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đặt A=1/2+(1/2)^2+(1/2)^3+...+(1/2)^98+(1/2)^99+(1/2)^99
=>A=1/2+12/22+13/23+...+198/298+199/299+199/299
=>A=1/2+1/22+1/23+...+1/298+1/299+1/299
=>2A-1/299=1+1/2+1/22+...+1/298
=>(2A-1/299)-(A-1/299)=(1+1/2+1/22+...+1/298)-(1/2+1/22+1/23+...+1/298+1/299)
=>(2A-1/299)-(A-1/299)=1-1/299
=>A=1-1/299 +1/299=1
vậy A=1
chắc thế
Nhân 1/2B lên, ta được:
1/2B=(1/2)^2+(1/2)^3+...+(1/2)^100
lấy B-1/2B, ta được: 1/2B=1/2-(1/2)^100
=>B= (1/2-(1/2)^100)/2
\(=-\left(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9800}{9801}\right)\)
\(=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{98.100}{99.99}\right)\)
\(=-\left(\frac{1.2.3....98}{2.3...99}.\frac{3.4.5...100}{2.3...99}\right)\)
\(=-\left(\frac{1}{99}.50\right)\)
\(=-\frac{50}{99}\)
Chúc hok tốt!!!
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\left(\frac{1}{99^2}-1\right)\)
= \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}\)
= \(\frac{\left(1.2.3...98\right).\left(3.4.5...100\right)}{\left(2.3.4...99\right).\left(2.3.4...99\right)}\)
= \(\frac{100}{99.2}\)= \(\frac{50}{99}\)