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b)Ta có:\(A=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{16.\left(1+2+3+...+16\right)}\)
\(=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.3+\frac{1}{3}.6+...+\frac{1}{16}.136\)
\(=1+1,5+2+...+8,5\)
\(=\frac{\left(8,5+1\right).\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)
B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}<\)
B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
B=\(1-\frac{1}{8}=\frac{8}{8}-\frac{7}{8}=\frac{1}{8}<2\)
Vậy 1/8<2 hay 1/8<16/8
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a) \(A=1.2+2.3+3.4+...+98.97\)
\(\Rightarrow A=1.2+2.3+3.4+...+97.98\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+97.98.3\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+97.98.\left(99-96\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+...+97.98.99-96.97.98\)
\(\Rightarrow3A=97.98.99\)
\(\Rightarrow A=97.98.33\)