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Math Help - Squares of first n integers

  1. #1
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    Question Squares of first n integers

    Hello Friends,
    I am stuck at proving this problem.
    Prove that the sum of the squares of the first n integers is
    1/24 (2n) (2n+1) (2n+2)

    Thanks.
    Harshal.
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  2. #2
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    Hello, Harshal!

    How about an inductive proof?


    Prove: .1 + 2 + 3 + ... + n .= .n(n + 1)(2n + 1)/6

    Verify S(1): .1 .= .123/6 . . 1 = 1 ... true


    Assume S(k): .1 + 2 + 3 + ... k .= .k(k + 1)(2k + 1)/6


    Add (k + 1) to both sides:
    . . 1 + 2 + 3 + ... + (k + 1) .= .k(k + 1)(2k + 1)/6 + (k + 1)

    The left side is the left side of S(k + 1).


    Factor the right side: .(k + 1) [k(k + 1)/6 + (k + 1)]

    . . = .(k + 1) (k + k + 6k + 6)/6 .= .(k + 1)(k + 7k + 6)/6

    . . = .(k + 1)(k + 2)(2k + 3)/6 .= .
    (k + 1) ([k + 1] + 1)(2[k + 1] + 1)/6

    . . and this is the right side of S(k + 1).


    The inductive proof is complete.

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  3. #3
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    hints for solution to your problems

    We know that the sum of first n natural numbers is

    (n(n+1))/2

    Now we know again by assuming the sum of the squares of first n natural numbers as Sn

    x^3-(x-1)^3=3x^2-3x+1

    Put x = 1,2,3,.....,(n-1),n successively, we get:

    1^3-0^3 = 3*1^2 - 3*1+1

    2^3-1^3 = 3*2^2 - 3*2+1
    ....................................
    ....................................
    ....................................
    (n-1)^3 - (n-2)^3 =3*(n-1)^2 - 3*(n-1) +1
    n^3 - (n-1)^3 =3*n^2 - 3*n +1
    ----------------------------------------------------
    Add column wise and then simplify it a bit by applying the formula for addition of first n natural numbers to get the answer.


    ------------------------------
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  4. #4
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    Joined
    Mar 2007
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    hints for solution to your problems

    We know that the sum of first n natural numbers is

    (n(n+1))/2

    Now we know again by assuming the sum of the squares of first n natural numbers as Sn

    x^3-(x-1)^3=3x^2-3x+1

    Put x = 1,2,3,.....,(n-1),n successively, we get:

    1^3-0^3 = 3*1^2 - 3*1+1

    2^3-1^3 = 3*2^2 - 3*2+1
    ....................................
    ....................................
    ....................................
    (n-1)^3 - (n-2)^3 =3*(n-1)^2 - 3*(n-1) +1
    n^3 - (n-1)^3 =3*n^2 - 3*n +1
    ----------------------------------------------------
    Add column wise and then simplify it a bit by applying the formula for addition of first n natural numbers to get the answer.


    ------------------------------
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  5. #5
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    Hello, jagabandhu!

    I like that method . . . I've used it quite often.
    . . Let me try to format it.


    We know that the sum of first n natural numbers is: .∑k .= .n(n + 1)

    Now consider the difference of two consecutive cubes:
    . . x - (x - 1) .= .3x - 3x + 1


    Now let x = 1, 2, 3, ..., n and we get:

    . . 1 - 0 . . . .= . . 31 . - . 31 + 1
    . . 2 - 1 . . . .= . . 32 . - . 32 + 1
    . . 3 - 2 . . . .= . . 33 . - . 33 + 1
    . . . . . . . . . . . . . . . . . . . . . .. . . . .
    . . . . . . . . . . . . . . . . . . . . . .. . . . .
    . . . . . . . . . . . . . . . . . . . . . .. . . . .
    (n-1) - (n-2) .= .3(n-1) - 3(n-1) + 1
    . . .n - (n-1) .= . . 3n . - . .3n . + 1


    Now, add these equations.
    Nearly all of the left side cancels out.
    On the right, we have three sums.

    . . n .= .3∑k - 3∑k + ∑1

    . . n .= .3∑k - 3n(n+1) + n


    Then we have: .3∑k .= .n - n + 3n(n+1)/2

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

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


    Factor: . 3∑k .= .n(n + 1)/2 [(2(n - 1) + 3]

    . . . . . . .3∑k .= .n(n + 1)(2n + 1)/2


    Therefore: .∑k .= .n(n + 1)(2n + 1)/6

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