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\(\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+199}{199}\)
\(=\frac{\frac{3.2}{2}}{2}+\frac{\frac{4.3}{2}}{3}+...+\frac{\frac{200.199}{2}}{199}\)
\(=\frac{3.2}{2}.\frac{1}{2}+\frac{4.3}{2}.\frac{1}{3}+...+\frac{200.199}{2}.\frac{1}{199}\)
\(=\frac{3}{2}+\frac{4}{2}+...+\frac{200}{2}\)
\(=\frac{3+4+...+200}{2}=\frac{203.198}{2}.\frac{1}{2}=\frac{20097}{2}\)
\(\text{Đ}\text{ặt}S=1+\frac{1}{2}+\frac{1+2+3}{3}+...+\frac{1+2+...+199}{199}\)
\(\Rightarrow S=1+\frac{\left(2+1\right).2}{2}+\frac{\left(3+1\right)3}{3}+...+\frac{\left(199+1\right)199}{199}\)
\(S=1+\frac{2+1}{1}+\frac{3+1}{1}+...+\frac{199+1}{1}\)
\(\Rightarrow S=1+\left(3+4+...+200\right)\)
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
\(=1+\frac{2.3}{2.2}+\frac{3.4}{2.3}+\frac{4.5}{2.4}+....+\frac{199.200}{2.199}=\)
\(=1+\frac{1}{2}\left(3+4+5+...+200\right)=\)
\(1+\frac{1}{2}.\frac{198.203}{2}=10049,5\)
\(M=1+\frac{1}{199}+1+\frac{2}{198}+1+....+\frac{198}{2}+1=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}\)
\(=200.\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)=200 T
\(S=\frac{T}{200T}=\frac{1}{200}\)
Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)
= \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)
= \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)
= \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)
= 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)
= \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)
