[x] is the greatest integer less than or equal to x. What is

Evaluating the range of roots for

$$x=\ sqrt {1}-------------\ sqrt {50}$$
$$For\ x\ between\ 1\ and 3\ \left[\sqrt{x}\right]=1$$
$$For\ x\ between\ 4\ and\ 8\ \left[\sqrt{x}\right]=2$$
$$For\ x\ between\ 9\ and\ 15\ \left[\sqrt{x}\right]=3$$
$$For\ x\ between\ 16\ and\ 24\ \left[\sqrt{x}\right]=4$$
$$For\ x\ between\ 25\ and\ 35\ \left[\sqrt{x}\right]=5$$
$$For\ x\ between\ 36\ and\ 48\ \left[\sqrt{x}\right]=6$$
$$For\ x\ between\ 49\ and\ \ 50\ \left[\sqrt{x}\right]=7$$

Sum = [3*1] + [5*2] + [7*3] + [9*4] + [11*5] + [13*6] + [2*7]
$$=3+10+21+36+55+78+14\ =217$$

$$answer\ is\ Option\ C

=>

[√1] = 1
[√2] = 1
[√3] = 1
If 1 ≤ n < 4, then 1 ≤ √n < 2 and [√n] = 1 for n = 1, 2, 3 (3 terms).
If 4 ≤ n < 9, then 2 ≤ √n < 3 and [√n] = 2 for n = 4, 5, …, 8 (5(=8-4+1) terms).
If 9 ≤ n < 16, then 3 ≤ √n < 4 and [√n] = 3 for n = 9, 10, …, 15 (7 terms).
If 16 ≤ n < 25, then 4 ≤ √n < 5 and [√n] = 4 for n = 16, 17, …, 24 (9 terms).
If 25 ≤ n < 36, then 5 ≤ √n < 6 and [√n] = 5 for n = 25, 26, …, 35 (11 terms).
If 36 ≤ n < 49, then 6 ≤ √n < 7 and [√n] = 6 for n = 36, 37, …, 48 (13 terms).
If 49 ≤ n < 64, then 7 ≤ √n < 8 and [√n] = 7 for n = 49 and 50 (2 terms).
Therefore,
[√1] + [√2] + [√3] + … + [√49] + [√50]
= ( 1 + 1 + 1 ) + ( 2 + 2 + … + 2 ) + ( 3 + 3 + … + 3 ) + ( 4 + 4 + … + 4 ) + ( 5 + 5 + … + 5 ) + ( 6 + 6 + … + 6 ) + ( 7 + 7 )
= 1*3 + 2*5 + 3*7 + 4*9 + 5*11 + 6*13 + 7*2
= 3 + 10 + 21 + 36 + 55 + 78 + 14
= 217

Therefore, the answer is C.
Answer: C