avewright/proofwiki-math huggingface dataset:
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
We have that:
:$\ds \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + n$
Consider $\ds 2 \sum_{i \mathop = 1}^n i$.
Then:
{{begin-eqn}}
{{eqn | l = 2 \sum_{i \mathop = 1}^n i
      | r = 2 \paren {1 + 2 + \dotsb + \paren {n - 1} + n}
      | c = 
}}
{{eqn | r = \paren {1 + 2 + \dotsb + \paren {n - 1} + n} + \paren {n + \paren {n - 1} + \dotsb + 2 + 1}
      | c = 
}}
{{eqn | r = \paren {1 + n} + \paren {2 + \paren {n - 1} } + \dotsb + \paren {\paren {n - 1} + 2} + \paren {n + 1}
      | c = Integer Addition is Commutative, Integer Addition is Associative
}}
{{eqn | r = \paren {n + 1}_1 + \paren {n + 1}_2 + \dotsb + \paren {n + 1}_n
      | c = 
}}
{{eqn | r = n \paren {n + 1}
      | c = 
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = 2 \sum_{i \mathop = 1}^n i
      | r = n \paren {n + 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop = 1}^n i
      | r = \frac {n \paren {n + 1} } 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^{m - 1} \paren {a + i d}
      | r = m \paren {a + \frac {m - 1} 2 d}
      | c = Sum of Arithmetic Sequence
}}
{{eqn | l = \sum_{i \mathop = 0}^n \paren {a + i d}
      | r = \paren {n + 1} \paren {a + \frac n 2 d}
      | c = Let $n = m - 1$
}}
{{eqn | l = \sum_{i \mathop = 0}^n i
      | r = \paren {n + 1} \paren {\frac n 2}
      | c = Let $a = 0$ and $d = 1$
}}
{{eqn | l = 0 + \sum_{i \mathop = 1}^n i
      | r = \frac {n \paren {n + 1} } 2
      | c = 
}}
{{eqn | l = \sum_{i \mathop = 1}^n i
      | r = \frac {n \paren {n + 1} } 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
Proof by induction:
=== Basis for the Induction ===
When $n = 1$, we have:
:$\ds \sum_{i \mathop = 1}^1 i = 1$
Also:
:$\dfrac {n \paren {n + 1} } 2 = \dfrac {1 \cdot 2} 2 = 1$
This is our base case.
=== Induction Hypothesis ===
:$\forall k \in \N: k \ge 1: \ds \sum_{i \mathop = 1}^k i = \frac {k \paren {k + 1} } 2$
This is our induction hypothesis.
It is to be demonstrated that:
:$\ds \sum_{i \mathop = 1}^{k + 1} i = \frac {\paren {k + 1} \paren {k + 2} } 2$
=== Induction Step ===
This is our induction step:
Consider $n = k + 1$.
By the properties of summation:
:$\ds \sum_{i \mathop = 1}^{k + 1} i = \sum_{i \mathop = 1}^k i + k + 1$
Using the induction hypothesis this can be simplified to:
{{begin-eqn}}
{{eqn | l = \frac {k \paren {k + 1} } 2 + k + 1
      | r = \frac {k \paren {k + 1} + 2 k + 2} 2
      | c = 
}}
{{eqn | r = \frac {k^2 + 3 k + 2} 2
      | c = 
}}
{{eqn | r = \frac {\paren {k + 1} \paren {k + 2} } 2
      | c = 
}}
{{eqn | r = \frac {\paren {k + 1} \paren {\paren {k + 1} + 1} } 2
      | c = 
}}
{{end-eqn}}
Hence the result by induction.
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
Triangular numbers are $k$-gonal numbers where $k = 3$.
From Closed Form for Polygonal Numbers we have that:
:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
Hence:
{{begin-eqn}}
{{eqn | l = T_n
      | r = \frac n 2 \paren {\paren {3 - 2} n - 3 + 4}
      | c = Closed Form for Polygonal Numbers
}}
{{eqn | r = \frac n 2 \paren {n + 1}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
We have that:
:$\ds \map T n = 1 + 2 + \dotsb + n = \sum_{i \mathop = 1}^n i$
Thus:
{{begin-eqn}}
{{eqn | l = \map T n
      | r = n + \paren {n - 1} + \paren {n - 2} + \dotsb + 2 + 1
      | c =
}}
{{eqn | r = n + \paren {n - 1} + \paren {n - 2} + \dotsb + \paren {n - \paren {n - 2} } + \paren {n - \paren {n - 1} }
      | c =
}}
{{eqn | r = \underbrace {n + n + \dotsb + n}_{n \text { times} }
      | c = extracting $n$ from each term
}}
{{eqn | o = 
      | ro= +
      | r = \paren {-1} + \paren {-2} + \dotsb + \paren {-\paren {n - 2} } + \paren {-\paren {n - 1} }
      | c =
}}
{{eqn | r = n^2 - \paren {1 + 2 + \dotsb + \paren {n - 1} }
      | c =
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \map T n
      | r = n^2 - \map T {n - 1}
      | c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | n = 2
      | l = \map T n
      | r = n + \map T {n - 1}
      | c = {{Defof|Triangular Number|index = 1}}
}}
{{eqn | ll= \leadsto
      | l = 2 \, \map T n
      | r = n^2 + n
      | c = $(1) + (2)$
}}
{{eqn | ll= \leadsto
      | l = \map T n
      | r = \frac {n \paren {n + 1} } 2
      | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = S
      | o = :=
      | r = \sum_{i \mathop = 1}^n i
}}
{{eqn | l = u_r
      | o = :=
      | r = r
}}
{{eqn | l = v_r
      | o = :=
      | r = r \paren {r + 1}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = v_r - v_{r - 1}
      | r = r \paren {r + 1} - \paren {r - 1} r
      | c = 
}}
{{eqn | r = r \paren {\paren {r + 1} - \paren {r - 1} }
      | c = 
}}
{{eqn | r = r \times 2
      | c = 
}}
{{eqn | r = 2 u_r
      | c = 
}}
{{end-eqn}}
Thus we have:
{{begin-eqn}}
{{eqn | l = v_n - v_{n - 1}
      | r = 2 u_n
      | c = 
}}
{{eqn | l = v_{n - 1} - v_{n - 2}
      | r = 2 u_{n - 1}
      | c = 
}}
{{eqn | o = \vdots
      | c = 
}}
{{eqn | l = v_1 - v_0
      | r = 2 u_1
      | c = 
}}
{{end-eqn}}
Adding all these up, we get:
:$v_n - v_0 = 2 S$
Therefore:
:$S = \dfrac {v_n - v_0} 2$
We have that:
:$v_0 = 0 \times 1 = 0$
and so:
:$S = \dfrac {n \paren {n + 1} } 2$
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
Observe that:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^2 - i^2}
      | r = -\sum_{i \mathop = 1}^n \paren {i^2 - \paren {i + 1} ^2}
}}
{{eqn | r = -\paren {1 - \paren {n + 1}^2}
      | c = Telescoping Series
}}
{{eqn | r = \paren {n + 1}^2 - 1
}}
{{end-eqn}}
Moreover, we have:
:$\paren {i + 1}^2 - i^2 = 2 i + 1$
And also:
:$\paren {n + 1}^2 - 1 = n^2 + 2 n$
Combining these results, we obtain:
{{begin-eqn}}
{{eqn | l = n + 2 \sum_{i \mathop = 1}^n i
      | r = n^2 + 2 n
}}
{{eqn | ll= \leadsto
      | l = 2 \sum_{i \mathop = 1}^n i
      | r = n \left({n + 1}\right)
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop = 1}^n i
      | r = \frac {n \paren {n + 1} } 2
}}
{{end-eqn}}
This concludes the proof.
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
From Faulhaber's Formula:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
      | r = 1^p + 2^p + \cdots + n^p
      | c = 
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
      | c = 
}}
{{end-eqn}}
where $B_k$ are the Bernoulli numbers.
Setting $p = 1$:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^1
      | r = 1 + 2 + \cdots + n
      | c = 
}}
{{eqn | r = \frac {n^2} 2 + \frac {B_1 \, p^{\underline 0} n^1} {1!}
      | c = 
}}
{{eqn | r = \frac {n^2} 2 + \frac 1 2 \frac {1 \times n} 1
      | c = {{Defof|Bernoulli Numbers}}, Number to Power of Zero Falling is One
}}
{{eqn | r = \frac {n^2 + n} 2
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
From Binomial Coefficient with One:
:$\forall k \in \Z, k > 0: \dbinom k 1 = k$
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n k
      | r = \sum_{k \mathop = 1}^n \binom k 1
      | c = Binomial Coefficient with One
}}
{{eqn | r = \binom {n + 1} 2
      | c = Sum of k Choose m up to n
}}
{{eqn | r = \frac {\paren {n + 1} n} 2
      | c = {{Defof|Binomial Coefficient}}
}}
{{end-eqn}}
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
Let $\N_n^* = \set {1, 2, 3, \cdots, n}$ be the initial segment of natural numbers.
Let $A = \set {\tuple {a, b}: a \le b, a, b \in \N_n^*}$
Let $B = \set {\tuple {a, b}: a \ge b, a, b, \in \N_n^*}$
Let $\phi: A \to B$ be the mapping:
:$\map \phi {x, y} = \tuple {y, x}$
By definition of dual ordering, $\phi$ is a bijection:
:$(1): \quad \size A = \size B$
We have:
{{begin-eqn}}
{{eqn | l = A \cup B
      | r = \set {\tuple {a, b}: a, b \in \N_n^*}
}}
{{eqn | l = A \cap B
      | r = \set {\tuple {a, b}: a = b: a, b \in \N_n^*}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \size A + \size B
      | r = \size {A \cup B} + \size {A \cap B}
      | c = Inclusion-Exclusion Principle
}}
{{eqn | r = n^2 + n
      | c = Count of a finite set
}}
{{end-eqn}}
Combined with $\left({1}\right)$ this yields:
:$\size A = \dfrac {n^2 + n} 2 = \dfrac {n \paren {n + 1} } 2$
It remains to prove that:
:$T_n = \size A$
{{begin-eqn}}
{{eqn | l = T_n
      | r = \sum_{i \mathop = 1}^n i
      | c = Definition of $T_n$
}}
{{eqn | r = \sum_{i \mathop \in \N_n^*} i
      | c = 
}}
{{eqn | r = \sum_{i \mathop \in \N_n^*} \size {\set {a: a \in \N_i^*} }
      | c = Count of Finite Set,{{Defof|Initial Segment of One-Based Natural Numbers}}
}}
{{eqn | r = \size {\set {\tuple {a, i} :a\in \N_i^*, i \in \N_n^*} }
      | c = Inclusion-Exclusion Principle: the sets are pairwise disjoint as the second argument in the ordered pair is different
}}
{{eqn | r = \size {\set {\tuple {a, b}: a \le b, a, b \in \N_n^*} }
      | c = Change of Variable, {{Defof|Initial Segment of One-Based Natural Numbers}}
}}
{{eqn | r = \size A
      | c = Definition of $A$
}}
{{end-eqn}}
{{qed}}
Theorem: Closed Form for Triangular Numbers

The closed-form expression for the $n$th triangular number is:
:$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n \paren {2 j - 1}
      | r = n^2
      | c = Odd Number Theorem
}}
{{eqn | ll= \leadsto
      | l = \sum_{j \mathop = 1}^n \paren {2 j - 1} + \sum_{j \mathop = 1}^n 1
      | r = n^2 + n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{j \mathop = 1}^n \paren {2 j}
      | r = n \paren {n + 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{j \mathop = 1}^n j
      | r = \frac {n \paren {n + 1} } 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Sequence of Squares

:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

Proof:
From Binomial Coefficient with One:
:$\dbinom n 1 = n$
From Binomial Coefficient with Two:
:$\dbinom n 2 = \dfrac {n \paren {n - 1} } 2$
Thus:
{{begin-eqn}}
{{eqn | l = 2 \binom n 2 + \binom n 1
      | r = 2 \dfrac {n \paren {n - 1} } 2 + n
      | c = 
}}
{{eqn | r = n \paren {n - 1} + n
      | c = 
}}
{{eqn | r = n^2
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^2
      | r = \sum_{i \mathop = 1}^n \paren {2 \binom i 2 + \binom i 1}
      | c = 
}}
{{eqn | r = 2 \binom {n + 1} 3 + \binom {n + 1} 2
      | c = Sum of Binomial Coefficients over Upper Index
}}
{{eqn | r = \frac {2 \paren {n + 1} n \paren {n - 1} } {3 \times 2 \times 1} + \frac {\paren {n + 1} n} 2
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 1} } 6
      | c = after algebra
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Sequence of Squares

:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

Proof:
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
When $n = 0$, we see from the definition of vacuous sum that:
:$0 = \ds \sum_{i \mathop = 1}^0 i^2 = \frac {0 \paren 1 \paren 1} 6 = 0$
and so $\map P 0$ holds.
=== Basis for the Induction ===
When $n = 1$:
:$\ds \sum_{i \mathop = 1}^1 i^2 = 1^2 = 1$
Now, we have:
:$\ds \frac {n \paren {n + 1} \paren {2 n + 1} } 6 = \frac {1 \paren {1 + 1} \paren {2 \times 1 + 1} } 6 = \frac 6 6 = 1$
and $\map P 1$ is seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{i \mathop = 1}^k i^2 = \frac {k \paren {k + 1} \paren {2 k + 1} } 6$
Then we need to show:
:$\ds \sum_{i \mathop = 1}^{k + 1} i^2 = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 1} } 6$
=== Induction Step ===
This is our induction step:
Using the properties of summation, we have:
:$\ds \sum_{i \mathop = 1}^{k + 1} i^2 = \sum_{i \mathop = 1}^k i^2 + \paren {k + 1}^2$
We can now apply our induction hypothesis, obtaining:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^{k + 1} i^2
      | r = \frac {k \paren {k + 1} \paren {2 k + 1} } 6 + \paren {k + 1}^2
      | c = 
}}
{{eqn | r = \frac {k \paren {k + 1} \paren {2 k + 1} + 6 \paren {k + 1}^2} 6
      | c = 
}}
{{eqn | r = \frac {\paren {k + 1} \paren {k \paren {2 k + 1} + 6 \paren {k + 1} } } 6
      | c = 
}}
{{eqn | r = \frac {\paren {k + 1} \paren {2 k^2 + 7 k + 6} } 6
      | c = 
}}
{{eqn | r = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 1} } 6
      | c = 
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
{{qed}}
Theorem: Sum of Sequence of Squares

:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n 3 i \paren {i + 1}
      | r = n \paren {n + 1} \paren {n + 2}
      | c = Sum from $1$ to $n$ of $r \paren {r + 1}$
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop = 1}^n 3 i^2 + \sum_{i \mathop = 1}^n 3 i
      | r = n \paren {n + 1} \paren {n + 2}
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop = 1}^n 3 i^2
      | r = n \paren {n + 1} \paren {n + 2} - 3 \frac {n \paren {n + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop = 1}^n i^2
      | r = \frac {n \paren {n + 1} \paren {n + 2} } 3 - \frac {n \paren {n + 1} } 2
      | c = 
}}
{{eqn | r = \frac {2 n \paren {n + 1} \paren {n + 2} - 3 n \paren {n + 1} } 6
      | c = 
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {2 \paren {n + 2} - 3} } 6
      | c = 
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 4 - 3} } 6
      | c = 
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 1} } 6
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Sequence of Squares

:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^3 - i^3}
      | r = \sum_{i \mathop = 1}^n \paren {i^3 + 3 i^2 + 3 i + 1 - i^3}
      | c = Binomial Theorem
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {3 i^2 + 3 i + 1}
      | c =
}}
{{eqn | r = 3 \sum_{i \mathop = 1}^n i^2 + 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1
      | c = Summation is Linear
}}
{{eqn | r = 3\sum_{i \mathop = 1}^n i^2 + 3 \frac {n \paren {n + 1} } 2 + n
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \paren {\paren {i + 1}^3 - i^3}
      | r = \paren {n + 1}^3 - n^3 + n^3 - \paren {n - 1}^3 + \paren {n - 1}^3 - \cdots + 2^3 - 1^3
      | c = {{Defof|Summation}}
}}
{{eqn | r = \paren {n + 1}^3 - 1^3
      | c = Telescoping Series: Example 2
}}
{{eqn | r = n^3 + 3n^2 + 3n + 1 - 1
      | c = Binomial Theorem
}}
{{eqn | r = n^3 + 3n^2 + 3n
      | c = 
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = 3 \sum_{i \mathop = 1}^n i^2 + 3 \frac {n \paren {n + 1} } 2 + n
      | r = n^3 + 3 n^2 + 3 n
}}
{{eqn | ll= \leadsto
      | l = 3 \sum_{i \mathop = 1}^n i^2
      | r = n^3 + 3 n^2 + 3 n - 3 \frac {n \paren {n + 1} } 2 - n
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^2
      | r = \frac 1 3 \paren {n^3 + 3 n^2 + 3 n - 3 \frac {n \paren {n + 1} }2 - n}
}}
{{eqn | r = \frac 1 3 \paren {n^3 + 3 n^2 + 3 n - \frac {3 n^2} 2 - \frac {3 n} 2 - n}
}}
{{eqn | r = \frac 1 3 \paren {n^3 + \frac {3 n^2} 2 + \frac n 2}
}}
{{eqn | r = \frac 1 6 n \paren {2 n^2 + 3 n + 1}
}}
{{eqn | r = \frac 1 6 n \paren {n + 1} \paren {2 n + 1}
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Sequence of Squares

:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

Proof:
:File:Sum of Sequences of Squares.jpg
We can observe from the above diagram that:
:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = i}^n j}$
Therefore we have: 
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^2
      | r = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = i}^n j}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = 1}^n j - \sum_{j \mathop = 1}^{i - 1} j}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {\frac {n \paren {n + 1} } 2 - \frac {i \paren {i - 1} } 2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 \sum_{i \mathop = 1}^n i^2 
      | r = n^2 \paren {n + 1} - \sum_{i \mathop = 1}^n i^2 + \sum_{i \mathop = 1}^n i 
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 3 \sum_{i \mathop = 1}^n i^2
      | r = n^2 \paren {n + 1} + \sum_{i \mathop = 1}^n i 
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 3 \sum_{i \mathop = 1}^n i^2 
      | r = n^2 \paren {n + 1} + \frac {n \paren {n + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | ll= \leadsto
      | l = 6 \sum_{i \mathop = 1}^n i^2 
      | r = 2 n^2 \paren {n + 1} + n \paren {n + 1}
      | c = 
}}
{{eqn | r = n \paren {n + 1} \paren {2 n + 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{i \mathop = 1}^n i^2 
      | r = \frac {n \paren {n + 1} \paren {2 n + 1} } 6
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Sequence of Squares

:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

Proof:
From Sum of Consecutive Triangular Numbers is Square:
:$(1): \quad n^2 = T_n + T_{n - 1}$
where $T_n$ is the $n$th triangular number.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^2
      | r = 1 + \paren {T_1 + T_2} + \paren {T_2 + T_3} + \paren {T_3 + T_4} + \cdots + \paren {T_{n - 1} + T_n}
      | c = from $(1)$
}}
{{eqn | r = 1 + T_1 + 2 T_2 + 2 T_3 + 2 T_4 + \cdots + 2 T_{n - 1} + T_n
}}
{{eqn | r = 1 + 2 \paren {T_1 + T_2 + T_3 + T_4 + \cdots + T_n} - T_1 - T_n
}}
{{eqn | r = 1 + 2 \paren {\sum_{i \mathop = 1}^n T_i} - 1 - \frac {n \paren {n + 1} } 2
      | c = Closed Form for Triangular Numbers: $T_1 = 1$ and $T_n = \dfrac {n \paren {n + 1} } 2$
}}
{{eqn | r = 2 \paren {\frac {n \paren {n + 1} \paren {n + 2} } 6} - \frac {n \paren {n + 1} } 2
      | c = Sum of Sequence of Triangular Numbers and simplification
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {2n + 4} } 6 - \frac {n \paren {n + 1} 3} 6
      | c = common denominator
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 1} } 6
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Sequence of Squares

:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

Proof:
From Faulhaber's Formula:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
      | r = 1^p + 2^p + \cdots + n^p
      | c = 
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
      | c = 
}}
{{end-eqn}}
where $B_k$ are the Bernoulli numbers.
Setting $p = 2$:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^2
      | r = 1^2 + 2^2 + \cdots + n^2
      | c = 
}}
{{eqn | r = \frac {n^{2 + 1} } {2 + 1} + \frac {B_1 \, p^{\underline 0} \, n^2} {1!} + \frac {B_2 \, p^{\underline 1} \, n^1} {2!}
      | c = 
}}
{{eqn | r = \frac {n^3} 3 + B_1 n^2 + B_2 n
      | c = {{Defof|Falling Factorial}} and simplification
}}
{{eqn | r = \frac {n^3} 3 + \frac {n^2} 2 + \frac n 6
      | c = {{Defof|Bernoulli Numbers}}
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 1} } 6
      | c = after algebra
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Union is Associative

Set union is associative:
:$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = x \in A \cup \paren {B \cup C}
      | c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoandfrom
      | r = x \in A \lor \paren {x \in B \lor x \in C}
      | c = {{Defof|Set Union}}
}}
{{eqn | o = \leadstoandfrom
      | r = \paren {x \in A \lor x \in B} \lor x \in C
      | c = Rule of Association: Disjunction
}}
{{eqn | o = \leadstoandfrom
      | r = x \in \paren {A \cup B} \cup C
      | c = {{Defof|Set Union}}
}}
{{end-eqn}}
Therefore:
:$x \in A \cup \paren {B \cup C}$ {{iff}} $x \in \paren {A \cup B} \cup C$
Thus it has been shown that:
: $A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$
{{qed}}
Theorem: Pythagoras's Theorem

Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.
Then:
:$a^2 + b^2 = c^2$

Proof:
We start with the algebraic definitions for sine and cosine:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$
From these, we derive the proof that $\cos^2 x + \sin^2 x = 1$.
Then from the Equivalence of Definitions of Trigonometric Functions, we can use the geometric interpretation of sine and cosine:
:400px
:$\sin \theta = \dfrac {\text{Opposite}} {\text{Hypotenuse}}$
:$\cos \theta = \dfrac {\text{Adjacent}} {\text{Hypotenuse}}$
Let $\text{Adjacent} = a, \text{Opposite} = b, \text{Hypotenuse} = c$, as in the diagram at the top of the page.
Thus:
{{begin-eqn}}
{{eqn | l = \cos^2 x + \sin^2 x
      | r = 1
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | ll= \leadsto
      | l = \paren {\frac a c}^2 + \paren {\frac b c}^2
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^2 + b^2
      | r = c^2
      | c = multiplying both sides by $c^2$
}}
{{end-eqn}}
{{qed}}