝑥) 8! 1 = 4!4! 24 𝑥 8. 𝑥 4 8.7.6.5.4! = 24.4!. 16𝑥 4 = 1120𝑥 4 𝟏 𝟖 Problem : Find the middle term in the expansion of (𝒂𝒙 − 𝒂𝒙). Solution : 1 8 Here 𝑛 = 8 which is even. So, there is only one middle term in the expansion of (𝑎𝑥 − ) and it is 𝑎𝑥 8 𝑡ℎ ( + 1) , i.e., 5𝑡ℎ term which is given by 2 1 4 𝑡5 = 𝑡4+1 = 8𝐶4 (𝑎𝑥)8−4 (− 𝑎𝑥)  8! 1 = 4!4! 𝑎 4 𝑥 4. 𝑎4 𝑥 4 8.7.6.5.4! = 24.4! = 70 𝟏 𝟔 Problem : Find the middle term in the expansion of (𝒂𝒙 + 𝒂𝒙). Solution : 1 6 Here 𝑛 = 6 which is even. So, there is only one middle term in the expansion of (𝑎𝑥 + 𝑎𝑥) and it is 6 𝑡ℎ (2 + 1) , i.e., 4𝑡ℎ term which is given by 1 3 𝑡4 = 𝑡3+1 = 6𝐶3 (𝑎𝑥)6−3 ( ) 𝑎𝑥 6! 3 3 1 = 𝑎 𝑥. 3 3 3!3! 𝑎 𝑥 6.5.4.3! = 6.3! = 20 𝟏 𝟖 Problem : Find the middle term in the expansion of (𝟑𝒙 − 𝟐𝒙). Solution : 1 8 Here 𝑛 = 8 which is even. So, there is only one middle term in the expansion of (3𝑥 − 2𝑥) and it is 8 𝑡ℎ (2 + 1) , i.e., 5𝑡ℎ term which is given by 1 4 𝑡5 = 𝑡4+1 = 8𝐶4 (3𝑥)8−4 (− ) 2𝑥 8! 4 4 1 = 4!4! 3 𝑥. 24𝑥 4 8.7.6.5.4! 81 = × 24.4! 16 35×81 = 8 2835 = 8 𝟏 𝟖 Problem : Find the middle term in the expansion of (𝒙𝟐 + ). 𝒙𝟐 Solution : 1 8 Here 𝑛 = 8 which is even. So, there is only one middle term in the expansion of (𝑥 2 + 𝑥 2) and it is 8 𝑡ℎ (2 + 1) , i.e., 5𝑡ℎ term which is given by 1 4 𝑡5 = 𝑡4+1 = 8𝐶4 (𝑥 2 )8−4 (𝑥 2) 8! 8 1 = 𝑥. 8 4!4! 𝑥 8.7.6.5.4! = 24.4! = 70 Infinite Series If −1 < 𝑥 < 1 , 𝑖. 𝑒. , |𝑥| < 1,then (i) (1 − 𝑥)−1 = 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + ⋯ (ii) (1 + 𝑥)−1 = 1 − 𝑥 + 𝑥2 − 𝑥3 + 𝑥4 − 𝑥5 + ⋯ (iii) (1 − 𝑥)−2 = 1 + 2𝑥 + 3𝑥 2 + 4𝑥 3 + 5𝑥 4 + 6𝑥 5 + ⋯ (iv) (1 + 𝑥)−2 = 1 − 2𝑥 + 3𝑥 2 − 4𝑥 3 + 5𝑥 4 − 6𝑥 5 + ⋯ 𝟏 Problem : Find the values of 𝒙 for which (𝟏 − 𝟐𝒙)−𝟐 can be extended in a binomial series. Solution : 1 1 1 (1 − 2𝑥)−2 can be extended in a binomial series if −1 < 2𝑥 < 1 , i. e., if − < 𝑥 <. 2 2 Problem : If |𝒙| < 𝟏, then prove that (𝟏 + 𝒙 + 𝒙𝟐 + 𝒙𝟑 + ⋯ )(𝟏 − 𝒙 + 𝒙𝟐 − 𝒙𝟑 + ⋯ ) = 𝟏 + 𝒙𝟐 + 𝒙𝟒 + 𝒙𝟔 + ⋯. Solution : We have, (1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯ )(1 − 𝑥 + 𝑥 2 − 𝑥 3 + ⋯ ) = (1 − 𝑥)−1 (1 + 𝑥)−1 [∵ |𝑥| < 1 ] = [(1 − 𝑥)(1 + 𝑥)]−1 = (1 − 𝑥 2 )−1 = 1 + 𝑥 2 + (𝑥 2 )2 + (𝑥 2 )3 + ⋯ [∵ |𝑥| < 1 ⇒ |𝑥 2 | < 1] = 1 + 𝑥2 + 𝑥4 + 𝑥6 + ⋯ Problem If 𝒙 = 𝟏 + 𝒂 + 𝒂𝟐 + + ⋯ (|𝒂| < 𝟏) 𝐚𝐧𝐝 𝒚 = 𝟏 + 𝒃 + 𝒃𝟐 + ⋯ (|𝒃| < 𝟏), then prove that 𝒙𝒚 𝟏 + 𝒂𝒃 + 𝒂𝟐 𝒃𝟐 + ⋯ =. 𝒙+𝒚−𝟏 Solution : Given, 𝑥 = 1 + 𝑎 + 𝑎 2 + + ⋯ (|𝑎| < 1) and 𝑦 = 1 + 𝑏 + 𝑏2 + ⋯ (|𝑏| < 1) ⇒ 𝑥 = (1 − 𝑎)−1 ⇒ 𝑦 = (1 − 𝑏)−1 1 1 ⇒ 𝑥 = 1−𝑎 ⇒ 𝑦 = 1−𝑏 𝑥𝑦 Now, 𝑥+𝑦−1 1 1. 1−𝑎 1−𝑏 = 1 1 + 1−𝑎 1−𝑏 −1 1 (1−𝑎)(1−𝑏) = 1−𝑏+1−𝑎−(1−𝑎)(1−𝑏) (1−𝑎)(1−𝑏) 1 = 2−𝑎−𝑏−1+𝑏+𝑎−𝑎𝑏  1 = 1−𝑎𝑏 = (1 − 𝑎𝑏)−1 = 1 + (𝑎𝑏) + (𝑎𝑏)2 + ⋯ [∵ |𝑎| < 1 &|𝑏| < 1 ⇒ |𝑎𝑏| < 1] = 1 + 𝑎𝑏 + 𝑎 2 𝑏2 + ⋯ Exponential Series If 𝑥 is a real number, then 𝑥 𝑥2 𝑥3 𝑥4 𝑒𝑥 = 1 + + + + +⋯ 1! 2! 3! 4! Logarithmic Series (i) If −1 < 𝑥 ≤ 1, then 𝑥2 𝑥3 𝑥4 log𝑒 (1 + 𝑥) = 𝑥 − + − + ⋯ 2 3 4 (ii) If −1 ≤ 𝑥 < 1, then 𝑥2 𝑥3 𝑥4 log𝑒 (1 − 𝑥) = −𝑥 − − − − ⋯ 2 3 4