User:Ans/Sum of Sequence of Squares

Theorem

$\mathrm{\forall }n\in \mathbb{N}:\sum _{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$

Proof by Induction

{{:proofwiki:Sum of Sequence of Squares/Proof by Induction}}

Proof by Products of Consecutive Integers

{{:proofwiki:Sum of Sequence of Squares/Proof by Products of Consecutive Integers}}

Proof by Telescoping Series

{{:proofwiki:Sum of Sequence of Squares/Proof by Telescoping Series}}

Proof by Summation of Summations

File:Sum of Sequences of Squares.jpg

We can observe from the above diagram that:

$\mathrm{\forall }n\in \mathbb{N}:\sum _{i=1}^n{i}^2=\sum _{i=1}^n\left(\sum _{j=i}^{n}j\right)$

Therefore we have:

$\begin{array}{ccc}& \sum _{i=1}^n{i}^2& =\sum _{i=1}^n\left(\sum _{j=i}^{n}j\right)\\ & & =\sum _{i=1}^n\left(\sum _{j=1}^{n}j-\sum _{j=1}^{i-1}j\right)\\ & & =\sum _{i=1}^n\left(\frac{n\left(n+1\right)}{2}-\frac{i\left(i-1\right)}{2}\right)\\ ⟹& 2\sum _{i=1}^n{i}^2& =n^2\left(n+1\right)-\sum _{i=1}^n{i}^2+\sum _{i=1}^{n}i\\ ⟹& 3\sum _{i=1}^n{i}^2& =n^2\left(n+1\right)+\sum _{i=1}^{n}i\\ ⟹& 3\sum _{i=1}^n{i}^2& =n^2\left(n+1\right)+\frac{n\left(n+1\right)}{2}& \text{[[Closed Form for Triangular Numbers]]}\\ ⟹& 6\sum _{i=1}^n{i}^2& =2n^2\left(n+1\right)+n\left(n+1\right)\\ & & =n\left(n+1\right)\left(2n+1\right)\\ ⟹& \sum _{i=1}^n{i}^2& =\frac{n\left(n+1\right)\left(2n+1\right)}{6}\end{array}$

Proof by Sum of Differences of Cubes

$\begin{array}{ccc}\sum _{i=1}^n\left({\left(i+1\right)}^3-i^3\right)& =\sum _{i=1}^n\left(i^3+3i^2+3i+1-i^3\right)& \text{[[Binomial Theorem]]}\\ & =\sum _{i=1}^n\left(3i^2+3i+1\right)\\ & =3\sum _{i=1}^n{i}^2+3\sum _{i=1}^{n}i+\sum _{i=1}^{n}1& \text{[[Summation is Linear]]}\\ & =3\sum _{i=1}^n{i}^2+3\frac{n\left(n+1\right)}{2}+n& \text{[[Closed Form for Triangular Numbers]]}\\ \end{array}$

On the other hand:

$\begin{array}{ccc}\sum _{i=1}^n\left({\left(i+1\right)}^3-i^3\right)& ={\left(n+1\right)}^3-n^3+n^3-{\left(n-1\right)}^3+{\left(n-1\right)}^3-\cdots +2^3-1^3& \text{Definition of [[Definition:Summation!Summation]]}\\ & ={\left(n+1\right)}^3-1^3& \text{[[Telescoping Series/Example 2!Telescoping Series: Example 2]]}\\ & =n^3+3n^2+3n+1-1& \text{[[Binomial Theorem]]}\\ & =n^3+3n^2+3n\\ \end{array}$

Therefore:

$\begin{array}{ccc}& 3\sum _{i=1}^n{i}^2+3\frac{n\left(n+1\right)}{2}+n& =n^3+3n^2+3n\\ ⟹& 3\sum _{i=1}^n{i}^2& =n^3+3n^2+3n-3\frac{n\left(n+1\right)}{2}-n\end{array}$

Therefore:

$\begin{array}{cc}\sum _{i=1}^n{i}^2& =\frac{1}{3}\left(n^3+3n^2+3n-3\frac{n\left(n+1\right)}{2}-n\right)\\ & =\frac{1}{3}\left(n^3+3n^2+3n-\frac{3n^2}{2}-\frac{3n}{2}-n\right)\\ & =\frac{1}{3}\left(n^3+\frac{3n^2}{2}+\frac{n}{2}\right)\\ & =\frac{1}{6}n\left(2n^2+3n+1\right)\\ & =\frac{1}{6}n\left(n+1\right)\left(2n+1\right)\\ \end{array}$

Proof by Binomial Coefficients

{{:proofwiki:Sum of Sequence of Squares/Proof by Binomial Coefficients}}

Proof using Bernoulli Numbers

{{:proofwiki:Sum of Sequence of Squares/Proof using Bernoulli Numbers}}

Historical Note

{{:proofwiki:Sum of Sequence of Squares/Historical Note}}

Sources

• Template:BookReference: Exercise $18.10 \ \text{(b)}$
• Template:BookReference: $\text {1-1}$ Principle of Mathematical Induction: Exercise $1$
• Template:BookReference: $\S 1.17$: Finite Induction and Well-Ordering for Positive Integers: Exercise $3$
• Template:BookReference: Appendix $\text{A}.6$: Mathematical Induction: Problem Set $\text{A}.6$: $36$
• Template:BookReference: Chapter $\text {A}.5$: Archimedes (ca. Template:DateRange B.C.)