Question

1)find the first three term in the expansion in ascending power of x ,of (1-2x) power 5.
2)given that the coefficient of x square in the expansion of(1+ax+2x square)(1-2x)power 5 is 12, find the value of the constant a.

Answer 1

Answer: (1) 1-10x+40x² (2) a = 3

$(a)\;{(1-2x)}^5=1^5+С_5^1\times(-2x)+\\+C_5^2\times{(-2x)}^2+...=\\=1-10x+40x^2+...\\(b)\;(1+ax\;+\;2x^2){(1-2x)}^5=\\\;=(1+ax+2x^2)(1-10x+40x^2+...)\\It's\;obvious\;that\;if\;we\;consider\;the\;rest\;part\;\\of\;{(1-2x)}^5\;\;(except\;first\;three)\;the\;result\;degree\;\\of\;x\;will\;be\;greater\;than\;2.\;Because\;each\;of\;those\;\\part\;has\;degree\;at\;least\;of\;3.\\So,\;we\;should\;calculate\;coefficient\;of\;x^2\;in\;\;\\(1+ax+2x^2)(1-10x+40x^2)\\1\times40+a\times(-10)+2\times1=40-10a+2=12\\10a=30\Rightarrow a=3$