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{\displaystyle \displaystyle \forall n\in \mathbb {N} :\sum _{i\mathop {=} 1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}}
{{:proofwiki:Sum of Sequence of Squares/Proof by Induction}}
{{:proofwiki:Sum of Sequence of Squares/Proof by Products of Consecutive Integers}}
{{:proofwiki:Sum of Sequence of Squares/Proof by Telescoping Series}}
Proof by Summation of Summations
edit
File:Sum of Sequences of Squares.jpg We can observe from the above diagram that:
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{\displaystyle \displaystyle \forall n\in \mathbb {N} :\sum _{i\mathop {=} 1}^{n}i^{2}=\sum _{i\mathop {=} 1}^{n}\left({\sum _{j\mathop {=} i}^{n}j}\right)}
Therefore we have:
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[[Closed Form for Triangular Numbers]]
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{\displaystyle {\begin{array}{rrl}&\displaystyle \sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =\sum _{i\mathop {=} 1}^{n}\left({\sum _{j\mathop {=} i}^{n}j}\right)\\&&\displaystyle =\sum _{i\mathop {=} 1}^{n}\left({\sum _{j\mathop {=} 1}^{n}j-\sum _{j\mathop {=} 1}^{i-1}j}\right)\\&&\displaystyle =\sum _{i\mathop {=} 1}^{n}\left({{\frac {n\left({n+1}\right)}{2}}-{\frac {i\left({i-1}\right)}{2}}}\right)\\\implies &\displaystyle 2\sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =n^{2}\left({n+1}\right)-\sum _{i\mathop {=} 1}^{n}i^{2}+\sum _{i\mathop {=} 1}^{n}i\\\implies &\displaystyle 3\sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =n^{2}\left({n+1}\right)+\sum _{i\mathop {=} 1}^{n}i\\\implies &\displaystyle 3\sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =n^{2}\left({n+1}\right)+{\frac {n\left({n+1}\right)}{2}}&{\text{[[Closed Form for Triangular Numbers]]}}\\\implies &\displaystyle 6\sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =2n^{2}\left({n+1}\right)+n\left({n+1}\right)\\&&\displaystyle =n\left({n+1}\right)\left({2n+1}\right)\\\implies &\displaystyle \sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle ={\frac {n\left({n+1}\right)\left({2n+1}\right)}{6}}\\\end{array}}}
Template:Qed
Proof by Sum of Differences of Cubes
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[[Binomial Theorem]]
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[[Summation is Linear]]
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[[Closed Form for Triangular Numbers]]
{\displaystyle {\begin{aligned}\sum _{i\mathop {=} 1}^{n}\left({\left({i+1}\right)^{3}-i^{3}}\right)&=\sum _{i\mathop {=} 1}^{n}\left({i^{3}+3i^{2}+3i+1-i^{3}}\right)&{\text{[[Binomial Theorem]]}}\\&=\sum _{i\mathop {=} 1}^{n}\left({3i^{2}+3i+1}\right)\\&=3\sum _{i\mathop {=} 1}^{n}i^{2}+3\sum _{i\mathop {=} 1}^{n}i+\sum _{i\mathop {=} 1}^{n}1&{\text{[[Summation is Linear]]}}\\&=3\sum _{i\mathop {=} 1}^{n}i^{2}+3{\frac {n\left({n+1}\right)}{2}}+n&{\text{[[Closed Form for Triangular Numbers]]}}\\\end{aligned}}}
On the other hand:
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Definition of [[Definition:Summation!Summation]]
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{\displaystyle {\begin{aligned}\sum _{i\mathop {=} 1}^{n}\left({\left({i+1}\right)^{3}-i^{3}}\right)&=\left({n+1}\right)^{3}-n^{3}+n^{3}-\left({n-1}\right)^{3}+\left({n-1}\right)^{3}-\cdots +2^{3}-1^{3}&{\text{Definition of [[Definition:Summation!Summation]]}}\\&=\left({n+1}\right)^{3}-1^{3}&{\text{[[Telescoping Series/Example 2!Telescoping Series: Example 2]]}}\\&=n^{3}+3n^{2}+3n+1-1&{\text{[[Binomial Theorem]]}}\\&=n^{3}+3n^{2}+3n\\\end{aligned}}}
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{\displaystyle {\begin{array}{rrl}&\displaystyle 3\sum _{i\mathop {=} 1}^{n}i^{2}+3{\frac {n\left({n+1}\right)}{2}}+n&\displaystyle =n^{3}+3n^{2}+3n\\\implies &\displaystyle 3\sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =n^{3}+3n^{2}+3n-3{\frac {n\left({n+1}\right)}{2}}-n\\\end{array}}}
Therefore:
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{\displaystyle {\begin{aligned}\sum _{i\mathop {=} 1}^{n}i^{2}&={\frac {1}{3}}\left({n^{3}+3n^{2}+3n-3{\frac {n\left({n+1}\right)}{2}}-n}\right)\\&={\frac {1}{3}}\left({n^{3}+3n^{2}+3n-{\frac {3n^{2}}{2}}-{\frac {3n}{2}}-n}\right)\\&={\frac {1}{3}}\left({n^{3}+{\frac {3n^{2}}{2}}+{\frac {n}{2}}}\right)\\&={\frac {1}{6}}n\left({2n^{2}+3n+1}\right)\\&={\frac {1}{6}}n\left({n+1}\right)\left({2n+1}\right)\\\end{aligned}}}
Template:Qed
{{:proofwiki:Sum of Sequence of Squares/Proof by Binomial Coefficients}}
{{:proofwiki:Sum of Sequence of Squares/Proof using Bernoulli Numbers}}
{{:proofwiki:Sum of Sequence of Squares/Historical Note}}
![{\displaystyle {\begin{array}{rrl}&\displaystyle \sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =\sum _{i\mathop {=} 1}^{n}\left({\sum _{j\mathop {=} i}^{n}j}\right)\\&&\displaystyle =\sum _{i\mathop {=} 1}^{n}\left({\sum _{j\mathop {=} 1}^{n}j-\sum _{j\mathop {=} 1}^{i-1}j}\right)\\&&\displaystyle =\sum _{i\mathop {=} 1}^{n}\left({{\frac {n\left({n+1}\right)}{2}}-{\frac {i\left({i-1}\right)}{2}}}\right)\\\implies &\displaystyle 2\sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =n^{2}\left({n+1}\right)-\sum _{i\mathop {=} 1}^{n}i^{2}+\sum _{i\mathop {=} 1}^{n}i\\\implies &\displaystyle 3\sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =n^{2}\left({n+1}\right)+\sum _{i\mathop {=} 1}^{n}i\\\implies &\displaystyle 3\sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =n^{2}\left({n+1}\right)+{\frac {n\left({n+1}\right)}{2}}&{\text{[[Closed Form for Triangular Numbers]]}}\\\implies &\displaystyle 6\sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle =2n^{2}\left({n+1}\right)+n\left({n+1}\right)\\&&\displaystyle =n\left({n+1}\right)\left({2n+1}\right)\\\implies &\displaystyle \sum _{i\mathop {=} 1}^{n}i^{2}&\displaystyle ={\frac {n\left({n+1}\right)\left({2n+1}\right)}{6}}\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47528e22e776cbf511b7e53cbc486a9bc3a4b429)
![{\displaystyle {\begin{aligned}\sum _{i\mathop {=} 1}^{n}\left({\left({i+1}\right)^{3}-i^{3}}\right)&=\sum _{i\mathop {=} 1}^{n}\left({i^{3}+3i^{2}+3i+1-i^{3}}\right)&{\text{[[Binomial Theorem]]}}\\&=\sum _{i\mathop {=} 1}^{n}\left({3i^{2}+3i+1}\right)\\&=3\sum _{i\mathop {=} 1}^{n}i^{2}+3\sum _{i\mathop {=} 1}^{n}i+\sum _{i\mathop {=} 1}^{n}1&{\text{[[Summation is Linear]]}}\\&=3\sum _{i\mathop {=} 1}^{n}i^{2}+3{\frac {n\left({n+1}\right)}{2}}+n&{\text{[[Closed Form for Triangular Numbers]]}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e11cc53bcd2760bad8c2b8a2dea9e32775a4a7ed)
![{\displaystyle {\begin{aligned}\sum _{i\mathop {=} 1}^{n}\left({\left({i+1}\right)^{3}-i^{3}}\right)&=\left({n+1}\right)^{3}-n^{3}+n^{3}-\left({n-1}\right)^{3}+\left({n-1}\right)^{3}-\cdots +2^{3}-1^{3}&{\text{Definition of [[Definition:Summation!Summation]]}}\\&=\left({n+1}\right)^{3}-1^{3}&{\text{[[Telescoping Series/Example 2!Telescoping Series: Example 2]]}}\\&=n^{3}+3n^{2}+3n+1-1&{\text{[[Binomial Theorem]]}}\\&=n^{3}+3n^{2}+3n\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b9efa6c35a0d65f865e1384f5ec3098fb6cd277)
