JEE Advance - Mathematics (2022 - Paper 2 Online - No. 18)
For positive integer $n$, define
$$ f(n)=n+\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+\frac{48-3 n-3 n^{2}}{12 n+3 n^{2}}+\cdots+\frac{25 n-7 n^{2}}{7 n^{2}} . $$
Then, the value of $$\mathop {\lim }\limits_{n \to \infty } f\left( n \right)$$ is equal to :
$$ f(n)=n+\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+\frac{48-3 n-3 n^{2}}{12 n+3 n^{2}}+\cdots+\frac{25 n-7 n^{2}}{7 n^{2}} . $$
Then, the value of $$\mathop {\lim }\limits_{n \to \infty } f\left( n \right)$$ is equal to :
$3+\frac{4}{3} \log _{e} 7$
$4-\frac{3}{4} \log _{e}\left(\frac{7}{3}\right)$
$4-\frac{4}{3} \log _{e}\left(\frac{7}{3}\right)$
$3+\frac{3}{4} \log _{e} 7$
Explanation
$$
\begin{aligned}
& f(n)=n+\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+\ldots .+\frac{25 n-7 n^{2}}{7 n^{2}} \\\\
& =\left(\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+1\right)+\left(\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+1\right)+\ldots \ldots+\left(\frac{25 n-7 n^{2}}{7 n^{2}}+1\right) \\\\
& f(n)=\frac{9 n+16}{4 n+3 n^{2}}+\frac{9 n+32}{8 n+3 n^{2}}+\ldots . .+\frac{25 n}{7 n^{2}} \\\\
& =\sum_{r=1}^{n} \frac{9 n+16 r}{4 r n+3 n^{2}}=\frac{1}{n} \sum_{r=1}^{n} \frac{9+16\left(\frac{r}{n}\right)}{4\left(\frac{r}{n}\right)+3} \\\\
& \lim _{n \rightarrow \infty} f(n)=\int_{0}^{1} \frac{9+16 x}{4 x+3} d x \\\\
& =\int_{0}^{1} \frac{(16 x+12)-3}{4 x+3} d x \\\\
& =\left.\left(4 x-\frac{3}{4} \ln |4 x+3|\right)\right|_{0} ^{1} \\\\
& =4-\frac{3}{4} \ln \frac{7}{3}
\end{aligned}
$$
Comments (0)
