2. Use Mathematical Induction to prove that \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\square{ }^{1}+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1} \text { for all natural numbers, } n . a) Verify that

it is true for n=1. 1/1.3 = ________ 1/ 2.1+1 = __________ b) Assuming that it is true for n=k , you get that \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\square+\frac{1}{(2 k-1)(2 k+1)}=\frac{k}{2 k+1} \text { Add } \frac{1}{[2(k+1)-1][2(k+1)+1]}, \text { which equals } \frac{1}{(2 k+1)(2 k+3)}, \text { on both sides, } and show that it is true for n-k +1, by simplifying the right side of the following equation. \frac{1}{1 \cdot3}+\frac{1}{3 \cdot5}+\frac{1}{5 \cdot7}+?+\frac{1}{(2 k-1)(2 k+1)}+\frac{1}{(2 k+1)(2 k+3)} =\frac{k}{2 k+1}+\frac{1}{(2 k+1)(2 k+3)}