Question

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

Answer 1

`1/(2+x)^2=1/(4+4x+x^2)`

Explanation:

The binomial expansion allows us to take something in the form

`(a+b)^n`

and express it as a sum of terms:

`(a+b)^n=(C_(n,0))a^nb^0+(C_(n,1))a^(n-1)b^1+...+(C_(n,n))a^0b^n`

We can do this for the denominator in the fraction:

`(2+x)^2=1(2)^2(1)+2(2)(x)+1(1)(x^2)=4+4x+x^2`

and so:

`1/(2+x)^2=1/(4+4x+x^2)`

Answer 2

`1/(2+x)^2=1/4-1/4x+3/16x^2-1/8x^3+5/64x^4-3/64x^5+...............`

Explanation:

`1/(2+x)^2=(2+x)^(-2)=2^(-2)xx(1+x/2)^(-2)`

Now Binomial expansion of `(1+a)^n`

= `1+na+(n(n-1))/(1*2)a^2+(n(n-1)(n-2))/(1*2*3)a^3+(n(n-1)(n-2)(n-3))/(1*2*3*4)a^4+.............`

Hence `(1+x/2)^(-2)`

= `1+(-2)x/2+((-2)(-3))/(1*2)(x/2)^2+((-2)(-3)(-4))/(1*2*3)(x/2)^3+((-2)(-3)(-4)(-5))/(1*2*3*4)(x/2)^4+.............`

= `1-x+3/4x^2-1/2x^3+5/16x^4-3/16x^5+...............`

and `1/(2+x)^2=1/4xx(1-x+3/4x^2-1/2x^3+5/16x^4-3/16x^5+...............)`

= `1/4-1/4x+3/16x^2-1/8x^3+5/64x^4-3/64x^5+...............`