Tính B=1+1/2(1+2)+1/3(1+2+3)+....+1/500(1+2+3+...+500)
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a, \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{200}-1\right)\)
\(-A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{200}\right)\)
\(-A=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{199}{200}\)
\(-A=\frac{1}{200}\)
\(A=\frac{-1}{200}>\frac{-1}{199}\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...............+\frac{1}{500}\left(1+2+3+.........+500\right)\)
\(=1+\frac{1}{2}\frac{3.2}{2}+\frac{1}{3}\frac{4.3}{2}+.............+\frac{1}{500}\frac{501.500}{2}\)
\(=\frac{1}{2}\left(2+3+............+501\right)\)
\(=\frac{1}{2}.251000\)
\(=125500\)
S = 1-1/2 + 1/3-1/4 + 1/5-1/6 + ..... 1/499-1/500 = (1 + 1/3 + 1/5 + ..+ 1/499) - (1/2 + 1/4 + 1/6 + ...+ 1/500) - (1/2 + 1/4 + 1/6 + ...+ 1/500) + (1/2 + 1/4 + 1/6 + ...+ 1/500) S = (1 + 1/2 + 1/3 + 1/4 + ....+ 1/500) - 2.(1/2 + 1/4 + 1/6 + ...+ 1/500) = (1 + 1/2 + 1/3 + 1/4 + ....+ 1/500)- (1 + 1/2 + 1/3 + ...+1/250) = 1/251 + 1/252 + ...+ 1/500.
Vậy S = 1/251 + 1/252 + ...+ 1/500
S = 1-1/2 + 1/3-1/4 + 1/5-1/6 + ..... 1/499-1/500
= (1 + 1/3 + 1/5 + ..+ 1/499) - (1/2 + 1/4 + 1/6 + ...+ 1/500) - (1/2 + 1/4 + 1/6 + ...+ 1/500) + (1/2 + 1/4 + 1/6 + ...+ 1/500)
S = (1 + 1/2 + 1/3 + 1/4 + ....+ 1/500) - 2.(1/2 + 1/4 + 1/6 + ...+ 1/500)
= (1 + 1/2 + 1/3 + 1/4 + ....+ 1/500)- (1 + 1/2 + 1/3 + ...+1/250)
= 1/251 + 1/252 + ...+ 1/500.
Vậy S = 1/251 + 1/252 + ...+ 1/500