Misc 1 - Find a, b, n in expansion of (a + b)n if first three

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Question 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 + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn 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 ∴ an = 729 nC1 an – 1 b = 7290 nC2 an – 2 b2 = 30375 Dividing (1) by (2) 𝑎^𝑛/(𝑛𝐶1𝑎^(𝑛 − 1) 𝑏) = 729/7290 𝑎^𝑛/(𝑛!/1!(𝑛−1)! 𝑎^(𝑛 − 1) 𝑏) = 1/10 𝑎^𝑛/((𝑛(𝑛 − 1)!)/1(𝑛 − 1)! 𝑎^(𝑛 − 1) 𝑏) = 1/10 𝑎^𝑛/(𝑛𝑎^(𝑛 − 1) 𝑏) = 1/10 𝑎^(𝑛 − (𝑛 − 1))/𝑛𝑏 = 1/10 𝑎^1/𝑛𝑏 = 1/10 𝑎/𝑛𝑏 = 1/10 Dividing (2) by (3) (𝑛𝐶1𝑎^(𝑛−1) 𝑏)/(𝑛𝐶2𝑎^(𝑛−2) 𝑏^2 ) = 7290/30375 (𝑛!/(1!(𝑛−1)! ) 〖 𝑎〗^(𝑛−1).𝑏)/(𝑛!/2!(𝑛 −2)! 𝑎^(𝑛−2) 𝑏2) = 6/25 (𝑛!〖 𝑎〗^(𝑛 − 1) 𝑏)/(𝑛 − 1)! × 2!(𝑛 − 2)!/(𝑛!𝑎^(𝑛 − 2) 𝑏2) = 6/25 (𝑛! 2! (𝑛 − 2)!)/𝑛!(𝑛 − 1)! × 𝑎^(𝑛 − 1)/𝑎^(𝑛 − 2) × 𝑏/𝑏2 = 6/25 (2 𝑛! (𝑛 − 2)!)/𝑛!(𝑛 − 1)(𝑛 − 2)! × 𝑎^(𝑛−1)/𝑎^(𝑛−2) × 𝑏/𝑏2 = 6/25 2/((𝑛 − 1) ) × 𝑎^(𝑛 −1 − (𝑛 − 2))/1 × 𝑏/𝑏2 = 6/25 2/((𝑛 − 1)) × 𝑎^(𝑛 − 1 − 𝑛 + 2)/𝑏 = 6/25 2𝑎/(𝑛 − 1)𝑏 = 6/25 Now, our equations are 𝑎/𝑛𝑏 = 1/10 …(4) 2𝑎/(𝑛−1)𝑏 = 6/25 …(5) Dividing (4) by (5) (𝑎/𝑛𝑏)/(2𝑎/(𝑛−1)𝑏) = (1/10)/(6/25) 𝑎/𝑛𝑏 × (𝑛 − 1)𝑏/2𝑎 = 1/10 × 25/6 ((𝑛 − 1))/2𝑛 = 5/12 12 (n – 1) = 2n (5) 12n – 12 = 10n 12n – 10n = 12 2n = 12 n = 12/2 n = 6 Putting n = 6 in (1) an = 729 a6 = 729 a6 = (3)6 a = 3 Putting a = 3 , n = 6 in (5) 2𝑎/(𝑛−1)𝑏 = 6/25 3/(6 × 𝑏) = 1/10 1/2𝑏 = 1/10 10/2 = b 5 = b b = 5 Hence, a = 3 , b = 5 & n = 6

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.