Tinh B=1+1/2*(1+2)+1/3*(1+2+3)+1/4*91+2+3+4)+.........+1/20*(1+2+3+...+20)
Những câu hỏi liên quan
B=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3...+20) . Tinh B gium minh voi minh cam on .
tinh nhanh: 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/10(1+2+3+...+20)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{20}.\left(1+2+....+20\right)\)
\(=1+\frac{1}{2}\times\frac{2.3}{2}+\frac{1}{3}\times\frac{3.4}{2}+...+\frac{1}{20}\times\frac{20.21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(=\frac{\left(2+21\right).20:2}{2}=\frac{230}{2}=115\)
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Số cuối là
\(\frac{1}{10}.\left(1+2+3+...+10\right)\) hay \(\frac{1}{20}.\left(1+2+3+...+20\right)\) ??
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tinh tổng 1/1+2 +1/1+2+3 1/1+2+3+4 + ... + 1/1+2+3+...+20
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
TÍNH: B= 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3+4+...+20)
Tính B=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3+4+...+20)
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 \(B=\dfrac{209}{2}\)
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Tính B= 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3+...+20)
Tính B=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3+...+20)
tính B=1+1/2*(1+2)+1/3*(1+2+3)*1/4*(1+2+3+4)+...+1/20*(1+2+3+...+20)
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+...+\frac{1}{20}.\frac{20\left(20+1\right)}{2}\)
\(=\frac{2}{2}+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{20+1}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{20}{2}\)
\(=\frac{2+3+4+...+20}{2}=\frac{\frac{20\left(20+1\right)}{2}-1}{2}=\frac{209}{2}\)
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ở dòng thứ 4 của bài làm của Đinh Đức Hùng thì phân số 20/2 phải là 21/2 mới đúng.
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tính B= 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3+...+20)