We need to calculate the sum of an from n = 1 to 50, where an is the largest integer less than or equal to √n. This is equivalent to finding an = ⌊√n⌋ for each value of n from 1 to 50.
| n | √n | ⌊√n⌋ |
|---|---|---|
| 1 | 1.000 | 1Nbsp; |
| 2 | 1.414 | 1 |
| 3 | 1.732 | 1 |
| 4 | 2.000 | 2 |
| 5 | 2.236 | 2 |
| 6 | 2.449 | 2 |
| 7 | 2.646 | 2 |
| 8 | 2.828 | 2 |
| 9 | 3.000 | 3 |
| 10 | 3.162 | 3 |
| 11 | 3.317 | 3 |
| 12 | 3.464 | 3 |
| 13 | 3.606 | 3 |
| 14 | 3.742 | 3 |
| 15 | 3.873 | 3 |
| 16 | 4.000 | 4 |
| 17 | 4.123 | 4 |
| 18 | 4.243 | 4 |
| 19 | 4.359 | 4 |
| 20 | 4.472 | 4 |
| 21 | 4.583 | 4 |
| 22 | 4.690 | 4 |
| 23 | 4.796 | 4 |
| 24 | 4.899 | 4 |
| 25 | 5.000 | 5 |
| 26 | 5.099 | 5 |
| 27 | 5.196 | 5 |
| 28 | 5.291 | 5 |
| 29 | 5.385 | 5 |
| 30 | 5.477 | 5 |
| 31 | 5.568 | 5 |
| 32 | 5.657 | 5 |
| 33 | 5.745 | 5 |
| 34 | 5.831 | 5 |
| 35 | 5.916 | 5 |
| 36 | 6.000 | 6 |
| 37 | 6.083 | 6 |
| 38 | 6.164 | 6 |
| 39 | 6.245 | 6 |
| 40 | 6.325 | 6 |
| 41 | 6.403 | 6 |
| 42 | 6.481 | 6 |
| 43 | 6.557 | 6 |
| 44 | 6.633 | 6 |
| 45 | 6.708 | 6 |
| 46 | 6.782 | 6 |
| 47 | 6.856 | 6 |
| 48 | 6.928 | 6 |
| 49 | 7.000 | 7 |
| 50 | 7.071 | 7 |
Now, let's sum the values of an from a1 to a50 by grouping them:
The value 1 appears for n = 1, 2, 3 (3 times), contributing 3 × 1 = 3 to the sum.
The value 2 appears for n = 4 through 8 (5 times), contributing 5 × 2 = 10 to the sum.
The value 3 appears for n = 9 through 15 (7 times), contributing 7 × 3 = 21 to the sum.
The value 4 appears for n = 16 through 24 (9 times), contributing 9 × 4 = 36 to the sum.
The value 5 appears for n = 25 through 35 (11 times), contributing 11 × 5 = 55 to the sum.
The value 6 appears for n = 36 through 48 (13 times), contributing 13 × 6 = 78 to the sum.
The value 7 appears for n = 49 and 50 (2 times), contributing 2 × 7 = 14 to the sum.
The total sum is the sum of these contributions: 3 + 10 + 21 + 36 + 55 + 78 + 14 = 217.
Thus, the value of a1 + a2 + ... + a50 is 217.