Question 10 - General Term of Binomial Theorem - Chapter 7 Class 11 Binomial Theorem
Last updated at April 16, 2024 by Teachoo
General Term of Binomial Theorem
Last updated at April 16, 2024 by Teachoo
Question 10 The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio 1 : 3 : 5. Find n and r. Finding (r – 1)th term , rth & (r + 1)th term of (x + 1)n Writing (x + 1)n as (1 + x)n We know that general term of expansion (a + b)n is Tr+1 = nCr an – rbr For (1 + x)n , Putting a = 1 , b = x Tr+1 = nCr 1n – r xr Tr+1 = nCr xr ∴ Coefficient of (r + 1)th term = nCr For rth term of (1 + x)n Replacing r with r – 1 in (1) Tr – 1 + 1 = nCr – 1 xr – 1 Tr = nCr – 1 xr – 1 ∴ Coefficient of (r)th term = nCr – 1 For (r – 1)th term of (1 + x)n Replacing r with r – 2 in (1) Tr – 2 + 1 = nCr – 2 xr – 2 Tr – 1 = nCr – 2 xr – 2 ∴ Coefficient of (r – 1)th term = nCr-2 For rth term of (1 + x)n Replacing r with r – 1 in (1) Tr – 1 + 1 = nCr – 1 xr – 1 Tr = nCr – 1 xr – 1 ∴ Coefficient of (r)th term = nCr – 1 For (r – 1)th term of (1 + x)n Replacing r with r – 2 in (1) Tr – 2 + 1 = nCr – 2 xr – 2 Tr – 1 = nCr – 2 xr – 2 ∴ Coefficient of (r – 1)th term = nCr-2 Since the coefficient of (r – 1)th, rth and (r + 1)th terms are in ratio 1 : 3 : 5 (𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 〖(𝒓 − 𝟏)〗^𝒕𝒉 𝒕𝒆𝒓𝒎)/(𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒓^𝒕𝒉 𝒕𝒆𝒓𝒎) = 𝟏/𝟑 〖𝑛𝐶〗_(𝑟 − 2)/〖𝑛𝐶〗_(𝑟 − 1) = 1/3 (𝑛!/((𝑟 − 2)![𝑛 − (𝑟 − 2)]!))/(𝑛!/(𝑟 − 1)!(𝑛 − (𝑟 − 1))!) = 1/3 𝑛!/((𝑟. −2)![𝑛 − (𝑟 − 2)]!) × ((𝑟 − 1)![𝑛 − (𝑟 − 1)]!)/𝑛! = 1/3 (𝑛!(𝑟 − 1)(𝑟 − 2)![𝑛 − (𝑟 − 1)]!)/(𝑛!(𝑟 − 2)![𝑛 − (𝑟 − 2)]!) = 1/3 (𝑟 − 1)(𝑛 − (𝑟 − 1))!/((𝑛 − (𝑟 − 2))!) = 1/3 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)!) = 1/3 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)(𝑛 − 𝑟 + 2 −1)!) = 1/3 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)(𝑛 − 𝑟 +1)!) = 1/3 ((𝑟 − 1))/((𝑛 − 𝑟 + 2) ) = 1/3 3(r – 1) = n – r + 2 3r – 3 = n + 2 – r n + 2 – r – 3r + 3 = 0 n – 4r + 5 = 0 Also (𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒓^𝒕𝒉 𝒕𝒆𝒓𝒎)/(𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 〖(𝒓 + 𝟏)〗^𝒕𝒉 𝒕𝒆𝒓𝒎) = 𝟑/𝟓 〖𝑛𝐶〗_(𝑟 − 1)/〖𝑛𝐶〗_𝑟 = 3/5 (𝑛!/((𝑟 − 1)![𝑛 − (𝑟 − 1)]!))/(𝑛!/(𝑟! (𝑛 − 𝑟)!)) = 3/5 𝑛!/((𝑟 − 1)!(𝑛 − 𝑟 + 1)!) × (𝑟! (𝑛 − 𝑟)! )/𝑛! = 3/5 (𝑛! × 𝑟 × (𝑟 − 1)!(𝑛 − 𝑟)!)/(𝑛!(𝑟 − 1)! (𝑛 − 𝑟 + 1)!) = 3/5 𝑟(𝑛 − 𝑟)!/((𝑛 − 𝑟 +1)!) = 3/5 (𝑟 (𝑛 − 𝑟)!)/((𝑛 − 𝑟 +1) (𝑛 − 𝑟)!) = 3/5 𝑟/(𝑛 − 𝑟 + 1) = 3/5 5r = 3 (n – r + 1) 5r = 3n – 3r + 3 0 = 3n – 3r + 3 – 5r 0 = 3n – 8r + 3 3n – 8r + 3 = 0 Now our equations are n – 4r + 5 = 0 & 3n – 8r + 3 = 0 From (1) n – 4r + 5 = 0 n = 4r – 5 Putting n = 4r – 5 in (2) 3n – 8r + 3 = 0 3(4r – 5) – 8r + 3 = 0 12r – 15 – 8r + 3 = 0 12r – 8r – 15 + 3 = 0 4r – 12 = 0 4r = 12 r = 12/4 r = 3 Putting r = 3 in n = 4r – 5 n = 4(3) – 5 n = 12 – 5 n = 7 Hence, n = 7 & r = 3