Example 17

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Example 17 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 (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–6 T2r – 1 = 34C2r–2 . x2r–2 ∴ Coefficient of (2r – 1)th term = 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)