Check sibling questions

Misc 2 - Find a if coefficients of x2, x3 in (3 + ax)9 are equal

Misc 2 - Chapter 8 Class 11 Binomial Theorem - Part 2
Misc 2 - Chapter 8 Class 11 Binomial Theorem - Part 3 Misc 2 - Chapter 8 Class 11 Binomial Theorem - Part 4


Transcript

Misc 2 Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal. We know that General term of expansion (a + b)n is Tr + 1 = nCr (a)n – r . br For (3 + ax)9 , Putting a = 3 , b = 9x & n = 9 General term of (3 + ax)9 is Tr + 1 = 9Cr (3)9 – r . (ax)r = 9Cr 39 – r . ar . xr We need Coefficient of x2 & x3 Given that Coefficient of x2 = Coefficient of x3 9C2 (3)7 . a2 = 9C2 (3)6 . a3 Finding Coefficient of x2 Putting r = 2 in (1) T2 + 1 = 9C2 (3)9 – 2 . a2 . x2 = 9C2 (3)7 . a2 . x2 Coefficient of x2 = 9C2 (3)7 . a2 Finding coefficient of x3 Putting r = 3 in (1) T3+1 = 9C3 (3)9 – 3 . a3 . x3 ‘ = 9C3 (3)6 . a3 . x3 Coefficient of x3 = 9C3 (3)6 . a3 9!/2!(9 − 2)! 37 . a2 = 9!/3!(9 − 3)! 36 . a3 9!/2!(7)! 37 . a2 = 9!/3!(6)! 36 . a3 (9!/(2! 7!) . 37)/(9!/3!(9 −3)! . 36) = 𝑎3/𝑎2 9!/(2! 7!) . 37 × (3! 6!)/(9! . 36) = a (9! × 3! × 6!)/(9! × 2! × 7!) × 37/36 = a (3 × 2! × 6!)/(2! × 7 × 6!) . 3 = a 3/7 × 3 = a 9/7 = a Hence, a = 𝟗/𝟕

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.