It's probably a very elementary problem, but I can't seem to figure it out.
Proof by means of strong induction that for every natural number $n \geq 1$:
$$\sum_{i=1}^n \frac{1}{4k^2-1} = \frac{n}{2n+1}$$
I tried the following:
Base step: Let $n = 1$. In this case: $$\sum_{i=1}^1 \frac{1}{4-1} = \frac{1}{2+1} = \frac{1}{3}$$
Inductive step: Fix some $k \geq 1$, and assume that for every $t$ satisfying $1 \leq t \leq k$, the statement is true. We need to prove that:
\begin{equation} \begin{aligned} \sum_{i=1}^k \frac{1}{4(k+1)^2-1} &= \frac{k+1}{2(k+1)+1} \\ \sum_{i=1}^k \frac{1}{4k^2-1} + \frac{1}{4(k+1)^2-1} &= \frac{k}{2k+1}+ \frac{1}{4(k+1)^2-1} \\ &= \frac{k}{2k+1} + \frac{1}{4(k^2+2k+1)-1} \\ &= \frac{k}{2k+1} + \frac{1}{4k^2+8k+3} \\ &= \frac{4k^3-8k^2+3k}{(2k+1)(4k^2-8k+3)} + \frac{2k+1}{(2k+1)(4k^2-8k+3)} \\ &= \frac{4k^3-8k^2+5k+1}{(2k+1)(4k^2-8k+3)} \\ \end{aligned} \end{equation}
And this is pretty much where I get stuck. I was hoping that here at some point I'd be able to factor out the polynomials to get to $\frac{k+1}{2(k+1)+1}$, but I can't figure it out. I probably did something stupid. Any help is greatly appreciated!