Sequences & Series
Sum of Floor of Square Root
nta_pyq_2023_apr
Grade 11
Question:
$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\cdots+[\sqrt{120}]$ is equal to
Step-by-Step Solution
Key Concept: $[\sqrt{k}]=m$ for $m^2\leq k<(m+1)^2$, i.e., $2m+1$ values per $m$. Except the last group (from 100 to 120 gives 21 terms with value 10).
$825$.
Correct Answer: 825