Question

Let [α] denote the greatest integer ≤ α. Then \([\sqrt1] + [ \sqrt 2 ] + [ \sqrt3 ] + . . . + [ \sqrt120 ] [\sqrt1 ] + [ \sqrt2 ] + [ \sqrt3 ] + . . . + [ 120 ]\)  is equal to

Answer 1

Correct Answer: 825

Let: \[ S = \left\lfloor \sqrt{1} \right\rfloor + \left\lfloor \sqrt{2} \right\rfloor + \left\lfloor \sqrt{3} \right\rfloor + \cdots + \left\lfloor \sqrt{120} \right\rfloor \] We will group terms by the value of \( \left\lfloor \sqrt{x} \right\rfloor \): \[ \left\lfloor \sqrt{1} \right\rfloor = 1, \quad \left\lfloor \sqrt{2} \right\rfloor = 1, \quad \left\lfloor \sqrt{3} \right\rfloor = 1, \quad \left\lfloor \sqrt{4} \right\rfloor = 2, \quad \left\lfloor \sqrt{5} \right\rfloor = 2, \dots \] Thus: \[ S = 1 \times 3 + 2 \times 5 + 3 \times 7 + \cdots + 10 \times 21 \] We can calculate this sum as follows: \[ S = \sum_{r=1}^{10} r(2r + 1) \] The sum becomes: \[ S = 1 \times 3 + 2 \times 5 + 3 \times 7 + \dots + 10 \times 21 \] Solving the summation gives: \[ S = 770 + 55 = 825 \]