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1 ^(2) + 2^(2) + 3^(2) + . . . + n^(2) =...

`1 ^(2) + 2^(2) + 3^(2) + . . . + n^(2) = (n (n + 1) (2 n + 1))/( 6)`

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The correct Answer is:
`= (n(n + 1) (2 n + 1))/( 6)` for all `n in N`
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Knowledge Check

  • If 2^(3) + 4^(3) + 6^(3) + … + (2n)^(3) = kn^(2) ( n+1)^(2) then k=

    A
    `1//2`
    B
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    C
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    D
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  • If alpha in R, n in N " and " n + 2 (n - 1)+ 3 (n-2) + …… +(n-1)2 + n.1 = alpha n ( n + 1) (n + 2) , " then " alpha =

    A
    `1/2`
    B
    `1/3`
    C
    `1/5`
    D
    `1/6`

  • underset(n rarr infty)("lim") [(1)/(n)+ (n^(2))/((n + 1)^(3)) + (n^(2))/((n + 2)^(3)) + (n^(2))/((n + 3)^(3)) + .....+(1)/(125n)] =

    A
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    B
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    C
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    D
    `(35)/(72)`

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