[tex]\frac{1}{1\cdot3}+ \frac{7}{3\cdot 7} +...+ \frac{2n ^ 2 - 1}{(2n - 1)(2n + 1)} =\frac{n ^ 2}{2n + 1}[/tex]

1. Проверяем истинность утверждения для n = 1.

[tex]\frac{2\cdot 1 ^ 2 - 1}{(2\cdot 1 - 1)\cdot (2\cdot 1 + 1)} =\frac{2\cdot 1 - 1}{(2 - 1)(2 + 1)} =\frac{2 - 1}{1\cdot 3} =\frac{1}{3}[/tex]

[tex]\frac{1^ 2}{2\cdot 1 + 1}=\frac{1}{2+ 1}=\frac{1}{3}[/tex]

2. Предполагаем, что истинно для n = k (k - произвольное натуральное число).

[tex]\frac{1}{1\cdot3}+ \frac{7}{3\cdot 7} +...+ \frac{2k ^ 2 - 1}{(2k - 1)(2k + 1)} =\frac{k ^ 2}{2k + 1}[/tex]

3. Доказываем, что истинно, для n = k + 1.

[tex]\frac{1}{1\cdot3}+ \frac{7}{3\cdot 7} +...+ \frac{2(k+1) ^ 2 - 1}{(2(k+1) - 1)(2(k+1) + 1)} =\frac{(k+1) ^ 2}{2(k+1) + 1}[/tex]

[tex]\frac{1}{1\cdot3}+ \frac{7}{3\cdot 7} +...+\frac{2k ^ 2 - 1}{(2k - 1)(2k + 1)}+ \frac{2(k+1) ^ 2 - 1}{(2(k+1) - 1)(2(k+1) + 1)} =[/tex]

[tex]=\frac{k ^ 2}{2k + 1}+ \frac{2(k^2+2k+1)  - 1}{(2k+2 - 1)(2k+2 + 1)} =\frac{k ^ 2}{2k + 1}+ \frac{2k^2+4k+2  - 1}{(2k+ 1)(2k+3)} =[/tex]

[tex]=\frac{k ^ 2}{2k + 1}+ \frac{2k^2+4k+1}{(2k+ 1)(2k+3)} =\frac{k ^ 2(2k+3)}{(2k + 1)(2k+3)}+ \frac{2k^2+4k+1}{(2k+ 1)(2k+3)} =[/tex]

[tex]=\frac{2k ^3+3k^2+2k^2+4k+1}{(2k+ 1)(2k+3)} =\frac{2k ^3+4k^2+k^2+4k+1}{(2k+ 1)(2k+3)} =[/tex]

[tex]=\frac{2k ^3+2k^2+k^2+2k+1+2k^2+2k}{(2k+ 1)(2k+3)} =\frac{2k ^2(k+1)+(k+1)^2+2k(k+1)}{(2k+ 1)(2k+3)} =[/tex]

[tex]=\frac{(k+1)(2k^2+k+1+2k)}{(2k+ 1)(2k+3)} =\frac{(k+1)(2k^2+2k+k+1)}{(2k+ 1)(2k+3)} =[/tex]

[tex]\frac{(k+1)(2k(k+1)+(k+1))}{(2k+ 1)(2k+3)} =\frac{(k+1)(k+1)(2k+1)}{(2k+ 1)(2k+3)} =[/tex]

[tex]=\frac{(k+1)^2}{2k+3} =\frac{(k+1)^2}{2k+2+1} =\frac{(k+1) ^ 2}{2(k+1) + 1}[/tex]