QJEEAdv22Ma694c: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Ma694c Solution: For positive integer , define Then, the value of is equal to :

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

\begin{aligned}&amp; 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}} \\\\&amp; =\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) \\\\&amp; 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}} \\\\&amp; =\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} \\\\&amp; \lim _{n \rightarrow \infty} f(n)=\int_{0}^{1} \frac{9+16 x}{4 x+3} d x \\\\&amp; =\int_{0}^{1} \frac{(16 x+12)-3}{4 x+3} d x \\\\&amp; =\left.\left(4 x-\frac{3}{4} \ln |4 x+3|\right)\right|_{0} ^{1} \\\\&amp; =4-\frac{3}{4} \ln \frac{7}{3}\end{aligned}

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