# Example 11 - Chapter 8 Class 11 Binomial Theorem

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 11 If the coefficients of ar – 1, ar and ar + 1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n(4r + 1) + 4r2 – 2 = 0. We know that General term of (a + b)n Tr+1 = nCr an–r . br For (1 + a)n Putting a = 1 & b = a Tr+1 = nCr (1)n–r . ar Tr+1 = nCr ar Hence Coefficient of ar = nCr Similarly, Finding coefficient of ar–1 Putting r = r – 1 in (1) T(r – 1) + 1 = nCr – 1 ar – 1 Tr = nCr – 1 ar – 1 ∴ Coefficient of ar – 1 = nCr – 1 Similarly, For coefficient of ar + 1 Putting r = r + 1 in (1) T(r+1) + 1 = nCr + 1 ar + 1 Tr + 2 = nCr ar + 1 ∴ Coefficient of ar + 1 = nCr + 2 Given that coefficient of ar–1 , ar , ar+1 are in AP i.e. nCr–1 , nCr , & nCr+1 are in AP Therefore, common difference must be equal So, nCr – , nCr –1 = nCr+1 – nCr nCr + nCr = nCr+1 + nCr–1 2nCr = nCr+1 + nCr–1 nCr–1 + nCr+1 = 2nCr 𝑛!𝑟−1! 𝑛 −𝑟 −1! + 𝑛!𝑟+1! (𝑛−𝑟+1)! = 2 × 𝑛!𝑟!𝑛−𝑟! n! 1𝑟−1!𝑛−𝑟+1! + 1𝑟+1!(𝑛−𝑟 −1)! = n! 2𝑟!𝑛−𝑟! 𝑛!𝑛! 1𝑟−1!𝑛−𝑟+1! + 1𝑟+1!(𝑛−𝑟 −1)! = 2𝑟!𝑛−𝑟! 1r − 1! (n − r + 1)(n − r)(n − r − 1) + 1r + 1rr − 1! n − r − 1! = 2rr − 1! n − rn − r − 1! 1𝑟−1!𝑛−𝑟−1! 1𝑛−𝑟+1𝑛−𝑟 + 1𝑟+1𝑟 = 2r−1! n−r−1 ! [𝑟n−r] 𝑟−1!𝑛−𝑟+1!𝑟−1!𝑛−𝑟+1! 1𝑛−𝑟+1𝑛−𝑟 + 1𝑟+1𝑟 = 2𝑟n−r 1n − 𝑟 n − r − 1! + 1r + 1 𝑟 = 2𝑟 n − 𝑟 r + 1𝑟+n − r(n − r + 1) n − rn − r + 1 r + 1r = 2 rn − r (r + 1) r + (n – r) (n – r + 1) = 2n−rn−r+1r+1(𝑟) 𝑟n−r (r + 1) r + (n – r) (n – r + 1)= 2(n – r + 1) (r + 1) r2 + r + n (n – r + 1) – r ( n – r + 1) = 2 [r (n – r + 1) + 1(n – r + 1 )] r2 + r + n2 – nr + n – nr + r2 – r = 2(rn – r2 + r + n – r + 1) r2 + r2 + r – r + n2 – nr – nr + n = 2rn – 2r2 + 2r + 2n – 2r + 1 2r2 + 0 + n2 – 2nr + n = – 2r2 + 2r – 2r + 2rn + 2n + 2 2r2 + n2 – 2nr + n = – 2r2 + 2rn + 2n + 2 2r2 + n2 – 2nr + n + 2r2 – 2rn – 2n – 2 = 0 2r2 + 2r2 + n2 – 2nr – 2rn + n – 2n – 2 = 0 4r2 – n2 – 4nr – n – 2 = 0 n2 – 4nr – n + 4r2 – 2 = 0 n2 – n (4r + 1) + 4r2 – 2 = 0 Hence proved

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.