Check sibling questions

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

Misc  1 - Chapter 8 Class 11 Binomial Theorem - Part 2
Misc  1 - Chapter 8 Class 11 Binomial Theorem - Part 3
Misc  1 - Chapter 8 Class 11 Binomial Theorem - Part 4
Misc  1 - Chapter 8 Class 11 Binomial Theorem - Part 5
Misc  1 - Chapter 8 Class 11 Binomial Theorem - Part 6
Misc  1 - Chapter 8 Class 11 Binomial Theorem - Part 7

Solve all your doubts with Teachoo Black (new monthly pack available now!)


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 + 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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.