Check sibling questions

Ex 8.2, 10 - Coefficients of (r - 1), r, (r + 1) in 1 : 3 : 5

Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem - Part 2
Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem - Part 3 Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem - Part 4 Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem - Part 5 Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem - Part 6 Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem - Part 7 Ex 8.2,10 - Chapter 8 Class 11 Binomial Theorem - Part 8


Transcript

Ex 8.2, 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

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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.