Can you easily simplify these terms?

Question

I need to simplify these terms step by step to prove they are equivalent

$$(100^{(2n+1)}-1+99×10^{(4n+2)}-99×10^{(2(n+1)-1)})/(11×10^{(2n+1)}-11) $$

and $$(100^{(2(n+1)+1)}-1)/(10^{(2(n+1)+1)}-1)×1/11$$

I know that they are, because I checked it with WolframAlpha, I think it`s using some sort of "brute force" algorithm to find these simplifications. Here are both terms simplifyed by WA: here and here

So, my questions are: What are the steps for these simplifications and are there any good books to read about this topic, so that I could find such steps easily by myself in future?

Answer 1

Two things to notice are that $100^{(2n+1)}=10^{(4n+2)}$ and $99=100-1$ $$\begin {align}\frac{100^{2n+1}-1+99×10^{4n+2}-99×10^{2(n+1)-1}}{11×10^{2n+1}-11}\\& =\frac{100^{2n+1}-1+99×100^{2n+1}-99×10^{2n+1}}{11(10^{2n+1}-1)}\\& =\frac{100^{2n+2}-1-10^{2n+3}+10^{2n+1}}{11(10^{2n+1}-1)}\\& =\frac{10^{4n+4}-10^{2n+3}+10^{2n+1}-1}{11(10^{2n+1}-1)}\\& =\frac{(10^{2n+1}-1)(10^{2n+3}+1)}{11(10^{2n+1}-1)}\\& =\frac{10^{2n+3}+1}{11}\\& =\frac{(10^{2n+3}+1)(10^{2n+3}-1)}{11(10^{2n+3}-1)}\\& =\frac{100^{2n+3}-1}{11(10^{2n+3}-1)}\end {align}$$

Answer 2

For the first one,

Notice that $100^{2n+1}=10^{4n+2}$,

Rearranging terms in the numerator gives :

$$100.10^{4n+2}-99.10^{2n+1}-1$$ I let $10^{2n+1}=t$

So, our fraction becomes:

$$\frac{100t^2-99t-1}{11t-11}$$ Notice that $(t-1)$ can be cancelled, giving the answer as:

$$\frac{100t+1}{11}$$ substituting t back,

the simplified answer is:

$$\frac{10^{2n+3}+1}{11}$$

For the second part:

again notice that $100^{2n+3}=10^{4n+6}$, making this change in the second part,we get:

$$\frac{10^{4n+6}-1}{(10^{2n+3}-1)11}$$

Now I put $10^{2n+3}=t$ so, the expression becomes:

$$\frac{t^2-1}{11(t-1)}$$ cancelling $t-1$,and substituting $t$ back, we get:

$$\frac{1+10^{2n+3}}{11}$$

Which is same as the first part.