# Ex 8.2,11 - Chapter 8 Class 11 Binomial Theorem

Last updated at Jan. 29, 2020 by Teachoo

Last updated at Jan. 29, 2020 by Teachoo

Transcript

Ex 8.2, 11 Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n – 1 . We know that General term of (a + b)n is Tr + 1 = nCr an – r . br For (1 + x)2n General term of (1 + x)2n Putting a = 1 , b = x, n = 2n Tr + 1 = 2nCr (1)2n – r . (x)r Tr + 1 = 2nCr. (x)r For coefficient of xn Putting r = n in (1) Tr+1 = 2nCn xn Coefficient of xn = 2nCn For (1 + x)2n – 1 General term of (1 + x)2n – 1 Putting a = 1 , b = x, n = 2n – 1 Tr + 1 = 2n – 1Cr .(1)2n – 1 – r . xr Tr + 1 = 2n – 1Cr xr For coefficient of xn Putting r = n in (2) Tr+1 = 2n – 1 Cn × xn Coefficient of xn = 2n – 1Cn We have to prove Coefficient of xn in (1 + x)2n = 2 × Coefficient of xn in (1 + x)2n – 1 i.e. 2nCn = 2 × 2n-1Cn 2nCn = 2𝑛!/𝑛!(2𝑛 −𝑛)! = 2𝑛!/𝑛!(𝑛)! 2 × 2n-1Cn = 2 × (2𝑛 − 1)!/𝑛!(2𝑛 − 1 − 𝑛)! = 2 × ((2𝑛 − 1)!)/𝑛!(𝑛 − 1)! Multiply & Divide by n = 2 × ((2𝑛 − 1)!)/𝑛!(𝑛 − 1)! × 𝑛/𝑛 = (𝟐𝒏 (𝟐𝒏 − 𝟏)!)/(𝑛! 𝒏(𝒏 − 𝟏)!) = (2𝑛)!/𝑛!𝑛! Hence L.H.S = R.H.S Hence Proved

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.