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\(\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{10}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot\left[\frac{1}{100}-\frac{1}{100}\right]\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)\(=\left[\frac{1}{100}-\left(\frac{1}{1}\right)^2\right]\cdot\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right]\cdot...\cdot0\cdot...\cdot\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)=0
\(\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).\left(\frac{1}{100}-\left(\frac{1}{2}\right)^2\right)......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)....\left(\frac{1}{100}-\left(\frac{1}{10}\right)^2\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right)...\left(\frac{1}{100}-\frac{1}{100}\right)...\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=\left(\frac{1}{100}-\left(\frac{1}{1}\right)^2\right).....0......\left(\frac{1}{100}-\left(\frac{1}{20}\right)^2\right)\)
\(=0\)
\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+.....+\left(1+2+3+4+......+100\right)}{\left(1.100+2.99+3.98+.......+99.2+100.1\right).2013}\)
\(=\frac{1.100+2.99+3.98+......+99.2+100.1}{\left(1.100+2.99+3.98+.....+99.2+100.1\right).2013}\)
\(=\frac{1}{2013}\)
A=(2/3+3/4+...+99/100)x(1/2+2/3+3/4+...+98/99)-(1/2+2/3+...+99/100)x(2/3+3/4+4/5+...98/99)
ta cho nó dài hơn như sau
A=(2/3+3/4+4/5+5/6+....+98/99+99/100)
ta thấy các mẫu số và tử số giống nhau nên chệt tiêu các số
2:3:4:5...99 vậy ta còn các số 2/100
ta làm vậy với(1/2+2/3+3/4+.....+98/99) thi con 1/99
làm vậy với câu (1/2+2/3+...+99/100) thì ra la 1/100
vậy với (2/3+3/4+...+98/99) ra 2/99
xùy ra ta có 2/100.1/99-1/100.2/99=1/50x1/99-1/100x2/99=tự tinh nhe mình ngủ đây
ta có: n2 - 1 = (n2 - n) + (n -1) = n(n-1) + (n-1) = (n-1).(n+1) ; n \(\in\) N
Áp dụng công thức tổng quát trên ta có:
A = (\(\dfrac{1}{2^2}\) - 1).(\(\dfrac{1}{3^2}\) - 1)...(\(\dfrac{1}{100^2}\) - 1)
A = \(\dfrac{2^2-1}{-2^2}\). \(\dfrac{3^2-1}{-3^2}\)......\(\dfrac{100^2-1}{-100^2}\)
A = \(\dfrac{\left(2-1\right)\left(2+1\right)}{-2^2}\).\(\dfrac{\left(3-1\right).\left(3+1\right)}{-3^2}\).....\(\dfrac{\left(100-1\right).\left(100+1\right)}{-100^2}\)
A = - \(\dfrac{1.3.2.4.3.5.......99.101}{2^2.3^2.4^2...100^2}\)
A = - \(\dfrac{101}{200}\)