Question : Given $\small 2^{2}+4^{2}+6^{2}+......+40^{2}=11480$, then the value of $\small 1^{2}+2^{2}+3^{2}+......+20^{2}$ is:
Option 1: 2870
Option 2: 2868
Option 3: 2867
Option 4: 2869
Answer (1)
25th Jan, 2024
Correct Answer: 2870
Solution :
Given: $2^{2}+4^{2}+6^{2}+......+40^{2}=11480$
$⇒2^{2}(1^{2}+2^{2}+3^{2}+......+20^{2})=11480$
$⇒1^{2}+2^{2}+3^{2}+......+20^{2}=\frac{11480}{4}$
$\therefore1^{2}+2^{2}+3^{2}+......+20^{2}=2870$
Hence, the correct answer is 2870.
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