Discussion

Question:

If [x] is the greater integer less than or equal to ‘x’, then find the value of the following series

[ $\sqrt{1}$ ] + [ $\sqrt{2}$ ] + [ $\sqrt{3}$ ] + [ $\sqrt{4}$ ] + ... + [ $\sqrt{361}$ ]

Explanation:

[ $\sqrt{1}$ ] = 1

[ $\sqrt{2}$ ] = 1

[ $\sqrt{3}$ ] = 1

[ $\sqrt{4}$ ] = 1

[ $\sqrt{5}$ ] = 1

[ $\sqrt{8}$ ] = 1

[ $\sqrt{9}$ ] = 1

and so on.

Thus, [ $\sqrt{n}$ ] = k

where k² is the greatest perfect square less than or equal to n.

Also, the difference between two consecutive perfect squares = (k + 1)² – k² = 2k + 1

∴ The required sum is

$[\sum_{k = 1}^{18} k (2 k + 1)]$ + $\sqrt{361}$

= $[\sum_{k = 1}^{18} 2 k^{2} + k]$ + 19

= 2 $\frac{(18) (18 + 1) (2 \times 18 + 1)}{6}$ + $\frac{18 (18 + 1)}{2}$ + 9

= 4408

Hence, option (a).

Note: Correct answer option was not present in actual paper.

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