Soái ca 2k6 Làm đi bạn !!
\(\frac{3^{2}+1}{3^{2}-1}+\frac{5^{2}+1}{5^{2}-1}+...+\frac{99^{2}+1}{99^{2}-1}=49+\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}=49+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}=49.49\)
\(\frac{3^2+1}{3^2-1}+\frac{5^2+1}{5^2-1}+\frac{7^2+1}{7^2-1}+...+\frac{99^2+1}{99^2-1}\)
\(=\frac{3^2-1+2}{3^2-1}+\frac{5^2-1+2}{5^2-1}+\frac{7^2-1+2}{7^2-1}+...+\frac{99^2-1+2}{99^2-1}\)
\(=1+\frac{2}{3^2-1}+1+\frac{2}{5^2-1}+1+\frac{2}{7^2-1}+...+1+\frac{2}{99^2-1}\)
\(=\left(1+1+...+1\right)+\left(\frac{2}{\left(3-1\right)\left(3+1\right)}+\frac{2}{\left(5-1\right)\left(5+1\right)}+...+\frac{2}{\left(99-1\right)\left(99+1\right)}\right)\)
\(=49+\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\right)\)
\(=49+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=49+\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=49+\frac{49}{100}\)
\(=\frac{4949}{100}\)
Ta có: \(\frac{3^2+1}{3^2-1}+\frac{5^2+1}{5^2-1}+\frac{7^2+1}{7^2-1}+...+\frac{99^2+1}{99^2-1}\)
\(=\frac{\left(3^2-1\right)+2}{3^2-1}+\frac{\left(5^2-1\right)+2}{5^2-1}+\frac{\left(7^2-1\right)+2}{7^2-1}+...+\frac{\left(99^2-1\right)+2}{99^2-1}\)
\(=\left(1+1+1+...+1\right)\left(\frac{2}{3^2-1}+\frac{2}{5^2-1}+\frac{2}{7^2-1}+...+\frac{2}{99^2-1}\right)\)
\(=49.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\right)\)
\(=49.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=49.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=49.\left(\frac{50-1}{100}\right)\)
\(=\frac{49.49}{100}\)
\(=\frac{2401}{100}=24,01\)
Xl nha . Nãy viết = tex nên viết nhầm đoạn cuối !!
Bạn ơi mình nhầm là dấu nhân
Kết quả phải là: \(\frac{4949}{100}\)
Các bn sửa cho mình nhé
