To prove the statement $\frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+2)(n+3)} + \cdots + \frac{1}{(2n-1)2n} = \frac{1}{2n}$ we’ll use mathematical induction.

Step 1: Base Case ($n = 1$)

For $n = 1$, the left side of the equation is: $\frac{1}{1 \cdot 2} = \frac{1}{2}$ The right side of the equation is also: $\frac{1}{2 \cdot 1} = \frac{1}{2}$ So, the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement is true for some integer $n = k$, i.e., $\frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + \cdots + \frac{1}{(2k-1)2k} = \frac{1}{2k}$

Step 3: Inductive Step

We need to show that the statement holds for $n = k + 1$, i.e., $\frac{1}{(k+1)(k+2)} + \frac{1}{(k+2)(k+3)} + \cdots + \frac{1}{(2k+1)(2k+2)} = \frac{1}{2(k+1)}$ Starting with the left side of the expression for $n = k + 1$, we add the next term $\frac{1}{(2k+1)(2k+2)}$ to both sides of the inductive hypothesis: $\frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + \cdots + \frac{1}{(2k-1)2k} + \frac{1}{(2k+1)(2k+2)}$ Using the inductive hypothesis, we replace the sum up to $\frac{1}{(2k-1)2k}$ with $\frac{1}{2k}$: $\frac{1}{2k} + \frac{1}{(2k+1)(2k+2)}$

Combining the Terms

Now we combine the two terms on the right side: $\frac{1}{2k} + \frac{1}{(2k+1)(2k+2)} = \frac{(2k+2) + 1}{2k(2k+2)}$ Simplify the numerator: $= \frac{2k+3}{2k(2k+2)}$