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

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

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

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

\(=\frac{\frac{201.202}{2}-1}{2}=10150\)

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

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

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

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

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

\(=10150\)

đmđmđmmt

đi mua đi mua đi mua mắm tôm

ko thèm trả lời
 

\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)

\(=\frac{1}{4}+\frac{1}{2}\)

\(=\frac{1}{4}+\frac{2}{4}\)

\(=\frac{3}{4}\)

\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)

\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)

\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)

\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)

\(=\frac{25}{20}-\frac{-1}{20}\)

\(=\frac{26}{20}\)

\(=\frac{13}{10}\)

1/2+3/4-(3/4-4/5)

1/2+3/4+3/4+4/5

1/2+6/4+4/5

10/20+30/20+16/20

56/20=14/5

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

đáp án = 10150 , bạn sai chỗ nào đấy

=>E=1+\(\frac{1}{2}\) .\(\frac{2.3}{2}\) +.......+\(\frac{1}{200}\) .\(\frac{201.200}{2}\)

=>E=1+\(\frac{2.3}{2^2}\) +.......+\(\frac{201.200}{2^2}\)

=>E=\(\frac{1.2}{2^2}\) +\(\frac{2.3}{2^2}\) +..................+\(\frac{201.200}{2^2}\) +\(\frac{1}{2}\)

=>E=\(\frac{1.2+2.3+....+201.200}{2^2}\) +\(\frac{1}{2}\)

=>3E=\(\frac{1.2.3+2.3.3+.....+201.200.3}{2^2}\) +\(\frac{1}{2}\)

=>3E=\(\frac{1.2.3+2.3.4-1.2.3+......+201.200.202-199.200.201}{2^2}\) +\(\frac{1}{2}\)

=>3E=\(\frac{200.201.202}{4}\) +\(\frac{4}{4}\)

=>3E=\(\frac{200.201.202+4}{4}\)

=>3E=50.201.202+1

=>E=\(\frac{50.201.202+1}{3}\)

Vậy E= \(\frac{50.201.202+1}{3}\)