How do you use the binomial series to expand $\frac{1}{{(2 + x)}^{3}}$ $1/(2+x)^3$ ?

Question

2114 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-23T10:11:22

The answer is $= \frac{1}{8} - \frac{3}{16} x + \frac{3}{16} x^{2} - \frac{5}{32} x^{3} + o (x^{3})$ $=1/8-3/16x+3/16x^2-5/32x^3+o(x^3)$

The binomial series is

$(a+b)^n=a^n+na^(n-1)b+((n)(n-1))/(1*2)a^(n-2)b^2+((n)(n-1)(n-2))/(1*2*3)a^(n-3)b^3+.....$

Here,

$a=2$

$b=x$

and

$n=-3$

Therefore,

$1/(2+x)^3=(2+x)^(-3)$

$=2^(-3)+(-3)*(2^(-4))*x+(-3*-4)/(2)*2^(-5)x^2+(-3*-4*-5)/(6)2^(-6)x^3+....$

$=1/8-3/16x+3/16x^2-5/32x^3+o(x^3)$

How do you use the binomial series to expand 1/(2+x)^31(2+x)3 1/(2+x)^3?