Tinh B=1+1/2*(1+2)+1/3*(1+2+3)+1/4*91+2+3+4)+.........+1/20*(1+2+3+...+20)
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NN
0
DT
2
8 tháng 1 2016
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{20}.\left(1+2+....+20\right)\)
\(=1+\frac{1}{2}\times\frac{2.3}{2}+\frac{1}{3}\times\frac{3.4}{2}+...+\frac{1}{20}\times\frac{20.21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(=\frac{\left(2+21\right).20:2}{2}=\frac{230}{2}=115\)
8 tháng 1 2016
Số cuối là
\(\frac{1}{10}.\left(1+2+3+...+10\right)\) hay \(\frac{1}{20}.\left(1+2+3+...+20\right)\) ??
DN
1
6 tháng 3 2020
Đặt \(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+..............+\frac{1}{1+2+3+...+20}\)
\(\Rightarrow A=\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+\frac{1}{\frac{\left(1+4\right).4}{2}}+.............+\frac{1}{\frac{\left(1+20\right).20}{2}}\)
\(\Rightarrow A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...........+\frac{2}{20.21}\)
\(\Rightarrow A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..............+\frac{1}{20.21}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{20}-\frac{1}{21}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\left(\frac{21}{42}-\frac{2}{42}\right)=2.\frac{19}{42}=\frac{19}{21}\)
Vậy \(A=\frac{19}{21}\)
Chúc bn học tốt
HH
0
LT
1
2 tháng 4 2017
Ta có:
\(B=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)
\(=1+\dfrac{1}{2}.\dfrac{2\left(2+1\right)}{2}+\dfrac{1}{3}.\dfrac{3\left(3+1\right)}{2}+...+\dfrac{1}{20}.\dfrac{20\left(20+1\right)}{2}\)
\(=\dfrac{2}{2}+\dfrac{2+1}{2}+\dfrac{3+1}{2}+...+\dfrac{20+1}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{20}{2}\)
\(=\dfrac{2+3+4+...+20}{2}=\dfrac{\dfrac{20\left(20+1\right)}{2}-1}{2}\)
\(=\dfrac{209}{2}\)
Vậy \(B=\dfrac{209}{2}\)
IM
0
H
0
TQ
1
2 tháng 3 2017
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+...+\frac{1}{20}.\frac{20\left(20+1\right)}{2}\)
\(=\frac{2}{2}+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{20+1}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{20}{2}\)
\(=\frac{2+3+4+...+20}{2}=\frac{\frac{20\left(20+1\right)}{2}-1}{2}=\frac{209}{2}\)
KT
0
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{20}.20.21:2\)
=\(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}=\frac{2+3+4+...+21}{2}=\frac{230}{2}=115\)
thank