Question 4 Write the general term in the expansion of (x2 – yx)12, x ≠ 0 We know that General term of expansion (a + b)n is Tr+1 = nCr an–r br For (x2 – yx), Putting n = 12, a = x2 , b = – yx Tr + 1 = 12Cr (x2)12 – r . (– yx)r = 12Cr (x)2(12 – r) . (–1)r . yr . xr = 12Cr x24 – 2r . (–1)r . yr . xr = 12Cr x24 – 2r .xr . yr . (–1)r = (–1)r 12Cr. (x)24 – 2r + r . yr = (–1)r .12Cr . x24 – r . yr Hence, General term = (−1)r . 12Cr x24 – r . yr

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Davneet Singh

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.

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