The distance covered in the nth second is proportional to:

n^{2}

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n

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2 n - 1

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2 n^{2} - 1

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Solution

The correct option is C $n^{2}$
Newton’s Second Equation
$h = u t + \frac{1}{2} (a t^{2})$
for free fall $u = 0$ $a = g$ and $t = n$ $g$ is gravity $t$ is time $u$ is initial velocity
$∴ h = 0 + \frac{1}{2} (g n^{2})$
h is directly proportional to $n^{2}$

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{lim}_{n \to \infty} (\frac{n}{n^{2}} + \frac{n}{n^{2} + 1} + \frac{n}{n^{2} + 2^{2}} + . . . . . . + \frac{n}{2 n^{2} - 2 n + 1})

The coefficient of $x^{n}$ in $\frac{(1 + x)^{2}}{(1 - x)^{3}}$ is :

lim n \to \infty [\frac{2 n}{2 n^{2} - 1} cos \frac{n + 1}{2 n - 1} - \frac{n}{1 - 2 n} \cdot \frac{n {(- 1)}^{n}}{n^{2} + 1}]

lim n \to \infty [\frac{2 n}{2 n^{2} - 1} cos \frac{n + 1}{2 n - 1} - \frac{n}{1 - 2 n} . \frac{n (- 1)^{n}}{n^{2} + 1}]

n^{th}

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