Additional Mathematics 0606 / Series / 1.121.12: Binomial expansion and term independent of xxx0606/12/O/N/17 — Question 3 · 6 marks · BinomialMark as done · Save for later · Show all solutions(i)Find, in ascending powers of xxx, the first 3 terms in the expansion of (2−x24)5\left(2 - \tfrac{x^{2}}{4}\right)^{5}(2−4x2)5.[3]▸ Answer(ii)Hence find the term independent of xxx in the expansion of (2−x24)5(1x−3x2)2\left(2 - \tfrac{x^{2}}{4}\right)^{5}\left(\tfrac{1}{x} - \tfrac{3}{x^{2}}\right)^{2}(2−4x2)5(x1−x23)2.[3]▸ Answer▸ Official mark scheme← 1.17: Binomial expansion and coefficient of x4x^4x41.20: Binomial expansion and coefficient of x2x^2x2 →
