I8
0
K
Khách
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11 tháng 9 2017
\(A=\frac{3}{1}+\frac{3}{\frac{\left(2+1\right).2}{2}}+\frac{3}{\frac{\left(3+1\right).3}{2}}+....+\frac{3}{\frac{\left(100+1\right).100}{2}}\)
\(\Rightarrow A=\frac{3}{1}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+\frac{6.99}{202}=\frac{297}{101}+\frac{3}{1}=\frac{600}{101}\)
kết quả k bik có sai k
D
12 tháng 5 2015
Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)
= \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)
= \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)
= \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)
= 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)
= \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)
VN
1
LM
15 tháng 2 2017
\(=\frac{1}{2\cdot3:2}+\frac{1}{3\cdot4:2}+\frac{1}{4\cdot5:2}+...+\frac{1}{50\cdot51:2}\)
\(=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{100\cdot101}\)
\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{50\cdot51}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{51}\right)=\frac{1}{2}\cdot\frac{49}{102}=\frac{49}{204}\)
VN
2
T
21 tháng 2 2017
\(\frac{1}{1+2}\)+ \(\frac{1}{1+2+3}\)+ \(\frac{1}{1+2+3+4}\)+ ....+ \(\frac{1}{1+2+3+4+...+99+100}\)= ?
= \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+...+ \(\frac{1}{5050}\)
= (\(\frac{1}{3}+\frac{1}{5050}\)) x \(\frac{2}{1}\)
= \(\frac{5050}{15150}\)+ \(\frac{3}{15150}\)x \(\frac{2}{1}\)
= \(\frac{5053}{15150}\)x \(\frac{2}{1}\)
= \(\frac{10106}{15150}\)
Vậy tổng là: \(\frac{10106}{15150}\)
k nha!Khó lắm đó mới giải được
1 tháng 3 2017
Xin lỗi bạn! Đáp án là bằng một vì dượng mình có chỉ nhưng dượng không chỉ mình cách giải.
NV
1
KN
13 tháng 7 2020
\(A=\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+3+...+100}\)
\(\Rightarrow A=\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+\frac{3}{10}+...+\frac{3}{5050}\)
\(\Rightarrow A=\frac{2}{2}\left(\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+\frac{3}{10}+...+\frac{3}{5050}\right)\)
\(\Rightarrow A=\frac{6}{2}+\frac{6}{6}+\frac{6}{10}+\frac{6}{20}+...+\frac{6}{10100}\)
\(\Rightarrow A=6\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{10}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(\Rightarrow A=6\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(\Rightarrow A=6\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A=6\left(1-\frac{1}{101}\right)\)
\(\Rightarrow A=6.\frac{100}{101}\)
\(\Rightarrow A=\frac{600}{101}\)
22 tháng 7 2015
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)++...+\left(1+\frac{98}{2}\right)1}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}}{\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}}{100\times\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)}\)
\(=\frac{1}{100}\)
LB
14 tháng 5 2017
\(A=5+\frac{5}{1+2}+\frac{5}{1+2+3}+...+\frac{5}{1+2+3+...+100}\)
N
14 tháng 5 2017
A = \(5+\frac{5}{1+2}+\frac{5}{1+2+3}+...+\frac{5}{1+2+3+..+100}\)
\(=5x\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(=5x\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{5050}\right)\)
\(=2x5x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(=10x\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{100x101}\right)\)
\(=10x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=10x\left(1-\frac{1}{101}\right)\)
\(=10x\frac{100}{101}\)
\(=\frac{1000}{101}\)