avewright/proofwiki-math huggingface dataset:
Theorem: Sum of Sequence of k x k!

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n j \times j!
      | r = 1 \times 1! + 2 \times 2! + 3 \times 3! + \dotsb + n \times n!
      | c = 
}}
{{eqn | r = \paren {n + 1}! - 1
      | c = 
}}
{{end-eqn}}

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n j \times j! = \paren {n + 1}! - 1$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = 1 \times 1!
      | r = 1
      | c = 
}}
{{eqn | r = \paren {1 + 1}! - 1
      | c = 
}}
{{end-eqn}}
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown 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 the induction hypothesis:
:$\ds \sum_{j \mathop = 1}^k j \times j! = \paren {k + 1}! - 1$
from which it is to be shown that:
:$\ds \sum_{j \mathop = 1}^{k + 1} j \times j! = \paren {k + 2}! - 1$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^{k + 1} j \times j!
      | r = \sum_{j \mathop = 1}^k j \times j! + \paren {k + 1} \times \paren {k + 1}!
      | c = 
}}
{{eqn | r = \paren {k + 1}! - 1 + \paren {k + 1} \times \paren {k + 1}!
      | c = Induction Hypothesis
}}
{{eqn | r = \paren {k + 1}! \paren {1 + \paren {k + 1} } - 1
      | c = factoring out $\paren {k + 1}!$
}}
{{eqn | r = \paren {k + 1}! \paren {k + 2} - 1
      | c = 
}}
{{eqn | r = \paren {k + 2}! - 1
      | c = 
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 1}: \ds \sum_{j \mathop = 1}^n j \times j! = \paren {n + 1}! - 1$
{{qed}}
Theorem: Sum of Sequence of k x k!

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n j \times j!
      | r = 1 \times 1! + 2 \times 2! + 3 \times 3! + \dotsb + n \times n!
      | c = 
}}
{{eqn | r = \paren {n + 1}! - 1
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn	| l = \sum_{j \mathop = 1}^n j \times j!
	| r = \sum_{j \mathop = 1}^n \paren {j + 1 - 1} \times j!
}}
{{eqn	| r = \sum_{j \mathop = 1}^n \paren {\paren {j + 1} j! - j!}
}}
{{eqn	| r = \sum_{j \mathop = 1}^n \paren {\paren {j + 1}! - j!}
}}
{{eqn	| r = \paren {n + 1}! - 1!
	| c = Telescoping Series: Example 1
}}
{{eqn	| r = \paren {n + 1}! - 1
}}
{{end-eqn}}
{{qed}}
Theorem: Condition for Increasing Binomial Coefficients

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\dbinom n k$ denote a binomial coefficient for $k \in \N$.
Then:
:$\dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \dbinom n k
      | o = <
      | r = \dbinom n {k + 1}
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac {n!} {\paren {n - k}! k!}
      | o = <
      | r = \frac {n!} {\paren {n - k - 1}! \paren {k + 1}!}
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | ll= \leadstoandfrom
      | l = k + 1
      | o = <
      | r = n - k
      | c = Multiplying both sides by $\dfrac {\paren {n - k}! \paren {k + 1}!} {n!}$
}}
{{eqn | ll= \leadstoandfrom
      | l = 2 k
      | o = <
      | r = n - 1
}}
{{eqn | ll= \leadstoandfrom
      | l = k
      | o = <
      | r = \frac {n - 1} 2
}}
{{end-eqn}}
{{qed}}
Theorem: Condition for Increasing Binomial Coefficients

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\dbinom n k$ denote a binomial coefficient for $k \in \N$.
Then:
:$\dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$

Proof:
The proof proceeds by induction on $n$.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$
First we investigate the edge case.
Let $n = 1$.
Then we have:
{{begin-eqn}}
{{eqn | l = \dbinom 1 0
      | r = 1
      | c = Binomial Coefficient with Zero
}}
{{eqn | l = \dbinom 1 1
      | r = 1
      | c = Binomial Coefficient with Self
}}
{{end-eqn}}
Thus we see:
:there are no $k$ such that $0 \le k < \dfrac {1 - 1} 2 = 0$
and:
:there are no $k$ such that $\dbinom 1 k < \dbinom 1 {k + 1}$
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
Let $n = 2$.
Then we have:
{{begin-eqn}}
{{eqn | l = \dbinom 2 0
      | r = 1
      | c = Binomial Coefficient with Zero
}}
{{eqn | l = \dbinom 2 1
      | r = 2
      | c = Binomial Coefficient with One
}}
{{eqn | l = \dbinom 2 2
      | r = 1
      | c = Binomial Coefficient with Self
}}
{{end-eqn}}
Thus we see:
:there is one $k$ such that $0 \le k < \dfrac {2 - 1} 2 = \dfrac 1 2$, and that is $k = 0$
and:
:$\dbinom 2 k < \dbinom 2 {k + 1}$ holds for exactly $k = 0$.
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\dbinom r k < \dbinom r {k + 1} \iff 0 \le k < \dfrac {r - 1} 2$
from which it is to be shown that:
:$\dbinom {r + 1} k < \dbinom {r + 1} {k + 1} \iff 0 \le k < \dfrac r 2$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = \dbinom {r + 1} k
      | r = \dbinom r k + \dbinom r {k - 1}
      | c = Pascal's Rule
}}
{{eqn | o = <
      | r = \dbinom r k + \dbinom r k
      | rr= \iff 0 \le {k - 1} < \dfrac {r - 1} 2
      | c = Induction Hypothesis
}}
{{eqn | o = <
      | r = \dbinom r {k + 1} + \dbinom r k
      | rr= \iff 0 \le {k - 1} < \dfrac {r - 1} 2 \text { and } 0 \le k < \dfrac {r - 1} 2
      | c = Induction Hypothesis
}}
{{eqn | o = <
      | r = \dbinom r {k + 1} + \dbinom r k
      | rr= \iff 0 \le k < \dfrac {r - 1} 2
      | c = 
}}
{{eqn | r = \dbinom {r + 1} {k + 1}
      | rr= \iff 0 \le k < \dfrac {r - 1} 2
      | c = Pascal's Rule
}}
{{end-eqn}}
{{finish|Can't get my head round the inequalities on $k$, needs more thought}}
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{> 0}: \dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$
Theorem: Condition for Equality of Adjacent Binomial Coefficients

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\dbinom n k$ denote a binomial coefficient for $k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {k + 1}$
{{iff}}:
:$n$ is an odd integer
:$k = \dfrac {n - 1} 2$

Proof:
=== Sufficient Condition ===
Let $n$ be odd and $k = \dfrac {n - 1} 2$.
Let $n = 2 m + 1$ for some $m \in \Z_{\ge 0}$.
We have:
{{begin-eqn}}
{{eqn | l = k
      | r = \dfrac {n - 1} 2
      | c = 
}}
{{eqn | r = \dfrac {\paren {2 m + 1} - 1} 2
      | c = 
}}
{{eqn | r = \dfrac {2 m} 2
      | c = 
}}
{{eqn | r = m
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \dbinom n k
      | r = \dbinom {2 m + 1} m
      | c = 
}}
{{eqn | r = \dbinom {2 m + 1} {\paren {2 m + 1} - m}
      | c = Symmetry Rule for Binomial Coefficients
}}
{{eqn | r = \dbinom {2 m + 1} {m + 1}
      | c = 
}}
{{eqn | r = \dbinom n {k + 1}
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Equivalence of Definitions of Component/Equivalence Class equals Union of Connected Sets

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \set {A \subseteq S: x \in A \land A \text{ is connected in } T}$
Let $C = \bigcup \CC_x$
Let $\sim$ be the equivalence relation defined by:
:$y \sim z$ {{iff}} $y$ and $z$ are connected in $T$.
Let $C'$ be the equivalence class of $\sim$ containing $x$.
Then $C = C'$.

Proof:
{{begin-eqn}}
{{eqn | l = y \in C'
      | o = \leadstoandfrom
      | r = \exists B \text{ a connected set of } T, x \in B, y \in B
      | c = Definition of $\sim$
}}
{{eqn | o = \leadstoandfrom
      | r = \exists B \in \CC_x : y \in B
      | c = equivalent definition
}}
{{eqn | o = \leadstoandfrom
      | r = y \in \bigcup \CC_x
      | c = {{Defof|Union of Set of Sets}}
}}
{{eqn | o = \leadstoandfrom
      | r = y \in C
      | c = Definition of $C$
}}
{{end-eqn}}
The result follows.
Theorem: Equivalence of Definitions of Component/Union of Connected Sets is Maximal Connected Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text{ is connected in } T}$
Let $\ds C = \bigcup \CC_x$
Then $C$ is a maximal connected set of $T$.

Proof:
Let $\tilde C$ be any connected set such that:
:$C \subseteq \tilde C$
Then $x \in \tilde C$.
Hence $\tilde C \in \CC_x$.
From Set is Subset of Union, 
:$\tilde C \subseteq C$.
Hence $\tilde C = C$.
It follows that $C$ is a maximal connected set of $T$ by definition.
Theorem: Equivalence of Definitions of Component/Maximal Connected Set is Union of Connected Sets

Let $\tilde C$ be a maximal connected set of $T$ that contains $x$.

Proof:
By definition:
:$\tilde C \in \CC_x$
From Set is Subset of Union:
:$\tilde C \subseteq C$
By maximality of $\tilde C$:
:$\tilde C = C$
Theorem: Binomial Coefficient n Choose j in terms of n-2 Choose r

Let $n \in \Z$ such that $n \ge 4$.
Let $\dbinom n k$ denote a binomial coefficient for $k \in \Z$.
Then:
:$\dbinom n k = \dbinom {n - 2} {k - 2} + 2 \dbinom {n - 2} {k - 1} + \dbinom {n - 2} k$
for $2 \le k \le n - 2$.

Proof:
{{begin-eqn}}
{{eqn | l = \dbinom n k
      | r = \dbinom {n - 1} k + \dbinom {n - 1} {k - 1}
      | c = Pascal's Rule
}}
{{eqn | r = \paren {\dbinom {n - 2} {k - 1} + \dbinom {n - 2} k} + \paren {\dbinom {n - 2} {k - 2} + \dbinom {n - 2} {k - 1} }
      | c = Pascal's Rule (twice)
}}
{{eqn | r = \dbinom {n - 2} {k - 2} + 2 \dbinom {n - 2} {k - 1} + \dbinom {n - 2} k
      | c = simplifying
}}
{{end-eqn}}
In the expression $\dbinom {n - 2} {k - 2} + 2 \dbinom {n - 2} {k - 1} + \dbinom {n - 2} k$ we note that:
:if $k < 2$ then $\dbinom {n - 2} {k - 2}$ has a negative coefficient on the bottom
:if $k > n - 2$ then $\dbinom {n - 2} k$ has a coefficient on the bottom that is greater than $n$.
Hence the usual comfortable range of $k$ is exceeded and so it cannot be guaranteed that the conditions are satisfied for the equation to be true.
If $n \le 3$ then $2 \le k \le n - 2$ cannot be fulfilled.
Hence the bounds on both $k$ and $n$.
{{qed}}
Theorem: Sum of Sequence of Binomial Coefficients by Powers of 2

{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n 2^j \binom n j
      | r = \dbinom n 0 + 2 \dbinom n 1 + 2^2 \dbinom n 2 + \dotsb + 2^n \dbinom n n
      | c = 
}}
{{eqn | r = 3^n
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = 3^n
      | r = \paren {2 + 1}^n
}}
{{eqn | r = \sum_{j \mathop = 0}^n 2^j 1^{n - j} \binom n j
      | c = Binomial Theorem
}}
{{eqn | r =\sum_{j \mathop = 0}^n 2^j \binom n j
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Sequence of Binomial Coefficients by Powers of 2

{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n 2^j \binom n j
      | r = \dbinom n 0 + 2 \dbinom n 1 + 2^2 \dbinom n 2 + \dotsb + 2^n \dbinom n n
      | c = 
}}
{{eqn | r = 3^n
      | c = 
}}
{{end-eqn}}

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^n 2^j \binom n j = 3^n$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^0 2^j \binom n j
      | r = \dbinom 0 0
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = 3^0
      | c = 
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^1 2^j \binom n j
      | r = \dbinom 1 0 + 2 \dbinom 1 1
      | c = 
}}
{{eqn | r = 1 + 2
      | c = 
}}
{{eqn | r = 3^1
      | c = 
}}
{{end-eqn}}
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown 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 the induction hypothesis:
:$\ds \sum_{j \mathop = 0}^k 2^j \binom k j = 3^k$
from which it is to be shown that:
:$\ds \sum_{j \mathop = 0}^{k + 1} 2^j \binom {k + 1} j = 3^{k + 1}$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^{k + 1} 2^j \binom {k + 1} j
      | r = \binom {k + 1} 0 + \sum_{j \mathop = 1}^k 2^j \binom {k + 1} j + 2^{k + 1} \dbinom {k + 1} {k + 1}
      | c = separating out top and bottom indices
}}
{{eqn | r = 1 + \sum_{j \mathop = 1}^k 2^j \binom {k + 1} j + 2^{k + 1}
      | c = Binomial Coefficient with Zero, Binomial Coefficient with Self
}}
{{eqn | r = 1 + \sum_{j \mathop = 1}^k 2^j \paren {\binom k j + \binom k {j - 1} } + 2^{k + 1}
      | c = Pascal's Rule
}}
{{eqn | r = 1 + \sum_{j \mathop = 1}^k 2^j \binom k j + 2 \sum_{j \mathop = 1}^k 2^{j - 1} \binom k {j - 1} + 2^{k + 1}
      | c = 
}}
{{eqn | r = 1 + \sum_{j \mathop = 1}^k 2^j \binom k j + 2 \sum_{j \mathop = 0}^{k - 1} 2^j \binom k j + 2^{k + 1}
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = 1 + \paren {\sum_{j \mathop = 0}^k 2^j \binom k j - 2^0 \binom k 0} + 2 \paren {\sum_{j \mathop = 0}^k 2^j \binom k j - 2^k \binom k k} + 2^{k + 1}
      | c = rectifying the end points
}}
{{eqn | r = \sum_{j \mathop = 0}^k 2^j \binom k j + 2 \sum_{j \mathop = 0}^k 2^j \binom k j - 2 \times 2^k + 2^{k + 1}
      | c = simplifying
}}
{{eqn | r = 3^k + 2 \times 3^k - 2 \times 2^k + 2^{k + 1}
      | c = Induction Hypothesis
}}
{{eqn | r = 3^{k + 1}
      | c = tidying up
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \ds \sum_{j \mathop = 0}^n 2^j \binom n j = 3^n$
{{qed}}
Theorem: Sum of Sequence of n Choose 2

Let $n \in \Z$ be an integer such that $n \ge 2$.
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 2}^n \dbinom j 2
      | r = \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2
      | c = 
}}
{{eqn | r = \dbinom {n + 1} 3
      | c = 
}}
{{end-eqn}}
where $\dbinom n j$ denotes a binomial coefficient.

Proof:
We can rewrite the {{LHS}} as:
:$\ds \sum_{j \mathop = 0}^m \dbinom {2 + j} 2$
where $m = n - 2$.
From Rising Sum of Binomial Coefficients:
:$\ds \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$
The result follows by setting $n = 2$ and changing the upper index from $m$ to $n - 2$.
{{qed}}
Theorem: Sum of Sequence of n Choose 2

Let $n \in \Z$ be an integer such that $n \ge 2$.
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 2}^n \dbinom j 2
      | r = \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2
      | c = 
}}
{{eqn | r = \dbinom {n + 1} 3
      | c = 
}}
{{end-eqn}}
where $\dbinom n j$ denotes a binomial coefficient.

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$
=== Basis for the Induction ===
$\map P 2$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 2}^2 \dbinom j 2
      | r = \dbinom 2 2
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = \dbinom 3 3
      | c = 
}}
{{end-eqn}}
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{j \mathop = 2}^k \dbinom j 2 = \dbinom {k + 1} 3$
from which it is to be shown that:
:$\ds \sum_{j \mathop = 2}^{k + 1} \dbinom j 2 = \dbinom {k + 2} 3$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 2}^{k + 1} \dbinom j 2
      | r = \sum_{j \mathop = 2}^k \dbinom j 2 + \dbinom {k + 1} 2
      | c = 
}}
{{eqn | r = \dbinom {k + 1} 3 + \dbinom {k + 1} 2
      | c = Induction Hypothesis
}}
{{eqn | r = \dbinom {k + 2} 3
      | c = Pascal's Rule
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 2}: \ds \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$
Theorem: If n is Triangular then so is 25n + 3

Let $n$ be a triangular number.
Then $25 n + 3$ is also triangular.

Proof:
Let $n$ be triangular.
Then:
:$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$
So:
{{begin-eqn}}
{{eqn | l = 25 n + 3
      | r = 25 \frac {k \paren {k + 1} } 2 + 3
      | c = 
}}
{{eqn | r = \frac {25 k^2 + 25 k + 6} 2
      | c = 
}}
{{eqn | r = \frac {\paren {5 k + 2} \paren {5 k + 3} } 2
      | c = 
}}
{{end-eqn}}
which is triangular.
{{qed}}
Theorem: If n is Triangular then so is 49n + 6

Let $n$ be a triangular number.
Then $49 n + 6$ is also triangular.

Proof:
Let $n$ be triangular.
Then:
:$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$
So:
{{begin-eqn}}
{{eqn | l = 49 n + 6
      | r = 49 \frac {k \paren {k + 1} } 2 + 6
      | c = 
}}
{{eqn | r = \frac {49 k^2 + 49 k + 12} 2
      | c = 
}}
{{eqn | r = \frac {\paren {7 k + 3} \paren {7 k + 4} } 2
      | c = 
}}
{{end-eqn}}
which is triangular.
{{qed}}
Theorem: Sum of Sequence of Triangular Numbers

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $T_n$ denote the $n$th triangular number.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n T_j
      | r = T_1 + T_2 + T_3 + \dotsb + T_n
      | c = 
}}
{{eqn | r = \dfrac {n \paren {n + 1} \paren {n + 2} } 6
      | c = 
}}
{{end-eqn}}

Proof:
From Sum of Sequence of n Choose 2 we have:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 2}^n \dbinom j 2
      | r = \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2
      | c = 
}}
{{eqn | r = \dbinom {n + 1} 3
      | c = 
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 2}^{n + 1} \dbinom j 2
      | r = \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2 + \dbinom {n + 1} 2
      | c = 
}}
{{eqn | r = \dbinom {n + 2} 3
      | c = 
}}
{{end-eqn}}
But we have that:
:$\dbinom {n + 2} 3 = \dfrac {\paren {n + 2} \paren {n + 1} n} {3 \times 2 \times 1}$
and from Binomial Coefficient with Two:
:$T_n = \dfrac {\paren {n + 1} n} 2 = \dbinom {n + 1} 2$
The result follows.
{{qed}}
Theorem: Sum of Sequence of Triangular Numbers

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $T_n$ denote the $n$th triangular number.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 1}^n T_j
      | r = T_1 + T_2 + T_3 + \dotsb + T_n
      | c = 
}}
{{eqn | r = \dfrac {n \paren {n + 1} \paren {n + 2} } 6
      | c = 
}}
{{end-eqn}}

Proof:
First let $n$ be even.
Thus we have:
:$n = 2 m$
Then:
{{begin-eqn}}
{{eqn | l = T_1 + T_2 + T_3 + \dotsb + T_{2 m}
      | r = \paren {T_1 + T_2} + \paren {T_3 + T_4} + \dotsb + \paren {T_{2 m - 1} + T_{2 m} }
      | c = 
}}
{{eqn | r = 2^2 + 4^2 + \dotsb + \paren {2 m}^2
      | c = Sum of Consecutive Triangular Numbers is Square
}}
{{eqn | r = 2^2 \times 1^2 + 2^2 \times 2^2 + \dotsb + 2^2 \times m^2
      | c = Sum of Consecutive Triangular Numbers is Square
}}
{{eqn | r = 2^2 \paren {1^2 + 2^2 + \dotsb + m^2}
      | c = 
}}
{{eqn | r = 4 \frac {m \paren {m + 1} \paren {2 m + 1} } 6
      | c = 
}}
{{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6
      | c = 
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6
      | c = 
}}
{{end-eqn}}
Now let $n$ be odd.
Thus we have:
:$n = 2 m + 1$
Then:
{{begin-eqn}}
{{eqn | l = T_1 + T_2 + T_3 + \dotsb + T_{2 m}
      | r = \paren {T_1 + T_2} + \paren {T_3 + T_4} + \dotsb + \paren {T_{2 m - 1} + T_{2 m} } + T_{2 m + 1}
      | c = 
}}
{{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6 + T_{2 m + 1}
      | c = from above
}}
{{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6 + \frac {\paren {2 m + 1} \paren {2 m + 2} } 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} + \paren {6 m + 3} \paren {2 m + 2} } 6
      | c = 
}}
{{eqn | r = \frac {\paren {2 m + 2} \paren {2 m \paren {2 m + 1} + \paren {6 m + 3} } } 6
      | c = 
}}
{{eqn | r = \frac {\paren {2 m + 2} \paren {4 m^2 + 8 m + 3} } 6
      | c = 
}}
{{eqn | r = \frac {\paren {2 m + 2} \paren {2 m + 1} \paren {2 m + 3} } 6
      | c = 
}}
{{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square of Odd Multiple of 3 is Difference between Triangular Numbers

Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $T_n$ denote the $n$th triangular number.
Let $m = 2 n + 1$ be an odd integer
Then:
:$\paren {3 m}^2 = T_{9 n + 4} - T_{3 n + 1}$

Proof:
{{begin-eqn}}
{{eqn | l = T_{9 n + 4} - T_{3 n + 1}
      | r = \dfrac {\paren {9 n + 4} \paren {9 n + 5} } 2 - \dfrac {\paren {3 n + 1} \paren {3 n + 2} } 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = \dfrac {\paren {81 n^2 + 81 n + 20} - \paren {9 n^2 + 9 n + 2} } 2
      | c = 
}}
{{eqn | r = \dfrac {72 n^2 + 72 n + 18} 2
      | c = 
}}
{{eqn | r = 36 n^2 + 36 n + 9
      | c = 
}}
{{eqn | r = 9 \paren {4 n^2 + 4 n + 1}
      | c = 
}}
{{eqn | r = 9 \paren {2 n + 1}^2
      | c = 
}}
{{eqn | r = 9 m^2
      | c = 
}}
{{eqn | r = \paren {3 m}^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square Sum of Three Consecutive Triangular Numbers

Let $T_n$ denote the $n$th triangular number for $n \in \Z_{>0}$ a (strictly) positive integer.
Let $T_n + T_{n + 1} + T_{n + 2}$ be a square number.
Then at least one value of $n$ fulfils this condition:
:$n = 5$

Proof:
Let $T_n + T_{n + 1} + T_{n + 2} = m^2$ for some $m \in \Z_{>0}$.
We have:
{{begin-eqn}}
{{eqn | l = T_n + T_{n + 1} + T_{n + 2}
      | r = \dfrac {n \paren {n + 1} } 2 + \dfrac {\paren {n + 1} \paren {n + 2} } 2 + \dfrac {\paren {n + 2} \paren {n + 3} } 2
      | c = 
}}
{{eqn | r = \dfrac {n \paren {n + 1} + \paren {n + 1} \paren {n + 2} + \paren {n + 2} \paren {n + 3} } 8
      | c = 
}}
{{eqn | r = \dfrac {n^2 + n + n^2 + 3 n + 2 + n^2 + 5 n + 6} 2
      | c = multiplying out
}}
{{eqn | r = \dfrac {3 n^2 + 9 n + 8} 2
      | c = multiplying out
}}
{{end-eqn}}
Thus we need to find $n$ such that $\dfrac {3 n^2 + 9 n + 8} 2$ is a square number.
We see that:
{{begin-eqn}}
{{eqn | l = \dfrac {3 \times 5^2 + 9 \times 5 + 8} 2
      | r = \dfrac {75 + 45 + 8} 2
      | c = 
}}
{{eqn | r = \dfrac {128} 2
      | c = 
}}
{{eqn | r = 64
      | c = 
}}
{{end-eqn}}
Thus $n$ appears to satisfy the conditions.
It remains for us to check:
{{begin-eqn}}
{{eqn | l = T_5 + T_6 + T_7
      | r = 15 + 21 + 28
      | c = 
}}
{{eqn | r = 64
      | c = 
}}
{{eqn | r = 8^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Square Product of Three Consecutive Triangular Numbers

Let $T_n$ denote the $n$th triangular number for $n \in \Z_{>0}$ a (strictly) positive integer.
Let $T_n \times T_{n + 1} \times T_{n + 2}$ be a square number.
Then at least one value of $n$ fulfils this condition:
:$n = 3$

Proof:
Let $T_n \times T_{n + 1} \times T_{n + 2} = m^2$ for some $m \in \Z_{>0}$.
We have:
{{begin-eqn}}
{{eqn | l = T_n \times T_{n + 1} \times T_{n + 2}
      | r = \dfrac {n \paren {n + 1} } 2 \dfrac {\paren {n + 1} \paren {n + 2} } 2 \dfrac {\paren {n + 2} \paren {n + 3} } 2
      | c = 
}}
{{eqn | r = \dfrac {n \paren {n + 1}^2 \paren {n + 2}^2 \paren {n + 3} } 8
      | c = 
}}
{{eqn | r = \paren {\dfrac {\paren {n + 1} \paren {n + 2} } 2}^2 \dfrac {n \paren {n + 3} } 2
      | c = 
}}
{{end-eqn}}
Thus we need to find $n$ such that $\dfrac {n \paren {n + 3} } 2$ is a square number.
We see that:
{{begin-eqn}}
{{eqn | l = 3 \times \paren {3 + 3}
      | r = 3 \times 6
      | c = 
}}
{{eqn | r = 2 \times 3^2
      | c = 
}}
{{end-eqn}}
Thus $n = 3$ appears to satisfy the conditions.
It remains for us to check:
{{begin-eqn}}
{{eqn | l = T_3 \times T_4 \times T_5
      | r = 6 \times 10 \times 15
      | c = 
}}
{{eqn | r = \paren {2 \times 3} \times {2 \times 5} \times {3 \times 5}
      | c = 
}}
{{eqn | r = \paren {2 \times 3 \times 5}^2
      | c = 
}}
{{eqn | r = 30^2
      | c = 
}}
{{eqn | r = 900
      | c = 
}}
{{end-eqn}}
{{qed}}