Question

tính :

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

Answer 1

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

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

\(=\frac{1^2}{1.3}.\frac{3^2}{3.5}...\frac{\left(2n+1\right)^2}{\left(2n+1\right)\left(2n+3\right)}\)

\(=\frac{2n+1}{2n+3}\)

P/S:  tham khảo nha, mk ko chắc là đúng