The value of _ n n 1 3n^2+8n+4 + 1 3n^2+16n+16 + ( terms) is equal to

MATHEMATICS · Single correct · AP · Calculus AB · Integration and Accumulation of Change · Integration and Accumulation of Change

Problem

The value of

\lim_{n\to\infty}n\left\{\frac{1}{3n^2+8n+4}+\frac{1}{3n^2+16n+16}+\dots\text{($n$ terms)}\right\}

is equal to (1)

\frac{1}{2}\log\left(\frac{9}{5}\right)

(2)

\frac{1}{3}\log\left(\frac{9}{5}\right)

(3)

\frac{1}{4}\log\left(\frac{9}{5}\right)

(4) None of these