Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+3+...+20\right)\)\(=1+\dfrac{1}{2}.2.3:2+\dfrac{1}{3}.3.4:2+...+\dfrac{1}{20}.20.21:2\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+...+\dfrac{21}{2}\)
\(=\dfrac{2+3+...+21}{2}\)
\(=\dfrac{230}{2}\)
\(=115\)
Tính B= 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/20(1+2+3+...+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)+....+1/20(1+2+3+4+....+20)
Tính B=1+1/2(1+2)+1/3.(1+2+3)+...+1/20.(1+2..+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 dùm mình nha thanks trước
tính hợp lý
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+\(\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+3+...+20\right)\)
Tính giá trị biểu thức :
A = 2/1+2 + (2+3)/1+2+3 + ....... + (2+3+...+20)/1+2+..+20
B = (1 - 1/2) (1 - 1/3) ..... (1 - 1/2005)(1 - 1/2006)