Question

Tính tổng:

S=\(\sqrt{1+\frac{8.1^2-1}{1^2.3^2}}++\sqrt{1+\frac{8.2^2-1}{3^2.5^2}}++\sqrt{1+\frac{8.3^2-1}{5^2.7^2}}+...++\sqrt{1+\frac{8.1009^2-1}{2017^2.2019^2}}\)

Answer 1

\(\sqrt{1+\frac{8n^2-1}{\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{1+\frac{8n^2-1}{\left(4n^2-1\right)^2}}=\sqrt{\frac{\left(4n^2-1\right)^2+8n^2-1}{\left(4n^2-1\right)^2}}\)

\(=\sqrt{\frac{16n^4-8n^2+1+8n^2-1}{\left(4n^2-1\right)^2}}=\frac{4n^2}{4n^2-1}=1+\frac{1}{4n^2-1}=1+\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)

\(\Rightarrow S=1009+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=1009+\frac{1}{2}\left(1-\frac{1}{2019}\right)=...\)