# Ex 8.2,10

Last updated at March 8, 2017 by Teachoo

Last updated at March 8, 2017 by Teachoo

Transcript

Ex8.2, 10 The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r. Finding (r – 1)th term , rth & (r + 1)th term of (x + 1)n Writing (x + 1)n as (1 + x)n We know that general term of expansion (a + b)n is Tr+1 = nCr an – rbr For (1 + x)n , Putting a = 1 , b = a Tr+1 = nCr 1n – r xr Tr+1 = nCr xr ∴ Coefficient of (r + 1)th term = nCr For rth term of (1 + x)n Replacing r with r – 1 in (1) Tr – 1 + 1 = nCr – 1 xr – 1 Tr = nCr – 1 xr – 1 ∴ Coefficient of (r)th term = nCr – 1 For (r – 1)th term of (1 + x)n Replacing r with r – 2 in (1) Tr – 2 + 1 = nCr – 2 xr – 2 Tr – 1 = nCr – 2 xr – 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 : 3 : 5 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 (𝑟 − 1)𝑡ℎ 𝑡𝑒𝑟𝑚𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑟𝑡ℎ 𝑡𝑒𝑟𝑚 = 13 𝑛𝐶𝑟 −2𝑛𝐶𝑟 −1 = 13 𝑛!𝑟−2![𝑛−𝑟−2)!𝑛!𝑟−1!𝑛−(𝑟−1)! = 13 𝑛!𝑟−2![𝑛−𝑟−2)! × 𝑟−1![𝑛−𝑟−1)!𝑛! = 13 𝑛!𝑟−1𝑟−2![𝑛−𝑟−1)!𝑛!𝑟−2![𝑛−𝑟−2)! = 13 𝑟−1𝑛−𝑟−1!(𝑛−𝑟−2)! = 13 𝑟 − 1𝑛− 𝑟 + 1!(𝑛 − 𝑟 + 2)! = 13 𝑟 − 1𝑛− 𝑟 + 1!(𝑛 − 𝑟 + 2)(𝑛 − 𝑟 +2 −1)! = 13 𝑟 − 1𝑛− 𝑟 + 1!(𝑛 − 𝑟 + 2)(𝑛 − 𝑟 +1)! = 13 𝑟 − 1(𝑛 − 𝑟 + 2) = 13 3(r – 1) = n – r + 2 3r – 3 = n + 2 – r n + 2 – r – 3r + 3 = 0 n – 4r + 5 = 0 Also 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑟𝑡ℎ 𝑡𝑒𝑟𝑚𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 (𝑟 + 1)𝑡ℎ 𝑡𝑒𝑟𝑚 = 35 𝑛!𝑟−1![𝑛−𝑟−1)!𝑛!𝑟! 𝑛−𝑟! = 35 𝑛!𝑟−1!(𝑛−𝑟+1)! × 𝑟! 𝑛−𝑟! 𝑛! = 35 𝑛! × 𝑟 × 𝑟−1!𝑛−𝑟!𝑛!𝑟 − 1! (𝑛 − 𝑟 +1)! = 35 𝑟𝑛−𝑟!(𝑛 − 𝑟 +1)! = 35 𝑟 (𝑛−𝑟)!(𝑛 − 𝑟 +1) (𝑛−𝑟)! = 35 𝑟𝑛 − 𝑟 + 1 = 35 5r = 3 (n – r + 1) 5r = 3n – 3r + 3 0 = 3n – 3r + 3 – 5r 0 = 3n – 8r + 3 3n – 8r + 3 = 0 Now our equations are n – 4r + 5 = 0 …(1) & 3n – 8r + 3 = 0 …(2) From (1) n – 4r + 5 = 0 n = 4r – 5 Putting n = 4r – 5 in (2) 3n – 8r + 3 = 0 3(4r – 5) – 8r + 3 = 0 12r – 15 – 8r + 3 = 0 12r – 8r – 15 + 3 = 0 4r – 12 = 0 4r = 12 r = 124 r = 3 Putting r = 3 in n = 4r – 5 n = 4(3) – 5 n = 12 – 5 n = 7 Hence n = 7 & r = 3

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.