Question : Given $2^{2} + 4^{2} + 6^{2} + . . . . . . + 40^{2} = 11480$ , then the value of $1^{2} + 2^{2} + 3^{2} + . . . . . . + 20^{2}$ is:
Option 1: 2870
Option 2: 2868
Option 3: 2867
Option 4: 2869

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Question : If $1^{2} + 2^{2} + 3^{2} + . . . . . . + p^{2} = \frac{p (p + 1) (2 p + 1)}{6}$ , then $1^{2} + 3^{2} + 5^{2} + . . . . . . + 17^{2}$ is equal to:

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Answer 1

Correct Answer: 2870

Solution : Given: $2^{2} + 4^{2} + 6^{2} + . . . . . . + 40^{2} = 11480$
$\Rightarrow 2^{2} (1^{2} + 2^{2} + 3^{2} + . . . . . . + 20^{2}) = 11480$
$\Rightarrow 1^{2} + 2^{2} + 3^{2} + . . . . . . + 20^{2} = \frac{11480}{4}$
$∴ 1^{2} + 2^{2} + 3^{2} + . . . . . . + 20^{2} = 2870$
Hence, the correct answer is 2870.

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Question : Given 22+42+62+......+402=11480, then the value of 12+22+32+......+202 is:Option 1: 2870Option 2: 2868Option 3: 2867Option 4: 2869

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Question : Given $2^{2} + 4^{2} + 6^{2} + . . . . . . + 40^{2} = 11480$ , then the value of $1^{2} + 2^{2} + 3^{2} + . . . . . . + 20^{2}$ is:
Option 1: 2870
Option 2: 2868
Option 3: 2867
Option 4: 2869