Question

[zZz Cool Kid_new zZz]

\(CMR:A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}\notinℕ\)

Answer 1

[tth_new]

Tổng trên có số số hạng là: \(\left(n-2\right):1+1=n-1\) số hạng

Suy ra \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n}\)

\(=\frac{\left(\frac{1}{n}+\frac{1}{2}\right)\left(n-1\right)}{2}=\frac{\frac{1}{n}\left(n-1\right)+\frac{1}{2}\left(n-1\right)}{2}\)

\(=\frac{1-\frac{1}{n}+\frac{n}{2}-\frac{1}{2}}{2}=\frac{\frac{1}{2}-\left(\frac{1}{n}-\frac{n}{2}\right)}{2}\)

\(=\frac{\left(\frac{1}{2}\right)}{2}-\frac{\left(\frac{2}{2n}\right)}{2}+\frac{\left(\frac{n^2}{2n}\right)}{2}=\frac{1}{4}-\frac{1}{2n}+\frac{n}{4}\)

Suy ra \(n\ne0\).Ta có: \(S=\frac{1}{4}-\frac{1}{2n}+\frac{n}{4}=\frac{1+n}{4}-\frac{1}{2n}\)

\(=\frac{2n^2+2n+4}{8n}=\frac{2\left(n+\frac{1}{2}\right)^2}{8n}+\frac{\left(\frac{7}{2}\right)}{8n}\)

\(=\frac{2\left(n+\frac{1}{2}\right)^2}{8n}+\frac{7}{16n}\)

Đến đây bí =)Alibaba!