1. **Problem Statement:** Prove by mathematical induction that for all integers $n \geq 2$, $$\frac{3}{4} + \frac{5}{36} + \cdots + \frac{2n - 1}{n^2 (n-1)^2} = 1 - \frac{1}{n^2}.$$ 2. **Base Case ($n=2$):** Calculate the left side for $n=2$: $$\frac{2(2)-1}{2^2 (2-1)^2} = \frac{3}{4}.$$ Calculate the right side for $n=2$: $$1 - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}.$$ Since both sides equal $\frac{3}{4}$, the base case holds. 3. **Inductive Hypothesis:** Assume the formula holds for some $k \geq 2$: $$\sum_{n=2}^k \frac{2n - 1}{n^2 (n-1)^2} = 1 - \frac{1}{k^2}.$$ 4. **Inductive Step:** We need to prove it holds for $k+1$: $$\sum_{n=2}^{k+1} \frac{2n - 1}{n^2 (n-1)^2} = 1 - \frac{1}{(k+1)^2}.$$ Start with the left side: $$\sum_{n=2}^k \frac{2n - 1}{n^2 (n-1)^2} + \frac{2(k+1) - 1}{(k+1)^2 k^2}.$$ By the inductive hypothesis, this equals: $$1 - \frac{1}{k^2} + \frac{2k + 1}{(k+1)^2 k^2}.$$ 5. **Simplify the expression:** Combine the terms: $$1 - \frac{1}{k^2} + \frac{2k + 1}{(k+1)^2 k^2} = 1 + \frac{- (k+1)^2 + 2k + 1}{(k+1)^2 k^2}.$$ Expand $-(k+1)^2$: $$-(k^2 + 2k + 1) = -k^2 - 2k - 1.$$ So numerator becomes: $$-k^2 - 2k - 1 + 2k + 1 = -k^2.$$ Therefore: $$1 + \frac{-k^2}{(k+1)^2 k^2} = 1 - \frac{1}{(k+1)^2}.$$ 6. **Conclusion:** The formula holds for $k+1$ if it holds for $k$. Since the base case is true, by mathematical induction, the formula is true for all integers $n \geq 2$. **Final answer:** $$\sum_{n=2}^n \frac{2n - 1}{n^2 (n-1)^2} = 1 - \frac{1}{n^2}.$$