Дорешайте 12 умоляю ! Распишите подробнее. 12. начинается 5/1*2 + 13/2*5+25/3*4...

Решите задачу:

$\frac{5}{1\cdot 2} +\frac{13}{2\cdot 3}+ \frac{25}{3\cdot 4}+...+\frac{2n^2+2n+1}{n(n+1)}=\\\\\star \; \; \frac{2n^2+2n+1}{n(n+1)}= \frac{2n(n+1)+1}{n(n+1)}= \frac{2n(n+1)}{n(n+1)}+\frac{1}{n(n+1)}=2+\frac{1}{n(n+1)}\; \; \star \\\\ =\underbrace {(2+\frac{1}{1\cdot 2})}_{1}+\underbrace {( 2+\frac{1}{2\cdot 3})}_{2}+\underbrace {(2+\frac{1}{3\cdot 4})}_{3}+...+\underbrace {(2+\frac{1}{n(n+1)})}_{n}=\\\\=2\cdot n+\Big (\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{n(n+1)}\Big )=$

$\star \; \; \frac{1}{n(n+1)} = \frac{1}{n} -\frac{1}{n+1} \; \; \star \\\\=2\cdot n+( \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1} )=\\\\=2n+1-\frac{1}{n+1}= \frac{(2n+1)(n+1)-1}{n+1} = \frac{2n^2+3n}{n+1} = \frac{n(2n+3)}{n+1}$