3/4x8/9x15/16x24/25x....x(n^2-1)/n^2)

=(1x2x3x4x...x(n-1))x(3x4x5x...x(n+1)):(1x2x3x4x...x n)^2

=(n+1)/2n)

Ta có công thức :

\(1-\frac{1}{k^2}=\frac{k^2-1^2}{k^2}=\frac{\left(k+1\right)\left(k-1\right)}{k^2}\)

Áp dụng công thức trên ta được :

\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)

\(=\frac{2^2-1^2}{2^2}.\frac{3^2-1^2}{3^2}.\frac{4^2-1^2}{4^2}....\frac{n^2-1^2}{n^2}\)

\(=\frac{\left(2+1\right)\left(2-1\right)}{2.2}.\frac{\left(3+1\right)\left(3-1\right)}{3.3}.\frac{\left(4+1\right)\left(4-1\right)}{4.4}...\frac{\left(n+1\right)\left(n-1\right)}{n.n}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{\left(n+1\right)\left(n-1\right)}{n.n}\)

\(=\frac{\left[1.2.3.....\left(n+1\right)\right].\left[3.4.5...\left(n-1\right)\right]}{\left(2.3.4....n\right)\left(2.3.4....n\right)}\)

\(=\left(n+1\right).\frac{1}{2n}=\frac{n+1}{2n}\)

\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)

\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)

\(=\frac{1}{4}\)

1) Đặt \(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)

\(\Rightarrow3D=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3D-D=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)

\(\Leftrightarrow2D=1-\frac{1}{3^{100}}\)

\(\Leftrightarrow D=\frac{3^{100}-1}{2\cdot3^{100}}\)

Vậy \(D=\frac{3^{100}-1}{2\cdot3^{100}}\)

2) Ta có: \(\frac{49}{58}\cdot\frac{2^5}{4^2}-\frac{7^2}{-58}\cdot3\)

\(=\frac{49}{58}\cdot2-\frac{49}{58}\cdot3\)

\(=-1\cdot\frac{49}{58}\)

\(=-\frac{49}{58}\)