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Greatest Integer Function

Consider the ...

Question

Consider the sequence of number $[n + \sqrt{2 n} + \frac{1}{2}]$ for $n \geq 1$ , where $[x]$ denotes the greatest integer not exceeding $x$ . If the missing integers in the sequence are $n_{1} < n_{2} < n_{3} < \dots \dots$ then find the $n_{12}$ .

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Solution

$S_{n} = n + [\sqrt{2 n} + 0.5], n \geq 1$
where, $[n] =$ Greatest integer less than $n$ .

$\begin{matrix} n & \sqrt{2 n} & [\sqrt{2 n} + 0.5] & n + [\sqrt{2 n} + 0.5] 1 & 1.4 & 1 & 2 2 & 2 & 2 & 4 3 & 2.4 & 2 & 5 4 & 2.8 & 3 & 7 5 & 3.1 & 3 & 8 6 & 3.4 & 3 & 9 7 & 3.7 & 4 & 11 8 & 4 & 4 & 12 9 & 4.2 & 4 & 13 10 & 4.4 & 4 & 14 11 & 4.7 & 5 & 16 12 & 4.9 & 5 & 17 13 & 5.1 & 5 & 18 14 & 5.3 & 5 & 19 15 & 5.4 & 5 & 20 16 & 5.6 & 6 & 22 17 & 5.8 & 6 & 23 18 & 6 & 6 & 24 19 & 6.1 & 6 & 25 20 & 6.3 & 6 & 26 21 & 6.4 & 6 & 27 22 & 6.6 & 7 & 29 \end{matrix}$
By observing pattern,
Missing numbers =
$∴$ 12th number in series $= 78$

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[n + \sqrt{2 n} + \frac{1}{2}]

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l i m x \to n {(- 1)}^{[x]}

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[S_{2012}]

l i m n \to \infty \frac{[1 . x] + [2 . x] + [3 . x] + . . . . . . . + [n . x]}{n^{2}}

[.]

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