# Example 9 - Chapter 8 Class 11 Binomial Theorem

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 9 The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio 1: 7 : 42. Find n. Let the three consecutive terms be (r – 1)th, rth and (r + 1)th terms. i.e. Tr-1 , Tr & Tr+1 We know that general term of expansion (a + b )n is Tr+1 = nCr an – r br For (1 + a)n , Putting a = 1 , b = a Tr+1 = nCr 1n – r ar Tr+1 = nCr ar ∴ Coefficient of (r + 1)th term = nCr For rth term of (1 + a)n Replacing r with r – 1 in (1) Tr – 1 + 1 = nCr – 1 ar – 1 Tr = nCr – 1 ar – 1 ∴ Coefficient of (r)th term = nCr-1 For (r – 1)th term of (1 + a)n Replacing r with r – 2 in (1) Tr – 2 + 1 = nCr – 2 ar – 2 Tr – 1 = nCr – 2 ar – 2 ∴ Coefficient of (r – 1)th term = nCr – 2 Since the coefficient of (r – 1)th, rth and (r + 1)th terms are in ratio 1 : 7 : 42 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 (𝑟−1)𝑡ℎ 𝑡𝑒𝑟𝑚𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑟𝑡ℎ 𝑡𝑒𝑟𝑚 = 17 𝑛𝐶𝑟 −2𝑛𝐶𝑟 −1 = 17 𝑛!𝑟 − 2![𝑛 − 𝑟 − 2)!𝑛!𝑟 − 1!𝑛 − (𝑟 − 1)! = 17 𝑛!𝑟 − 2![𝑛 − 𝑟 − 2)! × 𝑟 − 1![𝑛 − 𝑟 − 1)!𝑛! = 17 𝑟−1𝑛−𝑟−1!(𝑛−𝑟−2)! = 17 𝑟 − 1𝑛− 𝑟 + 1!(𝑛 − 𝑟 + 2)! = 17 𝑟 − 1𝑛− 𝑟 + 1!(𝑛 − 𝑟 + 2)(𝑛 − 𝑟 +2 −1)! = 17 𝑟 − 1𝑛− 𝑟 + 1!(𝑛 − 𝑟 + 2)(𝑛 − 𝑟 +1)! = 17 𝑟 − 1(𝑛 − 𝑟 + 2) = 17 7(r – 1) = n – r + 2 n – 8r + 9 = 0 Also 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑟𝑡ℎ 𝑡𝑒𝑟𝑚𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 (𝑟 + 1)𝑡ℎ 𝑡𝑒𝑟𝑚 = 742 𝑛!𝑟−1![𝑛−𝑟−1)!𝑛!𝑟! 𝑛−𝑟! = 742 𝑛!𝑟−1!(𝑛−𝑟+1)! × 𝑟! 𝑛−𝑟! 𝑛! = 16 𝑛! × 𝑟 × 𝑟−1!𝑛−𝑟!𝑛!𝑟 − 1! (𝑛 − 𝑟 +1)! = 16 𝑟𝑛−𝑟!(𝑛 − 𝑟 +1)! = 16 𝑟 (𝑛−𝑟)!(𝑛 − 𝑟 +1) (𝑛−𝑟)! = 16 𝑟𝑛+1−𝑟 = 16 6r = n + 1 – r n – 7r + 1 = 0 Now we have n – 8r + 9 = 0 …(1) n – 7r + 1 = 0 …(2) From (1) n – 8r + 9 = 0 n = 8r – 9 Putting n = 8r – 9 in (2) (8r – 9) – 7r + 1 = 0 8r – 9 – 7r + 1 = 0 r – 8 = 0 r = 8 Putting value of r in (1) n – 8r + 9 = 0 n – 8(8) + 9 = 0 n – 64 + 9 = 0 n – 55 = 0 n = 55 Hence n = 55

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.