Check sibling questions

Example 11 - If coefficients of a^r-1 , a^r, a^r+1 in (1 + a)^n are in

Example  11 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example  11 - Chapter 8 Class 11 Binomial Theorem - Part 3 Example  11 - Chapter 8 Class 11 Binomial Theorem - Part 4 Example  11 - Chapter 8 Class 11 Binomial Theorem - Part 5


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 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 For coefficient of ar + 1 Putting r = r + 1 in (1) T(r+1) + 1 = nCr + 1 ar + 1 Tr + 2 = nCr+1 ar + 1 ∴ Coefficient of ar + 1 = nCr + 1 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 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! (1/(𝑟−1)!(𝑛−𝑟+1)! " + " 1/((𝑟+1)!(𝑛−𝑟 −1)!)) = n! (2/𝑟!(𝑛 − 𝑟)!) 𝑛!/𝑛! (1/(𝑟−1)!(𝑛−𝑟+1)! " + " 1/((𝑟+1)!(𝑛−𝑟 −1)!)) = (2/𝑟!(𝑛−𝑟)!) 1/((r − 1)! (n − r + 1)(n − r)(n − r − 1)!) + 1/((r + 1)(r)(r − 1)! (n − r − 1)!) = 2/(r(r − 1)! (n − r)(n − r − 1)!) 1/(𝑟 − 1)!(𝑛 − 𝑟 − 1)! (1/((𝑛 − 𝑟 + 1)(𝑛 − 𝑟) ) "+" 1/((𝑟 + 1)(𝑟) )) = 2/((r − 1)! (n − r − 1) ! [𝑟(n − r)]) (𝑟−1)!(𝑛−𝑟+1)!/(𝑟−1)!(𝑛−𝑟+1)! (1/((𝑛 − 𝑟 + 1)(𝑛 − 𝑟) ) "+" 1/((𝑟 + 1)(𝑟) )) = 2/𝑟(n − r) 1/((𝑛 − 𝑟) (𝑛 − 𝑟 − 1) ) + 1/((𝑟 + 1) 𝑟) = 2/(𝑟 (𝑛 − 𝑟) ) ((𝑟 + 1)𝑟 + (𝑛 − 𝑟)(𝑛 − 𝑟 + 1))/( (𝑛 − 𝑟)(𝑛 − 𝑟 + 1) (𝑟 + 1)𝑟) = 2/( 𝑟(𝑛 − 𝑟) ) (r + 1) r + (n – r) (n – r + 1) = (2(n − r)(n − r + 1)(r + 1)(𝑟))/( 𝑟(n − 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 + 2 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

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.