Binomial Theorem PDF 2020 A Applied Math

Departmental Library I/III Department of Applied Mathematics Since 2011 Semester The Maharaja Sayajirao University of Baroda,Vadodara...

Departmental Library I/III Department of Applied Mathematics Since 2011 Semester The Maharaja Sayajirao University of Baroda,Vadodara 2020 Polytechnic Subject: Applied Mathematics-I/III Binomial Theorem Factorial Notation: (i) 𝑛! = 𝑛 × (𝑛 − 1) × (𝑛 − 2) × … … 3 × 2 × 1 (ii) 0! ! = 1! = 1 𝑛! (iii) 𝑛𝐶𝑟 = 𝑟! !(𝑛−𝑟)! (iv) 𝑛𝐶0 = 𝑛𝐶𝑛 = 1, 𝑛𝐶1 = n (v) 𝑛𝐶𝑟 = 𝑛𝐶𝑛−𝑟 (vi) 𝑛𝐶𝑟 + 𝑛𝐶𝑟−1 = 𝑛 + 1𝐶𝑟 Result: If 𝑛𝐶𝑥 = 𝑛𝐶𝑦 , then either 𝑥 = 𝑦 or 𝑥 + 𝑦 = 𝑛 Binomial Expression: An expression consisting of two terms is called a Binomial Expression + Ex: 𝑎 + 𝑏, 2𝑎 + 3𝑏, 𝑎2 + 𝑏 2 etc. Binomial Theorem: The general form of the Binomial Expression is (𝑎 + 𝑏) and the expansion of (𝑎 + 𝑏)𝑛 , 𝑛 ∈ 𝑁 is called the Binomial Theorem. Binomial Theorem for Positive Integers: Theorem: If a and b are real numbers , then for all n∈ 𝑁 (𝑎 + 𝑏)𝑛 = 𝑛𝐶0 𝑎𝑛 𝑏 0 + 𝑛𝐶1 𝑎𝑛−1 𝑏1 + 𝑛𝐶2 𝑎𝑛−2 𝑏 2 + … … … … + 𝑛𝐶𝑛−1 𝑎1 𝑏 𝑛−1 + 𝑛𝐶𝑛 𝑎0 𝑏 𝑛 i.e. (𝑎 + 𝑏)𝑛 = ∑𝑛𝑟=0 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟 The General Term in a Binomial Expansion of (𝒂 + 𝒃)𝒏 : The (𝑟 + 1)𝑡ℎ term in a Binomial Expansion of (𝑎 + 𝑏)𝑛 is given by 𝑇𝑟+1= 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟 Is called the General Term. (1) Find the fourth term in the expansion of (𝑥 − 2𝑦)12 Sol: comparing (𝑥 − 2𝑦)12 with (𝑎 + 𝑏)𝑛 , we get 𝑎 = 𝑥 , 𝑏 = −2𝑦 , 𝑛 = 12 The (𝑟 + 1)𝑡ℎ term in the expansion of (𝑎 + 𝑏)𝑛 is given by  Departmental Library I/III Department of Applied Mathematics Since 2011 Semester The Maharaja Sayajirao University of Baroda,Vadodara 2020 Polytechnic Subject: Applied Mathematics-I/III 𝑇𝑟+1= 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟 ∴ 𝑇𝑟+1= 12𝐶𝑟 (𝑥)12−𝑟 (−2𝑦)𝑟 ….(1) For the fourth term, put 𝑟 = 3 in eq.(1), we have ∴ 𝑇4 = 𝑇3+1 = 12𝐶3 (𝑥)12−3 (−2𝑦)3 12 ! = (𝑥)9 (−2𝑦)3 3 !(12−3)! 12 ! = (𝑥)9 (−8)𝑦 3 3 !9! 12×11×10×9! =- 𝑥9𝑦3 3×2×1××9! = -2× 11 × 10 × 8 × 𝑥 9 𝑦 3 = -1760 𝑥 9 𝑦 3 Middle Terms in a Binomial Expansion of (𝒂 + 𝒃)𝒏 : Since the binomial expansion of (𝑎 + 𝑏)𝑛 ℎas (n+1) terms.Therefore 𝑛 (1) If n is even, then the middle term is ( 2 + 1)th term. 𝑛+1 th 𝑛+3 th (2) If n is odd, then the middle terms are ( ) term and ( ) term. 2 2 𝑥 10 Ex: Find the middle term in the expansion of (3 + 9𝑦) 𝑥 10 comparing (3 + 9𝑦) with (𝑎 + 𝑏)𝑛 , we get 𝑥 𝑎= , 𝑏 = 9𝑦, 𝑛 = 10 3 The (𝑟 + 1)𝑡ℎ term in the expansion of (𝑎 + 𝑏)𝑛 is given by 𝑇𝑟+1= 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟 𝑥 ∴ 𝑇𝑟+1= 10𝐶𝑟 (3)10−𝑟 (9𝑦)𝑟 ……(1) Here, n is even. 𝑛 The middle term is ( 2 + 1)th term. 10 i.e. ( 2 + 1)th term. = (5+1)th term = 6𝑡ℎ term  Departmental Library I/III Department of Applied Mathematics Since 2011 Semester The Maharaja Sayajirao University of Baroda,Vadodara 2020 Polytechnic Subject: Applied Mathematics-I/III For the 6th term , put 𝑟 = 5 in eq.(1), 𝑥 𝑇6 =𝑇5+1= 10𝐶5 (3)10−5 (9𝑦)5 10 ! 𝑥 5 = ( ) (9)5 𝑦 5 5 !(10−5)! 3 10 ! 𝑥 5 = 5 !5! (3)5 (9)5 𝑦 5 10×9×8×7×6×5! (9)5 = 5×4×3×2×1××5! (3)5 𝑥 5 𝑦 5 = 61236 𝑥 5 𝑦 5 Ex: Find the co-efficient of x5 in the binomial expansion of (𝑥 + 3)8. Sol: comparing (𝑥 + 3)8 with (𝑎 + 𝑏)𝑛 , we get 𝑎 = 𝑥 , 𝑏 = 3, 𝑛 = 8 The General term in the expansion of (𝑎 + 𝑏)𝑛 is 𝑇𝑟+1 = 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟 ∴ 𝑇𝑟+1= = 8𝐶𝑟 𝑥 8−𝑟 (3)𝑟 ….(1) Putting 8 − 𝑟 = 5 in eq.(1), we get 𝑟 = 8−5 = 3 Putting 𝑟 = 3 in eq.(1) , we get 𝑇4 =𝑇3+1 = 8𝐶3 𝑥 8−3 (3)3 = 8𝐶3. 27 𝑥 5 The coefficient of x5 is = 8𝐶3. (27) 8! =. (27) 3 !(8−3)! 8! =. (27) 3 !5! 8×7×6×5! = 3×2×1××5!. (27) = 8× 7 × 27  Departmental Library I/III Department of Applied Mathematics Since 2011 Semester The Maharaja Sayajirao University of Baroda,Vadodara 2020 Polytechnic Subject: Applied Mathematics-I/III = 1512 1 14 Ex: Find the term independent of x , x≠ 0 in the expansion of (𝑥 − 𝑥) 1 14 Sol: comparing with (𝑥 − 𝑥) with (𝑎 + 𝑏)𝑛 , we get 1 𝑎 = 𝑥 , 𝑏 = − 𝑥 , 𝑛 = 14 The General term in the expansion of (𝑎 + 𝑏)𝑛 is 𝑇𝑟+1 = 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟 1 𝑟 ∴ 𝑇𝑟+1 = 14𝐶𝑟 (𝑥)14−𝑟 (− ) 𝑥 (𝑥)14−𝑟 = 14𝐶𝑟 (−1)𝑟 (𝑥)𝑟 ∴ 𝑇𝑟+1= 14𝐶𝑟 (𝑥)14−2𝑟 (−1)𝑟 …..(1) For this term to be independent of 𝑥 , we must have 14 − 2𝑟 = 0 ∴ 2𝑟 = 14 Putting 𝑟 = 7 in eq.(1) , we have 𝑇8 =𝑇7+1= 14𝐶7 (𝑥)14−14 (−1)7 = 14𝐶7 (−1)7 𝑥 0 = 14𝐶7 (−1)7. (1) = - 14𝐶7 14 ! =- 7 !(14−7)! 14 ! =- 7 !7! 14×13×12×11×10×9×8×7! =- 7×6×5×4×3×2×1××7! = - 13× 11 × 30 × 8 = - 3432  Departmental Library I/III Department of Applied Mathematics Since 2011 Semester The Maharaja Sayajirao University of Baroda,Vadodara 2020 Polytechnic Subject: Applied Mathematics-I/III Ex: If 20𝐶𝑥 = 20𝐶𝑥+8 , find the value of 𝑥. Sol: We have 20𝐶𝑥 = 20𝐶𝑥+8 ∴ 𝑥 + 𝑥 + 8 = 20 ∴ 2𝑥 + 8 = 20 ∴ 2𝑥 = 20 − 8 ∴ 2𝑥 = 12 𝑥= 6 Problems: (1) (a) Prove that nc = n c n – 1 (b) Obtain the value of 8c3 and 25c23 r (2) If nc10= n c 5 then calculate the value of n. (3) If 2nc3/n c 2 = 12 then find the value of n. (4) If ncr/n-1 c r-1 then prove that n=2r. (5) If 2nc3= 11( nc3) ,find n. ∝c ∝c (6) If 4=(7) 3 then find the value of α k k (7) Obtain the value of k if c3 = ( 6 ) c2.  Departmental Library I/III Department of Applied Mathematics Since 2011 Semester The Maharaja Sayajirao University of Baroda,Vadodara 2020 Polytechnic Subject: Applied Mathematics-I/III 25c 25cβ – 3 then find the value of (8) If β= 𝛽 Note : If ncx= n c y then either x = y or x + y = n (9) If 15cm= 15 c m+1 then find mc3. (10) If pc5= p c 15 then find pc4. (11) If 21ca= 21 c a+1 then find ac7. 12 𝑥2 (12) Find the coefficient of 𝑥 22 in the expansion of ( 2 − 2𝑥). 1 10 (13)Find the coefficient of 𝑦 −8 in the expansion of (2𝑦 − 2𝑦2 ). 11 2 𝑧2 (14) Find the coefficient of 𝑧10 in the expansion of (𝑧 + 2 ). 𝑘 11 (15)The term independent of 𝑥 in the expansion of (𝑥 3 + 𝑥 8 ) is 1320 find 𝑘. (16) Find the fifth term in the expansion of (2𝑧 − 𝑦)8. 3 𝑤 10 (17) Find the sixth term in the expansion of (𝑤 − 2 )  Departmental Library I/III Department of Applied Mathematics Since 2011 Semester The Maharaja Sayajirao University of Baroda,Vadodara 2020 Polytechnic Subject: Applied Mathematics-I/III 1 (18) Find the middle term in the expansion of (3𝑥 2 − 2𝑥)6 𝑘 (19) If the term free from y in the expansion of (√𝑦 + 𝑦2 )10 is 405 find k. 𝑎 (20) If the coefficient of 𝑎7 and 𝑎8 in the expansion of (2 + 3)𝑘 are the same, find k. 1 (21) Find the term independent of 𝑥 in (√𝑥 − 𝑥 2 )10. 1 10 (22) Find the term independent of 𝑥 in (3𝑥 + ). √𝑥 1 10 (23) Find the term independent of 𝑥 in (2𝑥 − ). 3𝑥 1 (24) Find the term independent of 𝑥 in (3𝑥 − 2𝑥 3 )8. 𝛼 (25) If the middle term in the expansion of (2 + 2 )8 is 1120, find α. 𝑥 2 14 (26) Find the middle term in the expansion of (3 − ) 2 1 9 (27) Find the coefficient of 𝑥 5 in the expansion of (2𝑧 + 3𝑥) 1 (28) Find the term involving 𝑥 5 in the expansion of (3𝑥 + 2𝑥)7  Departmental Library I/III Department of Applied Mathematics Since 2011 Semester The Maharaja Sayajirao University of Baroda,Vadodara 2020 Polytechnic Subject: Applied Mathematics-I/III 8 𝑥3 𝑥 𝑦2 (29) Find the coefficient of in the expansion of ( − ). 𝑦2 𝑦 2𝑥 3 1 11 (30) Find the coefficient of 𝑥 −7 in the expansion of (𝑎𝑥 − ). 𝑏𝑥 2