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)
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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:
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{201}{2}\)
\(=\frac{2+3+4+...+201}{2}=\frac{\frac{201.202}{2}-1}{2}=10150\)
1: \(347\cdot2^2-2^2\cdot\left(216+184\right):8\)
\(=347\cdot4-4\cdot400:8\)
\(=347\cdot4-4\cdot50=4\cdot\left(347-50\right)=4\cdot297=1188\)
2: \(132-\left\lbrack116-\left(132-128\right)^2\right\rbrack\)
\(=132-\left\lbrack116-4^2\right\rbrack\)
=132-(116-16)
=132-100
=32
3: \(16:\left\lbrace400:\left\lbrack200-\left(37+46\cdot3\right)\right\rbrack\right\rbrace\)
=16:{400:[200-(37+138)]}
=16:{400:[200-175]}
=16:{400:25}
=16:16
=1
4: \(\left\lbrace184:\left\lbrack96-124:31\right\rbrack-2\right\rbrace\cdot3651\)
\(=\left\lbrace\frac{184}{96-4}-2\right\rbrace\cdot3651\)
=(184:92-2)*3651
=0
5: \(46-\left\lbrack\left(16+71\cdot4\right):15\right\rbrack-2\)
=46-[(16+284):15]-2
=46-300:15-2
=46-20-2
=46-22
=24
6: \(3^3\cdot18+72\cdot4^2-41\cdot18\)
\(=18\left(3^3-41\right)+72\cdot16\)
\(=18\cdot\left(27-41\right)+18\cdot64\)
=18(-14+64)
=18*50
=900
7: \(\left(56\cdot46-25\cdot23\right):23\)
\(=56\cdot\frac{46}{23}-25\)
=112-25
=87
8: \(\left(28\cdot54+56\cdot36\right):21:2\)
\(=18\cdot28\left(3+2\cdot2\right):42\)
\(=18\cdot\frac{28}{42}\cdot7=18\cdot\frac23\cdot7=12\cdot7=84\)
9: \(\left(76\cdot34-19\cdot64\right):\left(38\cdot9\right)\)
\(=\frac{19\cdot\left(4\cdot34-64\right)}{38\cdot9}=\frac{4\cdot34-64}{9\cdot2}=\frac{4\cdot\left(34-16\right)}{18}=4\)
10: \(\left(2+4+6+\cdots+100\right)\cdot\left(36\cdot333-108\cdot111\right)\)
\(=\left(2+4+6+\cdots+100\right)\cdot36\cdot111\left(3-3\right)\)
=0
11: \(\left(5\cdot4^{11}-3\cdot16^5\right):4^{10}\)
\(=\left(5\cdot4^{11}-3\cdot4^{10}\right):4^{10}\)
\(=5\cdot\frac{4^{11}}{4^{10}}-3\cdot\frac{4^{10}}{4^{10}}\)
=20-3
=17
12: \(\frac{7256\cdot4375-725}{3650+4375\cdot7255}=\frac{7255\cdot4375+4375-725}{7255\cdot4375+3650}\)
\(=\frac{7255\cdot4375+3650}{7255\cdot4375+3650}\)
=1
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{200}\left(1+2+...+200\right)\)
\(E=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{200}.\frac{\left(1+200\right).200}{2}\)
\(E=1+\frac{1+2}{2}+\frac{1+3}{2}+...+\frac{1+200}{2}\)
\(E=1+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)
\(E=\frac{2+3+4+...+201}{2}=\frac{\left(201+2\right).200:2}{2}\)
\(E=10150\)
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:
\(E=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{200}\left(1+2+3+...+200\right)\)
\(E=1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+200}{200}\)
\(E=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{200+1}{2}\)
\(E=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{201}{2}\)
\(E=\frac{2+3+4+...+201}{2}=\frac{20300}{2}=10150\)