$\displaystyle\lim_{n\to\infty}\dfrac{3}{n}\left\{4+\left(2+\dfrac{1}{n}\right)^2+\left(2+\dfrac{2}{n}\right)^2+\cdots+\left(3-\dfrac{1}{n}\right)^2\right\}$ is equal to:
Step-by-Step Solution
Key Concept: Recognise as Riemann sum: $\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}3\left(2+\frac{k}{n}\right)^2=\int_2^3 3x^2\,dx\cdot1... $ Actually limit $=\int_2^3 3\cdot x^2\,dx$ — wait the $3/n$ factor and the sum from $k=0$ to $n-1$ of $(2+k/n)^2$ gives $3\int_2^3 x^2\,dx$? No: $(3/n)\sum_{k=0}^{n-1}(2+k/n)^2\to 3\int_0^1(2+t)^2dt$... hmm. Actually the step is $1/n$ and multiplied by 3: $=3\int_0^1(2+t)^2dt=3[x^3/3]_2^3=27-8=19$.
$19$.
Correct Answer: 4