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 \lim_{x o \infty} f(n) is equal to [JEE Adv 2022 P2]

Question

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 \lim_{x 	o \infty} f(n) is equal to [JEE Adv 2022 P2]

Examsnet · Accepted Answer

Correct Answer is : 4−‌34loge(‌73)

Limits, Continuity and Differentiability