1/1+4/200=?
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VD
1
PT
20 tháng 2 2017
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(E=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{200}.\frac{\left(1+200\right).200}{2}\)
\(E=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{1+200}{2}\)
\(E=1+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)
\(E=\frac{2+3+4+...+201}{2}=\frac{\left(201+2\right).200:2}{2}\)
\(E=10150\)
NT
0
SC
0
LT
Lương Thị Vân Anh
CTVHS
11 tháng 3 2023
Ta đã biết \(\dfrac{1}{a\cdot a}< \dfrac{1}{\left(a+1\right)\left(a-1\right)}\) ( a ϵ Z )
⇒ \(Q=\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+\dfrac{1}{4\cdot4}+...+\dfrac{1}{200\cdot200}\) < \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
Ta có \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{199\cdot201}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{201}\right)=\dfrac{1}{2}\cdot\dfrac{200}{201}=\dfrac{100}{201}< \dfrac{100}{200}=\dfrac{1}{2}< \dfrac{3}{4}\)
Vậy Q < \(\dfrac{3}{4}\)
PT
1
NT
11 tháng 2 2017
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+....+\frac{1}{200}\left(1+2+...+200\right)\\ \Rightarrow E=1+\frac{1}{2}\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{200}\frac{\left(1+200\right).200}{2}\\ \Rightarrow E=1+\frac{1+2}{2}+\frac{1+3}{2}+....+\frac{1+200}{2}\\ \Rightarrow E=1+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\\ \Rightarrow E=\frac{2+3+4+....+201}{2}=\frac{\left(201+2\right).200:2}{2}=10150\)
Chúc bạn học tốt !!!
PT
0
HQ
0
BP
0
NT
0
LN
0
Trả lời:
= 1,02
=1,02 nha