Check sibling questions

Example 9 - The coefficients of three consecutive terms of (1 + a)^n

Example  9 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example  9 - Chapter 8 Class 11 Binomial Theorem - Part 3
Example  9 - Chapter 8 Class 11 Binomial Theorem - Part 4
Example  9 - Chapter 8 Class 11 Binomial Theorem - Part 5
Example  9 - Chapter 8 Class 11 Binomial Theorem - Part 6
Example  9 - Chapter 8 Class 11 Binomial Theorem - Part 7

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Example 9 The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio 1: 7 : 42. Find n. Let the three consecutive terms be (r – 1)th, rth and (r + 1)th terms. i.e. Tr – 1 , Tr & Tr + 1 We know that general term of expansion (a + b)n is Tr + 1 = nCr an – r br For (1 + a)n , Putting a = 1 , b = a Tr+1 = nCr 1n – r ar Tr+1 = nCr ar ∴ Coefficient of (r + 1)th term = nCr For rth term of (1 + a)n Replacing r with r – 1 in (1) Tr – 1 + 1 = nCr – 1 ar – 1 Tr = nCr – 1 ar – 1 ∴ Coefficient of (r)th term = nCr – 1 For (r – 1)th term of (1 + a)n Replacing r with r – 2 in (1) Tr – 2 + 1 = nCr – 2 ar – 2 Tr – 1 = nCr – 2 ar – 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 : 7 : 42 (𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 〖(𝒓 − 𝟏)〗^𝒕𝒉 𝒕𝒆𝒓𝒎)/(𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒓^𝒕𝒉 𝒕𝒆𝒓𝒎) = 𝟏/𝟕 〖𝑛𝐶〗_(𝑟 − 2)/〖𝑛𝐶〗_(𝑟 − 1) = 1/7 (𝑛!/((𝑟 − 2)![𝑛 − (𝑟 − 2)]!))/(𝑛!/(𝑟 − 1)!(𝑛 − (𝑟 − 1))!) = 1/7 𝑛!/((𝑟 − 2)![𝑛 − (𝑟 − 2)]!) × ((𝑟 − 1)![𝑛 − (𝑟 − 1)]!)/𝑛! = 1/7 (𝑟 − 1)(𝑟 − 2)!(𝑛 − (𝑟 − 1))!/((𝑟 − 2)! (𝑛 − (𝑟 − 2))!) = 1/7 (𝑟 − 1)(𝑛 − 𝑟 + 1)!/((𝑛 − 𝑟 + 2)!) = 1/7 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)(𝑛 − 𝑟 + 2 −1)!) = 1/7 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)(𝑛 − 𝑟 + 1)!) = 1/7 ((𝑟 − 1))/((𝑛 − 𝑟 + 2) ) = 1/7 7(r – 1) = n – r + 2 n – 8r + 9 = 0 Also (𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒓^𝒕𝒉 𝒕𝒆𝒓𝒎)/(𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 〖(𝒓 + 𝟏)〗^𝒕𝒉 𝒕𝒆𝒓𝒎) = 𝟕/𝟒𝟐 (𝑛!/((𝑟 − 1)![𝑛 − (𝑟 − 1)]!))/(𝑛!/(𝑟! (𝑛 − 𝑟)!)) = 7/42 𝑛!/((𝑟 − 1)!(𝑛 − 𝑟 + 1)!) × (𝑟! (𝑛 − 𝑟)! )/𝑛! = 1/6 (𝑛! × 𝑟 × (𝑟 − 1)!(𝑛 − 𝑟)!)/(𝑛!(𝑟 − 1)! (𝑛 − 𝑟 + 1)!) = 1/6 𝑟(𝑛 − 𝑟)!/((𝑛 − 𝑟 + 1)!) = 1/6 (𝑟 (𝑛 − 𝑟)!)/((𝑛 − 𝑟 + 1) (𝑛 − 𝑟)!) = 1/6 𝑟/(𝑛 + 1 − 𝑟) = 1/6 6r = n + 1 – r n – 7r + 1 = 0 Now we have n – 8r + 9 = 0 …(1) n – 7r + 1 = 0 …(2) From (1) n – 8r + 9 = 0 n = 8r – 9 Putting n = 8r – 9 in (2) (8r – 9) – 7r + 1 = 0 8r – 9 – 7r + 1 = 0 r – 8 = 0 r = 8 Putting value of r in (1) n – 8r + 9 = 0 n – 8(8) + 9 = 0 n – 64 + 9 = 0 n – 55 = 0 n = 55 Hence, n = 55

Davneet Singh's photo - Co-founder, Teachoo

Made by

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.