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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

$\lim _{n \rightarrow \infty} \frac{1}{n^{k+1}}\left[2^k+4^k+6^k+\ldots+(2 n)^k\right]=$

For $a \in \mathbb{R}$ (the set of all real numbers),$a \neq -1$,if $\lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1}[(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}$,then $a$ is equal to:

$\mathop {Lim}\limits_{n \to \infty } \frac{\pi }{{6n}}\left[ {{{\sec }^2}\left( {\frac{\pi }{{6n}}} \right) + {{\sec }^2}\left( {2 \cdot \frac{\pi }{{6n}}} \right) + \dots + {{\sec }^2}\left( {(n - 1)\frac{\pi }{{6n}}} \right) + \frac{4}{3}} \right]$ has the value equal to

The value of $\lim _{n \rightarrow \infty} \left\{ \frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{2n-1}}{n^{3/2}} \right\}$ is

If $\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=k$,then $\log k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module