For any natural number n, let a_n be the largest integer not exceeding √n. Then the value of a_1+a_2+ .+a_50 is

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For any natural number $n$ , let $a_{n}$ be the largest integer not exceeding $\sqrt{n}$ . Then the value of $a_{1}+a_{2}+\ldots .+a_{50}$ is

Entered answer:

✅ Correct Answer: 217

Find the sum of $a_1 + a_2 + ... + a_{50}$ where $a_n$ is the largest integer not exceeding $\sqrt{n}$ .

This is also called the "floor function" of $\sqrt{n}$ , written as $\lfloor\sqrt{n}\rfloor$ . It simply means we take the square root of $n$ and round it down to the nearest whole number.

n	√n	⌊√n⌋
1	1	1
2	1.41	1
3	1.73	1
4	2	2
5	2.24	2
6	2.45	2
7	2.65	2
8	2.83	2
9	3	3
10	3.16	3

The bolded values (1, 4, 9, 16) represent perfect squares where the square root is a whole number, so √n = ⌊√n⌋ for these values.

Instead of calculating each term individually, let's group terms that have the same value.

The value $k$ appears for all $n$ where $k ≤ \sqrt{n} < k+1$

This gives us $k^2 ≤ n < (k+1)^2$

This means value $k$ appears from $n = k^2$ to $n = (k+1)^2 - 1$ .

Value (k)	Range	Count	Contribution
1	n = 1 to 3	3	1 × 3 = 3
2	n = 4 to 8	5	2 × 5 = 10
3	n = 9 to 15	7	3 × 7 = 21
4	n = 16 to 24	9	4 × 9 = 36
5	n = 25 to 35	11	5 × 11 = 55
6	n = 36 to 48	13	6 × 13 = 78
7	n = 49 to 50	2	7 × 2 = 14

$a_1 + a_2 + ... + a_{50} = 3 + 10 + 21 + 36 + 55 + 78 + 14 = 217$

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