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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
The sum $\sum\limits_{k=1}^{20}(1+2+3+\ldots+k)$ is
  • A
    $1496$
  • B
    $1690$
  • ✓
    $1540$
  • D
    $1560$
✓

Answer

Correct option: C.
$1540$
c
$\sum_{\mathrm{k}=1}^{20} \frac{\mathrm{k}(\mathrm{k}+1)}{2}=\frac{1}{2} \sum_{\mathrm{k}=1}^{20} \frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)-(\mathrm{k}-1) \mathrm{k}(\mathrm{k}+1)}{3}$

$=\frac{1}{6} \times 20 \times 21 \times 22=1540$

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