Example 17 - If coefficients of (r - 5)th and (2r - 1)th terms in

Example 17 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example 17 - Chapter 8 Class 11 Binomial Theorem - Part 3 Example 17 - Chapter 8 Class 11 Binomial Theorem - Part 4

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Question 13 If the coefficients of (r – 5)th and (2r – 1)th terms in the expansion of (1 + x)34 are equal, find r. We Know that General term of expansion (a + b)n is Tr + 1 = nCr an–r br General term for (1 + x)34 Putting a = 1, b = x, n = 34 Tr + 1 = 34Cr 1n – r xr Tr + 1 = 34Cr xr Coefficient of (r – 5)th term i.e. Tr – 5 i.e. Tr – 6 + 1 Putting r = r – 6 in (1) T(r– 6 +1) = 34Cr–6 (x)r – 6 Tr – 5 = 34Cr – 6 . xr – 6 ∴ Coefficient of (r – 5)th term = 34Cr – 6 Coefficient of (2r – 1)th term i.e. T2r – 1 term i.e. T2r – 2 + 1 term Putting r = 2r – 2 in (1) T(2r – 2 +1) = 34C2r – 2 (x)2r – 2 T2r – 1 = 34C2r – 2 . x2r – 2 ∴ Coefficient of (2r – 1)th term = 34C2r – 2 Given that Coefficient of (r – 5)th & (2r – 1)th term are equal i.e. 34Cr – 6 = 34C2r – 2 We know that if nCr = nCp , then r = p or r = n – p So, r – 6 = 2r – 2 or r – 6 = 34 – (2r – 2) r – 6 = 2r – 2 –6 + 2 = 2r – r –4 = r r = –4 r – 6 = 34 – (2r – 2) r – 6 = 34 – 2r + 2 r + 2r = 34 + 2 + 6 3r = 42 r = 42/3 = 14 Since r is a natural number So r = – 4 not possible Hence, r = 14

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.