JEE Mains · Maths · STD 11 - 8. sequence and series
Let \([\alpha]\) denote the greatest integer \(\leq \alpha\). Then \([\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots .+[\sqrt{120}]\) is equal to.
- A \(824\)
- B \(825\)
- C \(823\)
- D \(822\)
Answer & Solution
Correct Answer
(B) \(825\)
Step-by-step Solution
Detailed explanation
\([\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots \ldots .[\sqrt{120}]\) \(\Rightarrow 1+1+1+2+2+2+2+2+3+3+\ldots \ldots .+\) \(3=7 \text { times }\) \(+4+4+\ldots \ldots .+4=9 \text { times }+\ldots \ldots 10+10+\) \(\ldots \ldots+10=21 \text { times }\)…
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