For a positive integer n,define f(n) = n + su | vedclass

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(B) Given $f(n) = n + \sum_{r=1}^n \frac{16r + (9-4r)n - 3n^2}{4rn + 3n^2}$.We can rewrite the general term as $\frac{(16r + 9n) - (4rn + 3n^2)}{4rn +

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$4 - \frac{3}{4} \log_e \left(\frac{7}{3}\right)$

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(B) Given $f(n) = n + \sum_{r=1}^n \frac{16r + (9-4r)n - 3n^2}{4rn + 3n^2}$.We can rewrite the general term as $\frac{(16r + 9n) - (4rn + 3n^2)}{4rn + 3n^2} = \frac{16r + 9n}{4rn + 3n^2} - 1$.Thus,$f(n) = n + \sum_{r=1}^n \left( \frac{16r + 9n}{4rn + 3n^2} - 1 \right) = n + \sum_{r=1}^n \frac{16r + 9n}{4rn + 3n^2} - n = \sum_{r=1}^n \frac{16r + 9n}{4rn + 3n^2}$.Dividing numerator and denominator by $n^2$,we get $\sum_{r=1}^n \frac{16(r/n) + 9}{4(r/n) + 3} \cdot \frac{1}{n}$.Taking the limit as $n \rightarrow \infty$,this becomes the definite integral $\int_0^1 \frac{16x + 9}{4x + 3} dx$.We can rewrite the integrand as $\frac{4(4x + 3) - 3}{4x + 3} = 4 - \frac{3}{4x + 3}$.Integrating,we get $\int_0^1 (4 - \frac{3}{4x + 3}) dx = [4x - \frac{3}{4} \ln|4x + 3|]_0^1$.Evaluating at the limits: $(4 - \frac{3}{4} \ln 7) - (0 - \frac{3}{4} \ln 3) = 4 - \frac{3}{4} \ln \left(\frac{7}{3}\right)$.

For a positive integer $n$,define $f(n) = n + \sum_{r=1}^n \frac{16r + (9-4r)n - 3n^2}{4rn + 3n^2}$. Then,the value of $\lim_{n \rightarrow \infty} f(n)$ is equal to

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