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For all n ge 1 then 1^2 + 2^2 + 3^2 + ...

For all `n ge 1` then `1^2 + 2^2 + 3^2 + 4^2 + ….+ n^2` is equal to-

A

`(n (n+1) )/( 2)`

B

`(n (n+1) (2n+1) )/( 6)`

C

`n^(2) (2n^(2) -1)`

D

None of these

Text Solution

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The correct Answer is:
B
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