Ex 7.1,14 - Chapter 7 Class 11 Binomial Theorem
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 7.1, 14 (Method 1) By Binomial Theorem, Putting b = 3 and a = 1 in the above equation Prove that โ_(๐=0)^๐โใ3^๐ nCrใ โ_(๐=0)^๐โnCr ๐^(๐ โ ๐) ๐^๐ โ_(๐=0)^๐โnCr 1^(๐โ๐) 3^๐ Hence proved Ex 7.1, 14 (Method 2) โ Introduction For r = 0, 3^0 nC0 For r = 1, 3^1 nC1 For r = 2, 3^2 nC2 For r = 3, 3^3 nC3 โฆ โฆ. For r = n, 3^๐ nCn nC0 30 + nC1 31 + nC2 32 + โฆ โฆโฆโฆ + nCn โ 1 3n โ 1 + nCn 3n Prove that = nC0 30 + nC1 31 + nC2 32 + โฆโฆโฆโฆโฆโฆ + nCn-1 3n-1 + nCn 3n Ex 7.1, 14(Method 2) Solving L.H.S This is similar to nC0 an b0 + nC1 an-1 b1 + nC2 an-2 b2 + โฆโฆ .+ nCn-1 a1 bn-1 + nCn a0 bn Where a = 1 , b = 3 And we know that (a + b)n = nC0 an b0 + nC1 an-1 b1 + โฆโฆ.+ nCn-1 a1 bn-1 + nCn a0 bn = (1 + 3)n = (4)n = R.H.S Hence proved
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo