Question

For any natural Number 'n', let a_n be the largest number not exceeding \(\sqrt{n}\) , then a1 + a2 + a3... +a50 =

Answer 1

Correct Answer: 217

To solve the problem, we need to find the sum a₁ + a₂ + ... + a₅₀, where a_n is the largest integer not exceeding √n. This means a_n = ⌊√n⌋, the floor function of √n. We calculate ⌊√n⌋ for n = 1 to 50:

n	√n	⌊√n⌋
1	1.000	1
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

Let's sum the values from a₁ to a₅₀:

a₁ = 1, a₂ = 1, a₃ = 1 (3 times, so sum = 3)

a₄ = 2, a₅ = 2, a₆ = 2, a₇ = 2, a₈ = 2 (5 times, sum = 10)

a₉ = 3, ... a₁₅ = 3 (7 times, sum = 21)

a₁₆ = 4, ... a₂₄ = 4 (9 times, sum = 36)

a₂₅ = 5, ... a₃₅ = 5 (11 times, sum = 55)

a₃₆ = 6, ... a₄₈ = 6 (13 times, sum = 78)

a₄₉ = 7, a₅₀ = 7 (2 times, sum = 14)

Total = 3 + 10 + 21 + 36 + 55 + 78 + 14 = 217

The sum, 217, falls within the range 217—217, confirming its correctness.

Thus, the value of a₁ + a₂ + ... + a₅₀ is 217.