Задача 60748 Найдите сумму n первых членов...
Условие
Решение
⇒
[m]a_{1}=\frac{1+1+1^2}{1\cdot (1+1)}=\frac{3}{2}[/m]
[m]a_{2}=\frac{1+2+2^2}{2\cdot (2+1)}=\frac{7}{6}[/m]
[m]a_{3}=\frac{1+3+3^2}{3\cdot (3+1)}=\frac{13}{12}[/m]
[m]a_{4}=\frac{1+4+4^2}{4\cdot (4+1)}=\frac{21}{20}[/m]
...
[m]a_{n-1}=\frac{1+(n-1)+(n-1)^2}{(n-1)\cdot n}[/m]
[m]a_{n}=\frac{1+n+n^2}{n\cdot (n+1)}[/m]
[m]a_{1}+a_{2}+a_{3}+a_{4}+...+a_{n}= \frac{3}{2}+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+...+\frac{1+(n-1)+(n-1)^2}{(n-1)\cdot n}+\frac{1+n+n^2}{n\cdot (n+1)}=[/m]
[m]=1+\frac{1}{2}+1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+...+1+\frac{1}{(n-1)n}+1+\frac{1}{n(n+1)}=n+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{(n-1)n}+\frac{1}{n(n+1)}[/m]
[m]=n+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}=n+1-\frac{1}{n+1}=\frac{(n+1)^2-1}{n+1}=\frac{n^2+2n+1-1}{n+1}=\frac{n^2+2n}{n+1}[/m]
[youtube=https://www.youtube.com/watch?v=fpIIl5KpSTs]