standard Taylor series using substitution

Question

Find Taylor series using substitution about $0$ for $f(x)=\frac{125}{(5+4x)^3}$ by writing $\frac{125}{(5+4x)^3}=\frac{1}{(1+\frac{4}{5}x)^3}$?

Determine a range of validity for this series.

Answer 1

you have $(1+4/5x)^{-3}$=1-12/5x+64/25$x^2$.....
is the required exxpansion about x=0. for range, its true for |4/5x|<1

Answer 2

Hint

Start with Taylor expansion $$\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+O\left(y^6\right)$$ Rise to the third power to obtain $$\frac{1}{(1+y)^3}=1-3 y+6 y^2-10 y^3+15 y^4-21 y^5+O\left(y^6\right)$$ Now, replace $y$ by $\frac{4 x}{5}$ to get your result.

I am sure that you can take from here.

Answer 3

As Claude did, $$\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+...+(-1)^ny^n...$$ and take second derivative of this expension to get,

$$ \frac{2}{(1+y)^3}=(\frac{1}{1+y})^{''}=2-6y+12y^2-20y^3...n(n-1)(-1)^ny^{n-2}...$$

$$\frac{1}{(1+y)^3}=1-3y+6y^2-10y^3...{n\choose2}(-1)^{n+1}y^{n-2}... $$

Then replace $y$ by $4x\over5$ then you will get your series.

Since above expension convergent for $|y|<1$ your expension is convergent for $|{4x\over5}|<1$