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\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+.....+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}\)
\(=\frac{\frac{201.\left(201+1\right)}{2}-1}{2}\)
\(=10150\)
Ta có: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
Áp dụng vào tính tổng E:
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.....+\frac{1}{200}.\left(1+2+3+....+200\right)\)
\(E=1+\frac{1}{2}.\frac{2.\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+....+\frac{1}{200}.\frac{200.\left(201+1\right)}{2}\)
\(E=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+......+\frac{1}{200}.\frac{200.201}{2}\)
\(E=1+\frac{1.2.3}{2.2}+\frac{1.3.4}{3.2}+......+\frac{1.200.201}{200.2}\)
\(E=1+\frac{3}{2}+\frac{4}{2}+......+\frac{201}{2}=\frac{1}{2}.\left(2+3+4+...+201\right)\)
Từ 2->201 có:201-1+1=201(số hạng)
=>\(2+3+4+....+201=\frac{201.\left(201+1\right)}{2}=20301\)
=>E=1/2.20301=20301/2=10150,5
\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{2}{4}\)
\(=\frac{3}{4}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)
\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)
\(=\frac{25}{20}-\frac{-1}{20}\)
\(=\frac{26}{20}\)
\(=\frac{13}{10}\)
a, Ta có: \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+....+\left(1-\frac{99}{100}\right)\right]\)
\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-100+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)(đpcm)
b, Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(đpcm)
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\)\(+\frac{99}{100}\)
Xét: \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
= \(\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{100}\right)\)
= \(\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)( có 99 số hạng là 1 )
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(\left(99+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)( đpcm )
Vậy: ...
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+....+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{200}.\frac{200.201}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}\)
\(=\frac{\frac{201.202}{2}-1}{2}=10150\)