# Misc 1

Last updated at March 8, 2017 by Teachoo

Last updated at March 8, 2017 by Teachoo

Transcript

Misc 1 Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively. We know that (a + b)n = nC0 an b0 + nC1 an – 1 b1 + nC2 an – 2 b2 + …. …. … + nCn – 1 a1 bn – 1 + nCn a0 bn = an + nC1 an – 1 b1 + nC2 an – 2b2 +…. …. + nCn – 1 a1 bn – 1 + bn So first 3 terms are an , nC1 an–1b and nC2 an–2 b2 Also, it is given that their value are 729,7290 and 30375 Therefore, an = 729 nC1 an–1 b = 7290 nC2 an–2 b2 = 30375 Dividing (1) by (2) 𝑎𝑛𝑛𝐶1𝑎𝑛−1 𝑏 = 7297290 𝑎𝑛𝑛!1!𝑛−1!𝑎𝑛−1 𝑏 = 110 𝑎𝑛𝑛(𝑛−1)!1𝑛−1!𝑎𝑛−1 𝑏 = 110 𝑎𝑛𝑛𝑎𝑛−1 𝑏 = 110 𝑎𝑛 −(𝑛−1)𝑛 𝑏 = 110 𝑎1𝑛𝑏 = 110 𝑎𝑛𝑏 = 110 Dividing (2) by (3) 𝑛𝐶1𝑎𝑛−1 𝑏𝑛𝐶2𝑎𝑛−2𝑏2 = 729030375 𝑛!1!𝑛−1! 𝑎𝑛−1.𝑏𝑛!2!𝑛 −2! 𝑎𝑛−2 𝑏2 = 625 𝑛! 𝑎𝑛−1𝑏𝑛−1! × 2!𝑛−2!𝑛!𝑎𝑛−2 𝑏2 = 625 𝑛! 2! 𝑛−2!𝑛!𝑛−1! × 𝑎𝑛−1𝑎𝑛−2 × 𝑏𝑏2 = 625 2 𝑛! 𝑛−2!𝑛!𝑛−1𝑛−2! × 𝑎𝑛−1𝑎𝑛−2 × 𝑏𝑏2 = 625 2𝑛−1 × 𝑎𝑛−1 −(𝑛−2)1 × 𝑏𝑏2 = 625 2(𝑛−1) × 𝑎𝑛−1−𝑛+2𝑏 = 625 2𝑎𝑛−1𝑏 = 625 Now, our equations are 𝑎𝑛𝑏 = 110 …(4) 2𝑎𝑛−1𝑏 = 625 …(5) Dividing (4) by (5) 𝑎𝑛𝑏2𝑎𝑛−1𝑏 = 110625 𝑎𝑛𝑏 × 𝑛−1𝑏2𝑎 = 110 × 256 𝑛−12𝑛 = 512 12 (n – 1) = 2n (5) 12n – 12 = 10n 12n – 10n = 12 2n = 12 n = 122 n = 6 Putting n = 6 in (1) an = 729 a6 = 729 a6 = (3)6 a = 3 Putting a = 3 , n = 6 in (5) 2𝑎𝑛−1𝑏 = 625 36 × 𝑏 = 110 12𝑏 = 110 102 = b 5 = b b = 5 Hence, a = 3 , b = 5 & n = 6

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

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/.