JEE Advanced 2022 · Paper 2 · Mathematics Q18

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MathematicsSingle correct+3 / -1Question 18 of 54

For positive integer $n$ , define
$f(n) = n + \frac{16 + 5n - 3n^2}{4n + 3n^2} + \frac{32 + n - 3n^2}{8n + 3n^2} + \frac{48 - 3n - 3n^2}{12n + 3n^2} + \cdots + \frac{25n - 7n^2}{7n^2}.$
Then, the value of $\displaystyle\lim_{n\to\infty} f(n)$ is equal to

3 + \frac43\log_e 7

4 - \frac34\log_e\left(\frac73\right)

4 - \frac43\log_e\left(\frac73\right)

3 + \frac34\log_e 7

Answer and worked solution

Official answer: B

General term. The $k$ -th fraction is
$\frac{16k+(9-4k)n-3n^2}{4kn+3n^2},\qquad k=1,\dots,n.$
At $k=n$ this is $\frac{25n-7n^2}{7n^2}$ , the last term.

Simplify. The numerator is $-(4kn+3n^2)+16k+9n$ , so each fraction equals $-1+\dfrac{16k+9n}{n(4k+3n)}$ . The $n$ terms of $-1$ cancel the leading $n$ :
$f(n)=\frac1n\sum_{k=1}^n\frac{16\frac kn+9}{4\frac kn+3}.$

Riemann sum.
$\lim f(n)=\int_0^1\frac{16x+9}{4x+3}\,dx=\int_0^1\left(4-\frac3{4x+3}\right)dx=4-\frac34\log_e\frac73.$
(B).

Topics:Limit of a sum Limits

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JEE Advanced 2022 · Paper 2, Mathematics question 18