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1+2+3+............+n=(n(n+1))/2 forall n...

`1+2+3+............+n=(n(n+1))/2 forall n in N.`

Answer

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1^(2)+2^(2)+3^(2)+............+n^(2)=(n(n+1)(2n+1))/6 forall n in N.

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Knowledge Check

  • The value of ""2^n[1.3.5 ..........(2n-3)(2n-1)] is

    A
    `((2n)i)/(n!)`
    B
    `((2n)i)/(2^n)`
    C
    `(n!)/((2n)!)`
    D
    None of these

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    1^(3)+2^(3)+3^(3)+………….+n^(3)=(n^(2)(n+1)^(2))/4 forall n in N.

    1*2*3+2*3*4+...+n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/4 forall n in N.

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    (1+3/1)(1+5/4)(1+7/9)...(1+((2n+1))/n^(2))=(n+1)^(2) forall n in N.

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