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Ex 8.1, 14 - Prove that Sum r=0 -> n,  3^r nCr = 4^n - Ex 8.1

Ex 8.1,14 - Chapter 8 Class 11 Binomial Theorem - Part 2
Ex 8.1,14 - Chapter 8 Class 11 Binomial Theorem - Part 3 Ex 8.1,14 - Chapter 8 Class 11 Binomial Theorem - Part 4


Transcript

Ex 8.1, 14 (Method 1) By Binomial Theorem, Putting b = 3 and a = 1 in the above equation Prove that ∑_(𝑟=0)^𝑛▒〖3^𝑟 nCr〗 ∑_(𝑟=0)^𝑛▒nCr 𝑎^(𝑛 − 𝑟) 𝑏^𝑟 ∑_(𝑟=0)^𝑛▒nCr 1^(𝑛−𝑟) 3^𝑟 Hence proved Ex 8.1, 14 (Method 2) – 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 … …. For r = n, 3^𝑛 nCn nC0 30 + nC1 31 + nC2 32 + … ……… + nCn − 1 3n − 1 + nCn 3n Prove that = nC0 30 + nC1 31 + nC2 32 + ……………… + nCn-1 3n-1 + nCn 3n Ex 8.1, 14(Method 2) Solving L.H.S This is similar to nC0 an b0 + nC1 an-1 b1 + nC2 an-2 b2 + …… .+ 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 + …….+ 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.