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

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

Update:

please solve it completely

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  • 1 decade ago
    Best answer

    Can any one proove sum of the squares of the first n integers is 1/24 (2n) (2n+1) (2n+2) step by step instructions pls

    Comment by Sunny — April 1, 2007 @ 4:58 am

    Sunny: You can use induction to prove your formula. I suggest you do a bit more reading in your text or research online if you’re not sure what that means.

    Comment by mrc — April 1, 2007 @ 1:30 pm

    Source(s): =)
  • 4 years ago

    Start from the identities: (k+1)^2 - k^2 = 2k + 1 ........(1) (k+1)^3 - k^3 = 3k^2 + 3k + 1 ..........(2) Then: (2) * 2 - (1) * 3 gives: 2[ (k+1)^3 - k^3 ] - 3[ (k+1)^2 - k^2 ] = 6k^2 - 1 .........(3) Note that when summing the left hand side from k = 1 to k = n, all terms in each of the square brackets disappear, except for the first term when k = n and the last term when n = 1. If s is the sum you require, then summing (3) from k = 1 to k = n gives: 6s - 1 = 2[ (n + 1)^3 - 1 ] - 3[ (n + 1)^2 - 1 ] = 2(n + 1)^3 - 3(n +1)^2 + 1 6s = 2(n + 1)^3 - 3(n +1)^2 + 2 = 2n^3 + 6n^2 + 6n + 2 - 3n^2 - 6n + 1 = 2n^3 + 3n^2 + 1 = n(2n + 1)(n + 1) Multiplying by 4 to obtain the rather unusual form you have: 24s = (2n)(2n + 1)(2n + 2) s = (1/24)(2n)(2n + 1)(2n + 2).

  • 1 decade ago

    THis is correct

    n = 1 sum = 1

    RHS = 1/24(2)*(3)*(4) = 1

    proof by induction

    let is be true for n add n+1st term

    we need to show for for n+1 terms it is

    1/24(2(n+1))(2n+3)(2n+4)

    or 1/24(2n+2)(2n+3)(2n+4)

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

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

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

    = 1/24((2n+2)(4n^2+2n+12n+12)

    =1/24(2n+2)(4n^2+14+12)

    = 1/24(2n+2)(2n+3)(2n+4)

    which is true

    hence proved

  • 1 decade ago

    Let us put the sums

    S1=1+2+...+n=n*(n+1)/2

    and the unknown

    S2=1^2+2^2+...+n^2

    We know that

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

    We put 1,2,....n in the place of x

    2^3=1^3+3*1^2+3*1+1

    3^3=2^3+3*2^2+3*2+1

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

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

    Adding The n Equalities together

    and deleting same quantities

    on both sides we have

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

    +3*(1+2+...+n)+(n+1)

    (n+1)^3=3*S2+3*S1+(n+1)

    (n+1)^3=3*S2+3*n*(n+1)/2+(n+1)

    3*S2=(n+1)[(n+1)^2-3n/2-(n+1)

    3*S2=(n+1)[n^2+2n+1-3n/2-1]

    3*S2=(n+1)(n^2+n/2)

    6*S2=(n+1)(2n^2+n)

    6*S2=(n+1)*n*(2n+1)

    S2=n*(n+1)*(2n+1)/6

    which is the same result

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  • 1 decade ago

    Sum (X^2) = 2 Integral of Sum of (x)

    Sum of X = X(X+1)/2

    Integral is X^3/3 + X^2/2 + X

  • 1 decade ago

    Why don't you pay attention in class and do your own homework?

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