How to expand (3-x)∧-2 in ascending powers of 1/x , stating the first four non-zero terms and the value of x for which the expansion is valid?
2 Answers
Explanation:
#"using the following version of the "color(blue)"binomial theorem"#
#•color(white)(x)(1+x)^n=1+nx+(n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+#
#•color(white)(x)(3-x)^-2=[3(1-x/3)]^-2=1/9(1-x/3)^-2#
#"here "n=-2" and "x=-x/3#
#rArr(3-x)^-2#
#=1/9[1+(-2)(-x/3)+((-2)(-3))/(2!)(-x/3)^2+((-2)(-3)(-4))/(3!)(-x/3)^3]#
#=1/9[1+2/3x+1/3x^2+4/27x^3]#
#=1/9+2/27x+1/27x^2+4/243x^3#
#"expansion is valid for "|-x/3|<1rArr|x|<3#
Explanation:
#"using the following version of the "color(blue)"binomial theorem"#
#•color(white)(x)(1+x)^n=1+nx+(n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+...#
#•color(white)(x)(3-x)^-2=[-x(1-3/x)]^-2=1/x^2(1-3/x)^-2#
#"here "n=-2" and "x=-3/x#
#rArr(3-x)^-2#
#=1/x^2[1+(-2)(-3/x)+((-2)(-3))/(2!)(-3/x)^2+((-2)(-3)(-4))/(3!)(-3/x)^3+ ...]#
#=1/x^2[1+6/x+27/x^2+108/x^3+....]#
#=1/x^2+6/x^3+27/x^4+108/x^5+...#
#"expansion is valid for "|-3/x|<1rArr|x|>3#
