CAT 2024 Slot 1 · Quant · Question 18

{a}_{n} equals k exactly when k\le \sqrt{n}<k+1, that is when {k}^{2}\le n<{\left( k+1 \right)}^{2}.

So the value k runs from n={k}^{2} up to n={\left( k+1 \right)}^{2}-1, a block of {\left( k+1 \right)}^{2}-{k}^{2}=2k+1 terms.

The block widths are therefore the odd numbers 3, 5, 7, 9, 11, 13 for k=1 to 6.

Those blocks contribute 3\left( 1 \right)+5\left( 2 \right)+7\left( 3 \right)+9\left( 4 \right)+11\left( 5 \right)+13\left( 6 \right).

That is 3+10+21+36+55+78=203, covering n up to 48.

The last block would start at n=49, but the sum stops at 50, so only n=49 and n=50 are left, each giving {a}_{n}=7.

They add 2\times 7=14.

The total is 203+14=217.

Shortcut. Group by value rather than by term. Each value k appears exactly 2k+1 times, so the sum is ∑k\left( 2k+1 \right) over the complete blocks, plus a short tail — six products and one correction instead of fifty square roots.