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\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+....+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{200}.\frac{200.201}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}\)
\(=\frac{\frac{201.202}{2}-1}{2}=10150\)
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+.....+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}\)
\(=\frac{\frac{201.\left(201+1\right)}{2}-1}{2}\)
\(=10150\)
Áp dụng công thức \(1+2+...+n=\frac{n\left(n+1\right)}{2}\)ta có:
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}=\frac{\frac{201.202}{2}-1}{2}=10150\)
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(E=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{200}.\frac{\left(1+200\right).200}{2}\)
\(E=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{1+200}{2}\)
\(E=1+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)
\(E=\frac{2+3+4+...+201}{2}=\frac{\left(201+2\right).200:2}{2}\)
\(E=10150\)
Xét thừa số tổng quát:
\(\frac{1+2+...+n}{n}=\frac{n\left(n+1\right):2}{n}=\frac{n+1}{2}\)
Thay vào bài toán:
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+3+...+200\right)\)
\(E=1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+200}{200}\)
\(E=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{200+1}{2}\)
\(E=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)
\(E=\frac{2+3+4+...+201}{2}=\frac{20300}{2}=10150\)