IIT-JEE Advanced 2022 Maths Paper 2 Problem 18

Problem

For a positive integer n, define

f(n) = n + \( \frac{16 + 5n-3n^2}{4n+3n^2}+\frac{32+n-3n^2}{8n+3n^2} + \frac{48-3n-3n^2}{12n+3n^2} + … + \frac{25n-7n^2}{7n^2} \)

Then, the value of \( \displaystyle \lim_{n \to \infty} \)f(n) is equal to

A 3 + (4/3)loge7
B 4 – (3/4)loge(7/3)
C 4 – (4/3)loge(7/3)
D 3 + (3/4)loge7

Solution

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