Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Examples

Example 1

Example 2 Important

Example 3 Important

Example 4

Question 1 Important Deleted for CBSE Board 2024 Exams

Question 2 Important Deleted for CBSE Board 2024 Exams

Question 3 Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Question 7 Important Deleted for CBSE Board 2024 Exams You are here

Question 8 Deleted for CBSE Board 2024 Exams

Question 9 Important Deleted for CBSE Board 2024 Exams

Question 10 Important Deleted for CBSE Board 2024 Exams

Question 11 Important Deleted for CBSE Board 2024 Exams

Question 12 Deleted for CBSE Board 2024 Exams

Question 13 Important Deleted for CBSE Board 2024 Exams

Chapter 7 Class 11 Binomial Theorem

Serial order wise

Last updated at May 29, 2023 by Teachoo

Question 7 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