\(B=\dfrac{1}{2\left(n-1\right)^2+3}\)
Ta có: \(2\left(n-1\right)^2\ge0\forall n\)
\(\Rightarrow2\left(n-1\right)^2+3\ge3\forall n\)
\(\Rightarrow\dfrac{1}{2\left(n-1\right)^2+3}\le\dfrac{1}{3}\forall n\)
Dấu "=" xảy ra \(\Leftrightarrow n=1\)
Vậy MAX \(B=\dfrac{1}{3}\Leftrightarrow n=1\)