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Property 3 The sum of the cubes of first...

Property 3 The sum of the cubes of first n natural numbers is equal to the square of their sum. That is `1^3 + 2^3 + 3^3 +.........+ n^3 = (1+2+3+........+n)^2`

Answer

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Knowledge Check

  • The sum of the series sum_(n=8)^17 1/((n+2)(n+3)) is equal to

    A
    `1/17`
    B
    `1/18`
    C
    `1/19`
    D
    `1/20`

  • The sum to n terms of the series 1^2 + (1^2 + 3^3) + (1^2 + 3^3 + 5^2) + … is :

    A
    `1/3 (n^3 + n^2 + 1)`
    B
    `1/6 n(n+1) (2n^2 + 2n-1)`
    C
    `1/3 (2n^2 + 2n -1)`
    D
    none of these

  • The sum of the numbers of the n^(th) term of the series (1) +(1+2) +(1+2+3) +(1 +2+3+4) +… (1 +2+3+….n)

    A
    `""^(n+1)C_3`
    B
    `""^(n+1)C_2`
    C
    `""^(n)C_2`
    D
    `""^(n+2)C_3`

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    If the sum of the first n terms of a sequence is equal to √ (3/ 2) ( n^ 2 − n ) , then the sum of the squares of these terms is equal to

    The cubes of the natural numbers are grouped as 1^3, (2^3,3^3), (4^3,5^3,6^3) , ..... then the sum of the numbers in the nth group is:

    The sum of the series 1^(2)+1+2^(2)+2+3^(2)+3+ . . . . .. +n^(2)+n , is

    Sum to n terms of the series 1^(3) - (1.5)^(3) +2^(3)-(2.5)^(3) +…. is

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