Vtyjkyuk

So I am trying to use differentiation to find the power series representations for a series of variations of a function. They are f(x)=1/(8+x)^2 g(x)=1/(8+x)^3 h(x)=x^2/(8+x)^3. I was able to get the first representation to be Sum of x^n(-1)^n((8^-1-n)*(1+n)). I am struggling where to go from here, any help is appreciated.

Hint:

When you have the first one $f(x)=dfrac1(8+x)^2$ then for $g(x)$ and $h(x)$ you should compute $f'(x)$ and $x^2f'(x)$.

As a pattern (when $|x| <1$) you have the expansions with binomial coefficients:

• $frac11+x = 1 - x +x^2 -x^3 + x^4 - ldots$


• $frac1(1+x)^2 = 1 - 2x +3x^2 -4x^3 + 5x^4 - ldots$


• $frac1(1+x)^3 = 1 - 3x +6x^2 -10x^3 + 15x^4 - ldots$


• $frac1(1+x)^4 = 1 - 4x +10x^2 -20x^3 + 35x^4 - ldots$

and the coefficient of $x^n$ in the expansion of $frac1(1+x)^k$ is $(-1)^n n+k-1 choose n$

But you are actually interested in the expansion of $frac1(8+x)^k = frac18^ktimes frac1left(1+fracx8right)^k$, so the coefficient of $x^n$ in its expansion is $$frac1(8+x)^k = sum_n=0^infty x^n frac(-1)^n8^n+k n+k-1 choose k-1$$ for $|x| lt 8$

Note when $k=2$ as in your example this would give $frac(-1)^n8^n+2 (n+1)$ which is almost what you have: you might want to check whether your $8^-1-n$ should really be $8^-2-n$