1/2^2+1/3^2+...+1/200^2<1
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2.
E=1+1/2.(1+2)+1/3.(1+2+3)+...+1/200.(1+2+...+200)
\(E=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{200}\cdot\dfrac{200\cdot201}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{201}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+...+\dfrac{201}{2}\)
\(=\dfrac{\left(201-2+1\right)\cdot\left(201+2\right)}{4}=\dfrac{200\cdot203}{4}=50\cdot203=10150\)
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E=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+............+1/200(1+2+......+200)
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(E=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{200}.\frac{\left(1+200\right).200}{2}\)
\(E=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{1+200}{2}\)
\(E=1+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)
\(E=\frac{2+3+4+...+201}{2}=\frac{\left(201+2\right).200:2}{2}\)
\(E=10150\)
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tinh B=1+1/2(1+2)+1/3(1+2+3)+1/4+(1+2+3+4)+...+1/200(1+2+....+200)
thuc hien tinh :E=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+.......+1/200(1+2+3+.....+200)
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+....+\frac{1}{200}\left(1+2+...+200\right)\\ \Rightarrow E=1+\frac{1}{2}\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{200}\frac{\left(1+200\right).200}{2}\\ \Rightarrow E=1+\frac{1+2}{2}+\frac{1+3}{2}+....+\frac{1+200}{2}\\ \Rightarrow E=1+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\\ \Rightarrow E=\frac{2+3+4+....+201}{2}=\frac{\left(201+2\right).200:2}{2}=10150\)
Chúc bạn học tốt !!!
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thuc hien tinh :E=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+.......+1/200(1+2+3+.....+200)
Thực hiện tính:
E=1+1/2.(1+2)+1/3.(1+2+3)+1/4.(1+2+3+4)+.....+1/200(1+2+...+200)
1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)........+1/200(1+2+3+4+5+6+7+.....+200)
thực hiện phép tính
E=1+1/2(1+2)+1/2(1+2+3)+1/4(1+2+3+4)+.....+1/200(1+2+3+....+200)
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:
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+3+...+200\right)\)
\(E=1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+200}{200}\)
\(E=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{200+1}{2}\)
\(E=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)
\(E=\frac{2+3+4+...+201}{2}=\frac{20300}{2}=10150\)
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\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+...+200\right)\)
\(2A=2+\dfrac{1}{2}.6+\dfrac{1}{3}.12+\dfrac{1}{4}.20+...+\dfrac{1}{200}.40200=\)
\(=2+\dfrac{1}{2}.2.3+\dfrac{1}{3}.3.4+\dfrac{1}{4}.4.5+...+\dfrac{1}{200}.200.201=\)
\(=2+3+4+5+...+201=\dfrac{200\left(2+201\right)}{2}\)
\(=20300\Rightarrow A=\dfrac{20300}{2}=10150\)
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Tính : -1-1/2.(1+2)-1/3.(1+2+3)-...-1/200.(1+2+3+...+200)
Mong các bạn giúp mình nhiều nha !!!!
