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Find the sum of \( \Large 21^{2}+22^{2}+....+60^{2} \)
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A) 70940 |
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B) 2870 |
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C) 12810 |
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D) 9946 |
Correct Answer:
| A) 70940 |
Description for Correct answer:
\( \Large 21^{2}+22^{2}+....+60^{2}= \left(1^{2}+2^{2}+....+60^{2}\right) \)- \( \Large \left(a^{2}+2^{2}+....+20^{2}\right) \)
=\( \Large \frac{60 \left(60+1\right) \left(120+1\right) }{6} - \frac{20 \times 21 \times 41}{6} \)
= \( \Large \frac{60 \times 61 \times 121}{6} - \frac{20 \times 21 \times 41}{6} \)
= \( \Large 610 \times 121-70 \times 41=73810-2870 = 70940 \)
\( \Large 21^{2}+22^{2}+....+60^{2}= \left(1^{2}+2^{2}+....+60^{2}\right) \)- \( \Large \left(a^{2}+2^{2}+....+20^{2}\right) \)
=\( \Large \frac{60 \left(60+1\right) \left(120+1\right) }{6} - \frac{20 \times 21 \times 41}{6} \)
= \( \Large \frac{60 \times 61 \times 121}{6} - \frac{20 \times 21 \times 41}{6} \)
= \( \Large 610 \times 121-70 \times 41=73810-2870 = 70940 \)
Part of solved Factorisation questions and answers : >> Elementary Mathematics >> Factorisation
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