Sum of Squares of First $$n$$ Natural Numbers

Find $$[\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + \ldots + [\sqrt{120}]$$ where $$[x]$$ is the greatest integer function.

For $$n^2 \leq k < (n+1)^2$$, $$[\sqrt{k}] = n$$.

Count of integers k with $$[\sqrt{k}] = n$$: from $$n^2$$ to $$(n+1)^2 - 1$$, that's $$2n + 1$$ integers.

For $$n = 1$$: $$k = 1, 2, 3$$ (3 values), contribution = $$1 \times 3 = 3$$

For $$n = 2$$: $$k = 4, ..., 8$$ (5 values), contribution = $$2 \times 5 = 10$$

For $$n = 3$$: $$k = 9, ..., 15$$ (7 values), contribution = $$3 \times 7 = 21$$

For $$n = 4$$: $$k = 16, ..., 24$$ (9 values), contribution = $$4 \times 9 = 36$$

For $$n = 5$$: $$k = 25, ..., 35$$ (11 values), contribution = $$5 \times 11 = 55$$

For $$n = 6$$: $$k = 36, ..., 48$$ (13 values), contribution = $$6 \times 13 = 78$$

For $$n = 7$$: $$k = 49, ..., 63$$ (15 values), contribution = $$7 \times 15 = 105$$

For $$n = 8$$: $$k = 64, ..., 80$$ (17 values), contribution = $$8 \times 17 = 136$$

For $$n = 9$$: $$k = 81, ..., 99$$ (19 values), contribution = $$9 \times 19 = 171$$

For $$n = 10$$: $$k = 100, ..., 120$$ (21 values), contribution = $$10 \times 21 = 210$$

Total = $$3 + 10 + 21 + 36 + 55 + 78 + 105 + 136 + 171 + 210 = 825$$

The answer is $$\boxed{825}$$.

The given series up to 40 terms is:

$$1 + 3 + 5^2 + 7 + 9^2 + 11 + 13^2 + \dots$$

We can split this series into two separate progressions consisting of 20 terms each based on the odd and even positions.

The terms at even positions (2nd, 4th, 6th, ..., 40th terms) form an Arithmetic Progression:

$$S_{\text{even}} = 3 + 7 + 11 + \dots \text{ up to 20 terms}$$

The general term for this AP is $$T_{2k} = 4k - 1$$.

The sum is given by:

$$S_{\text{even}} = \sum_{k=1}^{20} (4k - 1) = 4\left(\frac{20 \times 21}{2}\right) - 20 = 840 - 20 = 820$$

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The terms at odd positions (1st, 3rd, 5th, ..., 39th terms) form a sequence of squared odd numbers (where $$1 = 1^2$$):

$$S_{\text{odd}} = 1^2 + 5^2 + 9^2 + 13^2 + \dots \text{ up to 20 terms}$$

The general term for this sequence is $$T_{2k-1} = (4k - 3)^2 = 16k^2 - 24k + 9$$.

The sum is given by:

$$S_{\text{odd}} = \sum_{k=1}^{20} (16k^2 - 24k + 9) = 16\sum_{k=1}^{20} k^2 - 24\sum_{k=1}^{20} k + \sum_{k=1}^{20} 9$$

Using the standard summation formulas:

$$\sum_{k=1}^{20} k^2 = \frac{20 \times 21 \times 41}{6} = 2870$$

$$\sum_{k=1}^{20} k = \frac{20 \times 21}{2} = 210$$

Substituting these values into the expression for $$S_{\text{odd}}$$:

$$S_{\text{odd}} = 16(2870) - 24(210) + 9(20) = 45920 - 5040 + 180 = 41060$$

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Combining both sums to find the total value of the series:

$$\text{Total Sum} = S_{\text{even}} + S_{\text{odd}} = 820 + 41060 = 41880$$

Therefore, the final sum is equal to 41880.

The given series is $$1,\,3,\,11,\,25,\,45,\,71,\ldots$$

First-order differences:
$$3-1 = 2,\; 11-3 = 8,\; 25-11 = 14,\; 45-25 = 20,\; 71-45 = 26,\ldots$$

Second-order differences:
$$8-2 = 6,\; 14-8 = 6,\; 20-14 = 6,\; 26-20 = 6,\ldots$$

The second difference is constant $$6$$, so the $$n^{\text{th}}$$ term $$a_n$$ is a quadratic expression:
$$a_n = An^{2}+Bn+C$$

For a quadratic sequence, the constant second difference equals $$2A$$, hence
$$2A = 6 \;\Longrightarrow\; A = 3$$

Thus $$a_n = 3n^{2}+Bn+C$$. Use the first three terms to find $$B,C$$.

Case 1: $$n = 1$$ gives $$3(1)^{2}+B(1)+C = 1 \;$$ ⇒ $$\;3 + B + C = 1$$ $$-(1)$$

Case 2: $$n = 2$$ gives $$3(2)^{2}+B(2)+C = 3 \;$$ ⇒ $$\;12 + 2B + C = 3$$ $$-(2)$$

Subtract $$(1)$$ from $$(2)$$:
$$(12 + 2B + C) - (3 + B + C) = 3 - 1$$
$$9 + B = 2 \;\Longrightarrow\; B = -7$$

Put $$B = -7$$ in $$(1)$$:
$$3 - 7 + C = 1 \;\Longrightarrow\; C = 5$$

Therefore the general term is
$$a_n = 3n^{2} - 7n + 5$$

We need the sum of the first $$20$$ terms:
$$S_{20} = \sum_{n=1}^{20} a_n = \sum_{n=1}^{20} \left(3n^{2} - 7n + 5\right)$$

Separate the sums:
$$S_{20} = 3\sum_{n=1}^{20} n^{2} \;-\; 7\sum_{n=1}^{20} n \;+\; 5\sum_{n=1}^{20} 1$$

Standard formulas:
$$\sum_{n=1}^{N} n = \frac{N(N+1)}{2}, \qquad \sum_{n=1}^{N} n^{2} = \frac{N(N+1)(2N+1)}{6}$$

For $$N = 20$$:
$$\sum n = \frac{20\cdot21}{2} = 210$$
$$\sum n^{2} = \frac{20\cdot21\cdot41}{6} = \frac{17220}{6} = 2870$$
$$\sum 1 = 20$$

Substitute these into $$S_{20}$$:
$$S_{20} = 3(2870) - 7(210) + 5(20)$$

Compute each term:
$$3(2870) = 8610$$
$$7(210) = 1470$$
$$5(20) = 100$$

Hence
$$S_{20} = 8610 - 1470 + 100 = 7240$$

Therefore, the required sum is $$7240$$.
Option A is correct.