? asked in Science & MathematicsMathematics · 10 years ago

Prove that the sum of the squares of the first n integers is 1/24 (2n) (2n+1) (2n+2)?

3 Answers

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  • 10 years ago
    Best answer

    refer to http://www.cut-the-knot.org/htdocs/dcforum/DCForum...

    now sum = 1/6(n*(n+1)* (2n+1) = 1/24(2n* (2n+2) * (2n+ 1) = 1/24 (2n) (2n+1) (2n+2)

    Proved

    extending it we get ((2n+2)P3)/4

  • Raj K
    Lv 7
    10 years ago

    ∑n² = 1² +2²+3²+4²+................+n²

    Using the identity

    n³ −(n−1)³= n³ −(n ³−3n²+3n−1)

    →n³ −(n−1)³= 3n²−3n+1

    →(n−1)³ −(n−2)³= 3(n−1)²−3(n−1)+1

    ....................................

    .................................

    .......................................

    →(1)³ −(0)³= 3(1)²−3(1)+1

    Adding both sides, we have

    n³ = 3∑n²−3∑n+∑1

    → 3∑n²= n³+3∑n−∑1 Since

    =n³+3n(n+1)/2−n Since ∑n=n(n+1)/2 and ∑1=n

    =n{n²+3(n+1)/2−1} Simplifying

    =n(2n²+3n+3−1)/2 Simplifying

    =n(2n²+3n+2)/2

    =n(2n+1)(n+1)/2

    or ∑n²=n(2n+1)(n+1)/6

    or ∑n²=4×n(2n+1)(n+1)/(6×4) Multiply and divide by 4

    =2n(2n+1)(2n+2)/(24)

    =(2n)(2n+1)(2n+2)/(24)

    =(1/24)(2n)(2n+1)(2n+2)/(24)

  • 10 years ago

    As before, we set up as follows:

    sumSquares1.gif

    Saying the sum to n is one term less than the sum to (n+1)

    We expand the sums:

    sumSquares2.gif

    As expected, the cubic terms cancel, and we rearrange the formula to have the sum of the squares on the left:

    sumSquares3.gif

    Expanding the cube and summing the sums:

    sumSquares4.gif

    Adding like terms:

    sumSquares5.gif

    Dividing throughout by 3 gives us the formula for the sum of the squares:

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