1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3+4+...+20)
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3 tháng 5 2016 lúc 20:01
1+ 1/2(1+2) +1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3+4+5+...+20)
`Answer:`
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.3+...+\frac{1}{2}.210\)
\(=1+1,5+2+...+10,5\)
\(=\frac{\left(10,5+1\right)[\left(10,5-1\right):0,5+1]}{2}\)
\(=\frac{230}{2}\)
\(=115\)
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:
1+1/2.(1+2)+1/3.(1+2+3)+1/4.(1+2+3+4)+...+1/20.(1+2+3+4+5+...+20)
tính A= 1+1/2*(1+2)+1/3*(1+2+3)+1/4*(1+2+3+4)+...+1/20*(1+2+3+...+20)
Tính A=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)
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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