MathematicsEasy210×since 2002Q3296limn→∞3n{4+(2+1n)2+(2+2n)2+…+(3−1n)2}\lim\limits_{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}n→∞limn3{4+(2+n1)2+(2+n2)2+…+(3−n1)2} is equal to :A0B193\frac{19}{3}319C19D12Check answerSkip