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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
Let $[\alpha]$ denote the greatest integer $\leq \alpha$. Then $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots .+[\sqrt{120}]$ is equal to.
  • A
    $824$
  • ✓
    $825$
  • C
    $823$
  • D
    $822$
✓

Answer

Correct option: B.
$825$
b
$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots \ldots .[\sqrt{120}]$

$\Rightarrow 1+1+1+2+2+2+2+2+3+3+\ldots \ldots .+$

$3=7 \text { times }$

$+4+4+\ldots \ldots .+4=9 \text { times }+\ldots \ldots 10+10+$

$\ldots \ldots+10=21 \text { times }$

$\Rightarrow \sum_{r=1}^{10}(2 r+1) . r$

$\Rightarrow 2 \sum_{r=1}^{10} r^2+\sum_{r=1}^{10} r$

$\Rightarrow 2 \times \frac{10 \times 11 \times 21}{6}+\frac{10 \times 11}{2}$

$\Rightarrow 770+55$

$\Rightarrow 825$

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