Question

How do you use differentiation to find a power series representation for `1/(6+x)^2`?

Answer 1

Basically:
1. Integrate to get a form of `1/(1-x)`.
2. Modify the equation to achieve getting precisely `1/(1-X)` where X is some variant of x, whether it's `x/6` (here), `-y/2`, `theta/pi`, etc.
3. Write out the power series with `x` = some substituted value, like `x/6`, here.
4. Reverse what you did to re-acquire the original function. i.e. re-multiply by what you divided by (-1 and 1/6, here), then re-differentiate. Whatever returns your original function.

Notice how the power series `1/(1-x)` can be written as the power series:
`1 + x + x^2 + x^3 + ...`
`= sumx^n`

Similarly, use -x instead of x. Every odd power is negative, and every even power is positive by virtue of squaring to some order of magnitude (e.g. `(x^2)^n`).

`= sum(-1)^nx^n = 1 - x + x^2 - x^3 + x^4 - ...`

Integrate `1/(6+x)^2` to get `-1/(6+x)`. Divide by -(1/6) to get `1/(1 + x/6)`.

x/6 is your new x. Plug it in, use this alternating series from a few lines up, factor in the 1/6 to get back to `1/(6+x)`, and incorporate the negative to get back to `-1/(6+x)`.

`=> -(1/6)[(-1)^0(x/6)^0 + (-1)^1(x/6)^1 + (-1)^2(x/6)^2 + (-1)^3(x/6)^3 + ...]`
`=> -1/6 + x/36 - x^2/216 + x^3/1296 - ...`

Then, re-differentiate the result to get back to `1/(6+x)^2`.

`=> 1/36 - x/108 + (x^2)/432 - ...`