\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\left[\frac{1}{199}+1\right]+\left[\frac{2}{198}+1\right]+\left[\frac{3}{197}+1\right]+...+\left[\frac{198}{2}+1\right]}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{200\left[\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right]}=\frac{1}{200}\)

\(\text{Đ}\text{ặt}S=1+\frac{1}{2}+\frac{1+2+3}{3}+...+\frac{1+2+...+199}{199}\)

\(\Rightarrow S=1+\frac{\left(2+1\right).2}{2}+\frac{\left(3+1\right)3}{3}+...+\frac{\left(199+1\right)199}{199}\)

\(S=1+\frac{2+1}{1}+\frac{3+1}{1}+...+\frac{199+1}{1}\)

\(\Rightarrow S=1+\left(3+4+...+200\right)\)

Dãy (3+4+..+200 ) có số số hạng là :

(200-3):1+1=198 ( số )

Tổng của dãy (3+4+..+200 ) là :

(200+1)x198:2=19899

=> S=1+(3+4+...+200)

=> S=1+19899

=> S=19900

Bạn ơi, của bạn giống của mình mà, đăng làm gì, nếu cùng chung 1 kết quả thì bạn tick đúng đi.

Bạn ơi, bài này có người hỏi rồi và mình cũng trả lời rồi

\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+199}{199}\)\(=1+\frac{\frac{2.3}{2}}{2}+\frac{\frac{3.4}{2}}{3}+...+\frac{\frac{199.200}{2}}{199}\)\(=1+\frac{2.3}{2.2}+\frac{3.4}{3.2}+...+\frac{199.200}{199.2}\)\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{200}{2}\)\(=\frac{2+3+4+...+200}{2}\)\(=\frac{\frac{200.201}{2}}{2}\)\(=\frac{200.201}{2.2}\)\(=10050\)

trà my sai rồi,

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+...+\left(\frac{198}{2}+1\right)+1}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+\frac{200}{200}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{200\cdot\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right)}\)

\(=\frac{1}{200}\)