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Anonymous asked in Science & MathematicsMathematics · 9 years ago

Prove the relation: 1² + 2² + 3² ... n² = 1/6 n(n+1)(2n+1)?

I know that the above statement can be proved by Mathematical Induction. Proving this way only proves correctness of the formula. I was wondering how people found out that the sum of perfect squares is equal to n(n+1)(2n+1)/6.

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  • 9 years ago
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    One can prove sum of all powers of natural numbers one by one first taking power 1, i.e., 1 + 2 + 3 + ..., then power 2, i.e., 1^2 + 2^2 + 3^2 + ... and so on by a general method described as under.

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

    => n^3 - (n - 1)^3 = 3n^2 - 3n + 1

    Plugging n = 1, 2, 3, ... n

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

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

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

    .....

    ,.....

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

    ------------------------------------------------------

    Adding,

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

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

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

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

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

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

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

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

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

    = (1/6) n (n + 1) (2n + 1).

    The sum 1^3 + 23 + 3^3 + ... + n^3

    can similarly be proved taking

    n^4 - (n - 1)^4 = 4n^3 - 6n^2 + 4n - 1

    Thus, one after another, formula for sum of any power of the natural numbers can be derived.

    Refer to my answer as under where the sum 1^4 + 2^4 + 3^4 + ... + n^4 is derived.

    http://in.answers.yahoo.com/question/index;_ylt=Ag...

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  • 9 years ago

    One possible way is to consider the relation (r+1)^3 - r^3 =3r^2+3r+1

    If you sum each side from r=1 to r=n you get (n+1^3 - 1=3S+3n(n+1)/2+n , where S is the required sum

    and you get 3S=(n+1)^3-1-3n(n+1)/2-n=n^3+(3/2)n(n+1)-n=(n/2)(2n^2+3n+1)=(1/2)n(n+1)(2n+1`)

    so S=(1/6)n(n+1)(2n+1)

    This depends on knowing the sum of r from 1 to n but this is an arithmetic series and is well known.

    You also should be aware that the sum of [u(r+1)-u(r)] from r=1 to r=n is u(n+1)-u(n)

    You can then use the same idea to obtain the sum of r^3 using ((r+1)^4 -r^4)) etc.

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  • 9 years ago

    Form a recurrence relation and find the formula by comparing coefficients:

    a(n + 1) = a(n) + (n + 1)²

    Guess a(n) = an³ + bn² + cn + d

    a(n + 1)³ + b(n + 1)² + c(n + 1) + d = an³ + bn² + cn + d + (n + 1)²

    a(n³ + 3n² + 3n + 1) + b(n² + 2n + 1) + c(n + 1) + d = an³ + bn² + cn + d + n² + 2n + 1

    an³ + 3an² + 3an + a + bn² + 2bn + b + cn + c + d = an³ + bn² + cn + d + n² + 2n + 1

    3an² + (3a + 2b)n + (a + b + c) = n² + 2n + 1

    3a = 1

    a = ⅓

    3a + 2b = 2

    2b = 2 - 3a

    b = 1 - 3a / 2

    b = 1 - ½

    b = ½

    a + b + c = 1

    c = 1 - a - b

    c = 1 - ⅓ - ½

    c = 1 / 6

    Guess a(n) = n³ / 3 + n² / 2 + n / 6 + d

    When n = 0, a(n) = 0

    d = 0

    a(n) = n³ / 3 + n² / 2 + n / 6

    a(n) = (2n³ + 3n² + n) / 6

    a(n) = n(2n² + 3n + 1) / 6

    a(n) = n(n + 1)(2n + 1) / 6

    What is with the thumbs down, I used a perfectly valid method.

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  • 9 years ago

    2³=(1+1)³=1³+3*1²+3*1+1

    3³=(2+1)³=2³+3*2²+3*2+1

    4³=(3+1)³=3³+3*3²+3*3+1

    5³=(4+1)³=4³+3*4²+3*4+1

    ...

    n³=((n-1)+1)³=(n-1)³+3(n-1)²+3(n-1)+1

    -----------------------------

    SUM

    2³+3³+4³+5³+...+n³ =[1³+2³+3³+4³+5³+...+(n-1)³]+ 3[1²+2²+3²+4²+5²+...+(n-1)²]+ 3[1+2+3+4+5+...+(n-1)]+ [1+1+1+1+1+...+1]

    n³ =[1³]+ 3[1²+2²+3²+4²+5²+...+(n-1)²]+ 3[((n-1)+1)(n-1)/2]+ [n-1]

    n³-1 -3n(n-1)/2 -[n-1]= 3[1²+2²+3²+4²+5²+...+(n-1)²+n²] - 3n²

    [1²+2²+3²+4²+5²+...+(n-1)²+n²]= [n³-1 -3n(n-1)/2 -n+1+ 3n²]/3=

    = [2n³ +3n²+n]/6= n[2n² +3n+1]/6=

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

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  • 4 years ago

    applying induction instruct for the organic numbers n a million^2+2^2+3^2+...n^2=(n(n+a million)(2n+a million))/6 pf if n=a million then for sure a million^2 = (a million)(2)(3)/6 =a million so the assertion is genuine whilst n=a million. assume assertion is genuine for n=ok. it truly is we assume a million^2+2^2+3^2+...ok^2=(ok(ok+a million)(2k+a million))/6 Now, we instruct the assertion is genuine for n=ok+a million. a million^2+2^2+3^2+...+ok^2+(ok+a million)^2 by using induction hypothesis all of us comprehend that a million^2+2^2+3^2+...+ok^2 = (ok(ok+a million)(2k+a million))/6 if we upload (ok+a million)^2 to the two factors we get: a million^2+2^2+3^2+...+ok^2 + (ok+a million)^2 = (ok(ok+a million)(2k+a million))/6 + (ok+a million)^2 Now we simplify the best hand element. (ok(ok+a million)(2k+a million))/6 + (ok+a million)^2 = ( ok(ok+a million)(2k+a million))/6 + 6(ok+a million)^2/6= ( ok(ok+a million)(2k+a million) + 6(ok+a million)^2 ) /6= [ (ok+a million)(ok(2k+a million) + 6(ok+a million) ) ]/6= [(ok+a million)(2k^2 +7k +6) ]/6= [(ok+a million)(2k + 3)(ok + 2) ]/6= [(ok+a million)(2(ok+a million)+a million)((ok+a million)+a million)]/6 this exhibits the assertion is genuine for n =ok+a million so the assertion is genuine for all n by using induction.

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