Question

Rút gọn: \(A=\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^4+1}+\frac{4.3}{4.3^4+1}+...+\frac{4.k}{4.k^4+1}\)

Answer 1

\(\frac{4k}{4k^4+1}=\frac{4k}{4k^4+4k^2+1-4k^2}=\frac{4k}{\left(2k^2+1\right)^2-\left(2k\right)^2}=\frac{4k}{\left(2k^2+2k+1\right)\left(2k^2-2k+1\right)}=\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1}\)

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

\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)

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