Pearson Edexcel IAL Mathematics P4.4.1 Binomial Series for any rational nPractise expanding rational powers into binomial series, controlling coefficients, validity ranges and small-x approximations.SyllabusFirst assessment 2019CourseMathematics YMA01LevelA2
Exam pointsrewrite (a+bx)^n as a(1+kx)^n before expanding in ascending powers of xstate the interval of validity from |kx|<1 for the rewritten expressioncompare coefficients in a binomial series to find constants or an x^3 coefficient
P4.4.1 - Binomial Series for any rational n question 1[Maximum number: 4]f(x)=(8−3x)430<x<83\mathrm{f}(x)=(8-3 x)^{\frac{4}{3}} \quad 0<x<\frac{8}{3}f(x)=(8−3x)340<x<38Show that the binomial expansion of f(x) in ascending powers of x up to and including the term in x3x^{3}x3 isA−8x+x22+Bx3+…A-8 x+\frac{x^{2}}{2}+B x^{3}+\ldotsA−8x+2x2+Bx3+…where A and B are constants to be found.Show Answer(8−3x)43=16(1−38x)43(8-3 x)^{\frac{4}{3}}=16\left(1-\frac{3}{8} x\right)^{\frac{4}{3}}(8−3x)34=16(1−83x)34 but condone 16(1−kx)43,k≠316(1-k x)^{\frac{4}{3}}, k \neq 316(1−kx)34,k=3M1"Correct" term 3 or term 4 in(1−kx)43=1±43kx+43×132(±kx)2+43×13×(−23)6(±kx)3+…(1-k x)^{\frac{4}{3}}=1 \pm \frac{4}{3} k x+\frac{\frac{4}{3} \times \frac{1}{3}}{2}( \pm k x)^{2}+\frac{\frac{4}{3} \times \frac{1}{3} \times\left(-\frac{2}{3}\right)}{6}( \pm k x)^{3}+\ldots(1−kx)34=1±34kx+234×31(±kx)2+634×31×(−32)(±kx)3+…M1Two terms correct of 16−8x+x22+x324+…16-8 x+\frac{x^{2}}{2}+\frac{x^{3}}{24}+\ldots16−8x+2x2+24x3+… following M1, M1 and k=±38k= \pm \frac{3}{8}k=±83A1=16−8x+x22+x324+…=16-8 x+\frac{x^{2}}{2}+\frac{x^{3}}{24}+\ldots=16−8x+2x2+24x3+…A1(4)Add to Test