JEE Mains · Maths · STD 11 - 8. sequence and series
The sum \(\sum\limits_{k=1}^{20}(1+2+3+\ldots+k)\) is
- A \(1496\)
- B \(1690\)
- C \(1540\)
- D \(1560\)
Answer & Solution
Correct Answer
(C) \(1540\)
Step-by-step Solution
Detailed explanation
\(\sum_{\mathrm{k}=1}^{20} \frac{\mathrm{k}(\mathrm{k}+1)}{2}=\frac{1}{2} \sum_{\mathrm{k}=1}^{20} \frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)-(\mathrm{k}-1) \mathrm{k}(\mathrm{k}+1)}{3}\) \(=\frac{1}{6} \times 20 \times 21 \times 22=1540\)
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