A small block slides without friction down an inclined plane starting from rest. Let S_n be the dis†an ce travelled from t=n-1 to t=n. Then, S_n/S_{n+1} is (1) 2n-1/2n (2) 2n+1/2n-1 (3) 2n-1/2n+1 (4) 2n/2n+1

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S_{n} = \frac{1}{2 n} + \frac{1}{\sqrt{4 n^{2} - 1}} + \frac{1}{\sqrt{4 n^{2} - 4}} + . . . . . . . . + \frac{1}{\sqrt{3 n^{2} + 2 n - 1}}, n \in N

{lim}_{n \to \infty} S_{n} = α

S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}

t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}},

\frac{t_{n}}{S_{n}}

S_{n}

t = (n - 1)

t = n

\frac{S_{n}}{S_{n + 1}}

n

S_{n}

t = (n - 1) t o t = n

\frac{S_{n}}{S_{n + 1}}

S_{n} = [\frac{1}{1 + \sqrt{n}} + \frac{1}{2 + \sqrt{2 n}} + \dots + \frac{1}{n + \sqrt{n^{2}}}]

lim n \to \infty S_{n} =

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