Question

Let $a_n$ be the largest integer not exceeding $\sqrt{n}$. Then the value of $a_1 + a_2 + \dots + a_{50}$ is

Answer 1

Correct Answer: 217

We are asked to find the sum of $a_1 + a_2 + \dots + a_{50}$, where $a_n = \lfloor \sqrt{n} \rfloor$.
The value of $a_n$ is the greatest integer less than or equal to $\sqrt{n}$. To find the sum, we can break the sum into intervals where $\lfloor \sqrt{n} \rfloor$ remains constant. The value of $\lfloor \sqrt{n} \rfloor$ will stay constant for values of $n$ within certain intervals.

- For $n = 1$ to 3, $\lfloor \sqrt{n} \rfloor = 1$ (3 terms).
- For $n = 4$ to 8, $\lfloor \sqrt{n} \rfloor = 2$ (5 terms).
- For $n = 9$ to 15, $\lfloor \sqrt{n} \rfloor = 3$ (7 terms).
- For $n = 16$ to 24, $\lfloor \sqrt{n} \rfloor = 4$ (9 terms).
- For $n = 25$ to 35, $\lfloor \sqrt{n} \rfloor = 5$ (11 terms).
- For $n = 36$ to 48, $\lfloor \sqrt{n} \rfloor = 6$ (13 terms).
- For $n = 49$ and 50, $\lfloor \sqrt{n} \rfloor = 7$ (2 terms).

Now, calculate the total sum:

Total sum $= 1 \times 3 + 2 \times 5 + 3 \times 7 + 4 \times 9 + 5 \times 11 + 6 \times 13 + 7 \times 2$
$= 3 + 10 + 21 + 36 + 55 + 78 + 14 = 217$.

Thus, the value of $a_1 + a_2 + \dots + a_{50} = 217$.