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\(Taco\):
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right).......................\left(1-\frac{1}{1+2+3+.............+2018}\right)\)
\(A=\left(\frac{1+2}{1+2}-\frac{1}{1+2}\right).............\left(\frac{1+2+3+......+2018}{1+2+3+.......+2018}-\frac{1}{1+2+3+......+2018}\right)\)
\(A=\left(\frac{2}{1+2}\right)...........\left(\frac{2+3+.......+2018}{1+2+3+......+2018}\right)\)
\(\Rightarrow A+2017.\left(\frac{1}{3}\right).....\frac{2+3+.....+2018}{1+2+3+...+2018}=1.1.1......1=1\)
\(.................................\)
Xét số hạng tổng quát:
\(k^4+\frac{1}{4}=\left(k^4+2\cdot\frac{1}{2}\cdot k^2+\frac{1}{4}\right)-k^2\)=\(\left(k^2+\frac{1}{2}\right)^2-k^2\)
= \(\left(k^2+\frac{1}{2}-k\right)\left(k^2+\frac{1}{2}+k\right)\)
Thay k từ 1 đến 12 ta được:
A=\(\frac{\frac{1}{2}\cdot\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(110+\frac{1}{2}\right)\left(132+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)...\left(132+\frac{1}{2}\right)\left(152+\frac{1}{2}\right)}\)=\(\frac{\frac{1}{2}}{152+\frac{1}{2}}=\frac{1}{305}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
=\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
= \(\frac{1}{x}\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)\left(-\frac{3}{4}\right)...\left(-\frac{2017}{2018}\right)\)
Tích trên là tích của các thừa số âm và có (2018-2)+1=2017 thừa số nên có kq âm
\(=-\frac{1}{2018}\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{2018}-1\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2016}{2017}.\frac{2017}{2018}\)
\(=\frac{1}{2018}\)