Example 11 - If coefficients ar-1, ar and ar+1 of (1 + a)n - Coefficient

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise
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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! ﷐﷐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 + 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

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