Proof with mathematical induction that $ (\frac{n}{n+1})^2 + (\frac{n+1}{n+2})^2 + ... + (\frac{2n - 1}{2n})^2 \le n - 0.7 $ Proof with mathematical induction.
I have the following induction problem:
$ (\frac{n}{n+1})^2 + (\frac{n+1}{n+2})^2 + ... + (\frac{2n - 1}{2n})^2 \le n - 0.7 $
This property applies to all $n \ge 1$. I've shown that it is true for $n = 1$ and $n = k$, however I can not show for $n = k+1$. 
This is how I start for $n = k+1$:
$ (\frac{k+1}{k+2})^2 + (\frac{k+2}{k+3})^2 + ... + (\frac{2k - 1}{2k})^2 + (\frac{2k}{2k + 1})^2 + (\frac{2k + 1}{2k + 2})^2 \le k + 1 - 0.7 $
Any ideas?


You are almost there. At the end, you wrote the difference between the two sums as a single complicated fraction. Maybe you have miscalculated something there, maybe not, anyway, it is not easy to see through. It becomes much easier if we proceed in steps, pairing like things with like things. The difference is
$$\begin{align}
\left(\frac{2n}{2n+1}\right)^2 + \left( \frac{2n+1}{2n+2}\right)^2 - \left(\frac{n}{n+1}\right)^2
&= \left(\frac{2n}{2n+1}\right)^2 + \left( \frac{2n+1}{2n+2}\right)^2 - \left(\frac{2n}{2n+2}\right)^2\\
&= \left(\frac{2n}{2n+1}\right)^2 + \frac{(2n+1)^2-(2n)^2}{(2n+2)^2}\\
&= 1 - \frac{(2n+1)^2 - (2n)^2}{(2n+1)^2} + \frac{(2n+1)^2-(2n)^2}{(2n+2)^2}\\
&= 1 - (4n+1)\left(\frac{1}{(2n+1)^2}-\frac{1}{(2n+2)^2}\right)\\
&< 1.
\end{align}$$
