\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{200^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)

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

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

\(=\frac{199}{200}\)

vậy \(\frac{99}{200}< \frac{199}{200}< 1\left(đpcm\right)\)

rồi sao

A=1+\(\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+........+\dfrac{1+2+.......+200}{200}\)

A=1+\(\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+.......+\dfrac{\dfrac{\left(1+200\right).200}{2}}{200}\)

A=\(\dfrac{2}{2}\)+\(\dfrac{3}{2}\)+......+\(\dfrac{200}{2}\)=\(\dfrac{2+3+.......+200}{2}\)=\(\dfrac{\dfrac{\left(2+200\right).\text{[}\left(200-2\right):1+1\text{]}}{2}}{2}\)=\(\dfrac{19701}{2}\)

đmđmđmmt

đi mua đi mua đi mua mắm tôm

ko thèm trả lời
 

\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)

\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+.....+\frac{1}{200}.\frac{200.201}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)

\(=\frac{2+3+4+...+201}{2}\)

\(=\frac{\frac{201.\left(201+1\right)}{2}-1}{2}\)

\(=10150\)

Ta co :

E=\(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{201}{2}\)

   =\(\frac{2+3+4+5+...+201}{2}\)

   =\(\frac{\left[\left(201+2\right)\left(201-2\right):1+1\right]:2}{2}\)

   =\(\frac{40398:2}{2}\)

=\(\frac{20199}{2}\)

Đúng thì k không thì giúp tớ với 

kết quả ra sai rồi

\(E=\frac{2+3+4+...+201}{2}=\frac{\frac{\left[\left(201-2\right):1+1\right].\left(201+2\right)}{2}}{2}=\frac{\frac{200.203}{2}}{2}=\frac{100.203}{2}\)=10150