JEE Advanced 2022 — Paper 2 — MATH-III solution

JEE Advanced2022Paper 2MATH-III

For positive integer $n$ , define

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

Then, the value of $n \to \infty lim f (n)$ is equal to

A
$3 + \frac{4}{3} lo g_{e} 7$
B
$4 - \frac{3}{4} lo g_{e} (\frac{7}{3})$
C
$4 - \frac{4}{3} lo g_{e} (\frac{7}{3})$
D
$3 + \frac{3}{4} lo g_{e} 7$

Solution

Each of the $n$ summed terms has the form $\frac{16 r + 5 n - 3 n ^{2}}{4 r n + 3 n ^{2}}$ for $r = 1, 2, \dots, n$ (the absorbed $n$ at the front comes from absorbing the constant term). Add $1$ to each summand to get

$f (n) = r = 1 \sum n \frac{9 n + 16 r}{4 r n + 3 n ^{2}} = \frac{1}{n} r = 1 \sum n \frac{9 + 16 ( r / n )}{4 ( r / n ) + 3}$ .

This is a Riemann sum, converging to $\int_{0}^{1} \frac{9 + 16 x}{4 x + 3} d x$ .

Rewrite the integrand: $\frac{16 x + 9}{4 x + 3} = \frac{4 ( 4 x + 3 ) - 3}{4 x + 3} = 4 - \frac{3}{4 x + 3}$ .

So the integral is $[4 x - \frac{3}{4} ln ∣4 x + 3∣]_{0}^{1} = 4 - \frac{3}{4} (ln 7 - ln 3) = 4 - \frac{3}{4} ln \frac{7}{3}$ .

This matches option (B).

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