Question

How do you find the Maclaurin series for `3/(1-2x)`?

Answer 1

The series is `3(1+2x+4x^2+8x^3+16x^4+......)`

Explanation:

Rewriting the expression
`3/(1-2x)=3(1-2x)^(-1)`
The development of a binomial is
`(1+x)^n=1+nx+(n(n-1))/(1.2)x^2+(n(n-1)(n-2))/(1.2.3)x^3+(n(n-1)(n-3)(n-4))/(1.2.3.4)x^2+.....`

So we have
`(1-2x)^(-1)=1+2x+(-1(-2))/(1.2)(-2x)^2+(-1(-2)(-3))/(1.2.3)(-2x)^3+(-1(-2)(-3)(-4))/(1.2.3.4)(-2x)^2+.....`

`=1+2x+4x^2+8x^3+16x^4+.....`

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