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