Tính:
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{200}\left(1+2+3+...+200\right)\)
K
Khách
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Những câu hỏi liên quan
25 tháng 2 2017
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+.....+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}\)
\(=\frac{\frac{201.\left(201+1\right)}{2}-1}{2}\)
\(=10150\)
19 tháng 3 2017
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+....+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{200}.\frac{200.201}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}\)
\(=\frac{\frac{201.202}{2}-1}{2}=10150\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 5 2019
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow E=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{1}{2}\left(3+4+5+...+201\right)\)
\(=1+\frac{1}{2}\left(1+2+3+...+201-1-2\right)\)
\(=1+\frac{1}{2}\left(\frac{201.202}{2}-3\right)=10150\)
\(\frac{21}{5}\left|x\right|< 2019\Rightarrow\left|x\right|< 2019\div\frac{21}{5}=\frac{3365}{7}\)
\(\Rightarrow-480\le x\le480\)
\(\Rightarrow\sum x=-480+480-479+479+...+-1+1+0=0\)
\(\frac{2^{24}\left(x-3\right)}{\frac{81}{35}.\left(6.2^{24}-2^{26}\right)}=\frac{25}{9}\)
\(\Leftrightarrow\frac{2^{24}\left(x-3\right)}{2^{24}\left(6-2^2\right)}=\frac{25}{9}.\frac{81}{35}\)
\(\Leftrightarrow\frac{x-3}{2}=\frac{45}{7}\)
\(\Leftrightarrow x-3=\frac{90}{7}\)
\(\Rightarrow x=\frac{111}{7}\)
DB
9 tháng 4 2017
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
27 tháng 12 2018
A = \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{200}-1\right)\)
= \(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-199}{200}\)
= \(\frac{-1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{199}{200}\)
= \(\frac{-1}{200}\)> \(\frac{-1}{199}\)( vì 1/200 < 1/999 => - 1 / 200 > -1/199 )