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

  1. Class 11
  2. Important Question for exams Class 11
Ask Download

Transcript

Ex8.2, 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 = a 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 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﷯ ﷐﷐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 …(1) & 3n – 8r + 3 = 0 …(2) 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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail