nyuuzyou/wikis huggingface dataset:
Sum of Sequence of Squares/Proof by Summation of Summations

== Theorem ==
== Proof ==
:
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: 
Category:Sum of Sequence of Squares
Sum of Sequence of Squares/Proof by Telescoping Series

== Theorem ==
== Proof ==
Observe that:
:$3 i \paren {i + 1} = i \paren {i + 1} \paren {i + 2} - i \paren {i + 1} \paren {i - 1}$
That is:
:$(1): \quad 6 T_i = \paren {i + 1} \paren {\paren {i + 1} + 1} \paren {\paren {i + 1} - 1} - i \paren {i + 1} \paren {i - 1}$
Then:
 
where $T_n$ is the $n$th Triangular number.
Then:
Category:Sum of Sequence of Squares
Sum of Sequence of Squares/Proof using Bernoulli Numbers

== Theorem ==
== Proof ==
From Faulhaber's Formula:
where $B_k$ are the Bernoulli numbers.
Setting $p = 2$:
Hence the result.
== Also see ==
* Faulhaber's Formula
== Sources ==
* : $\S 19$: Sums of Powers of Positive Integers: $19.10$
* : $\S 21$: Series of Constants: Sums of Powers of Positive Integers: $21.10.$
Category:Sum of Sequence of Squares
Sum of Sequence of Squares of Fibonacci Numbers

== Theorem ==
Let $F_k$ be the $k$th Fibonacci number.
Then:
:$\forall n \ge 1: \ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
That is:
:${F_1}^2 + {F_2}^2 + {F_3}^2 + \cdots + {F_n}^2 = F_n F_{n + 1}$
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
=== Basis for the Induction ===
$\map P 1$ is the case ${F_1}^2 = 1 = F_3 - 1$, which holds from the definition of Fibonacci numbers.
demonstrating that $\map P 1$ holds.
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:
:$\ds \sum_{j \mathop = 1}^k {F_j}^2 = F_k F_{k + 1}$
Then we need to show:
:$\ds \sum_{j \mathop = 1}^{k + 1} {F_j}^2 = F_{k + 1} F_{k + 2}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \ge 1: \ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
== Sources ==
* : $5$
* : $5$
Category:Fibonacci Numbers
Category:Sums of Sequences
Category:Proofs by Induction
Sum of Sequence of Squares of Primes

== Theorem ==
Let $S = \sequence {s_n}$ be the integer sequence defined as:
:$\ds s_n = \sum_{i \mathop = 1}^n {p_i}^2$
where $P_i$ denotes the $i$th prime number.
Then $S$ begins:
:$4, 13, 38, 87, 208, 377, 666, 1027, 1556, 2397, 3358, 4727, 6408, 8257, 10466, \ldots$
== Sources ==
* : $666$
Category:Prime Numbers
Category:Square Numbers
Sum of Sequence of Triangular Numbers/Proof 1

== Theorem ==
== Proof ==
From Sum of Sequence of n Choose 2 we have:
and so:
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.
Category:Sum of Sequence of Triangular Numbers
Sum of Sequence of Triangular Numbers/Proof 2

== Theorem ==
== Proof ==
First let $n$ be even.
Thus we have:
:$n = 2 m$
Then:
Now let $n$ be odd.
Thus we have:
:$n = 2 m + 1$
Then:
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $3$
Category:Sum of Sequence of Triangular Numbers
Sum of Sequence of k x k!

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $7$
Category:Factorials
Category:Sum of Sequence of k x k!
Sum of Sequence of k x k!/Proof 1

== Theorem ==
== 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:
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:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 1}: \ds \sum_{j \mathop = 1}^n j \times j! = \paren {n + 1}! - 1$
Category:Sum of Sequence of k x k!
Sum of Sequence of n Choose 2/Proof 1

== Theorem ==
== Proof ==
We can rewrite the 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 $n - 2$.
Category:Sum of Sequence of n Choose 2
Sum of Sequence of n Choose 2/Proof 2

== Theorem ==
== 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:
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:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 2}: \ds \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $4 \ \text {(a)}$
Category:Sum of Sequence of n Choose 2
Sum of Sequence of n by 2 to the Power of n

== Theorem ==
:$\ds \sum_{j \mathop = 0}^n j \, 2^j = n \, 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $15$
Category:Arithmetic-Geometric Sequences
Category:Sum of Sequence of n by 2 to the Power of n
Sum of Sequence of n by 2 to the Power of n/Proof 1

== Theorem ==
== Proof ==
From Sum of Arithmetic-Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^n \paren {a + j d} r^j = \frac {a \paren {1 - r^{n + 1} } } {1 - r} + \frac {r d \paren {1 - \paren {n + 1} r^n + n r^{n + 1} } } {\paren {1 - r}^2}$
Hence:
Hence the result.
Category:Sum of Sequence of n by 2 to the Power of n
Sum of Sequence of n by 2 to the Power of n/Proof 2

== Theorem ==
== Proof ==
From Sum of Sequence of Power by Index:
:$\ds \sum_{j \mathop = 0}^n j x^j = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}$
Hence:
Hence the result.
Category:Sum of Sequence of n by 2 to the Power of n
Sum of Sequences of Fifth and Seventh Powers

== Theorem ==
:$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$
=== Basis for the Induction ===
So $\map P 1$ has been demonstrated 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:
:$\ds \sum_{i \mathop = 1}^k i^5 + \sum_{i \mathop = 1}^k i^7 = 2 \paren {\sum_{i \mathop = 1}^k i}^4$
Then we need to show:
:$\ds \sum_{i \mathop = 1}^{k + 1} i^5 + \sum_{i \mathop = 1}^{k + 1} i^7 = 2 \paren {\sum_{i \mathop = 1}^{k + 1} i}^4$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N_{>0}: \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Exercise $18.10 \ \text{(d)}$
Category:Sums of Sequences
Category:Fifth Powers
Category:Seventh Powers
Sum of Sequentially Computable Real-Valued Functions is Sequentially Computable

== Theorem ==
Let $D \subseteq \R^n$ be a subset of real cartesian $n$-space.
Let $f, g : D \to \R$ be sequentially computable.
Then, $h : D \to \R$ defined as:
:$\map h \bsx = \map f \bsx + \map g \bsx$
is sequentially computable.
== Proof ==
Follows immediately from:
* Real Addition is Sequentially Computable
* Composition of Sequentially Computable Real-Valued Functions is Sequentially Computable
Category:Computability Theory
Sum of Series of Product of Power and Cosine

== Theorem ==
Let $r \in \R$ such that $\size r < 1$.
Then:
== Proof ==
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
== Also see ==
* Sum of Series of Product of Power and Sine
* Sum of Infinite Series of Product of Power and Cosine
* Sum of Infinite Series of Product of Power and Sine
== Sources ==
* : $\S 19$: Miscellaneous Series: $19.43$
* : $\S 21$: Series of Constants: Miscellaneous Series: $21.43.$
Category:Cosine Function
Category:Trigonometric Series
Sum of Series of Product of Power and Sine

== Theorem ==
Let $r \in \R$ such that $\size r < 1$.
Then:
== Proof ==
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
== Also see ==
* Sum of Series of Product of Power and Cosine
* Sum of Infinite Series of Product of Power and Sine
* Sum of Infinite Series of Product of Power and Cosine
== Sources ==
* : $\S 19$: Miscellaneous Series: $19.44$
* : $\S 21$: Series of Constants: Miscellaneous Series: $21.44.$
Category:Sine Function
Category:Trigonometric Series
Sum of Set and Open Set in Topological Vector Space is Open

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $A$ and $B$ be subsets of $X$, with $A$ an open set.
Then the linear combination $A + B$ is open.
== Proof ==
It can be shown that: 
:$\ds A + B = \bigcup_{b \mathop \in B} \paren {A + b}$
If: 
:$\ds v \in \bigcup_{b \mathop \in B} \paren {A + b}$
then $v = a + b$ for some $a \in A$ and $b \in B$.
So $v \in A + B$, giving: 
:$\ds \bigcup_{b \mathop \in B} \paren {A + b} \subseteq A + B$
Conversely, suppose that $v \in A + B$.
Then there exists $a \in A$ and $b \in B$ such that $v = a + b$. 
Then, we have $v \in A + b$, and so $v \in A + B$.
We therefore obtain: 
:$\ds A + B \subseteq \bigcup_{b \mathop \in B} \paren {A + b}$
So, by the definition of set equality, we have: 
:$\ds A + B = \bigcup_{b \mathop \in B} \paren {A + b}$
From Translation of Open Set in Topological Vector Space is Open:
:$A + b$ is open for each $b \in B$.
So $A + B$ is the union of open sets. 
Since topologies are closed under union, we therefore have that $A + B$ is open.
Category:Topological Vector Spaces
Sum of Sides of Triangle in terms of Circumradius and Half Angle Cosines

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$a + b + c = 8 R \cos \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2$
where $R$ denotes the circumradius of $\triangle ABC$.
== Proof ==
From Law of Sines:
:$\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$
Hence:
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $14$.
Category:Triangles
Category:Cosine Function
Sum of Sines of Angles in Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\sin A + \sin B + \sin C = 4 \cos \dfrac A 2 \cos \dfrac B 2 \cos \dfrac C 2$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(32)$
Category:Sum of Sines of Angles in Triangle
Category:Triangles
Category:Sine Function
Sum of Sines of Arithmetic Sequence of Angles

== Theorem ==
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
=== Formulation 1 ===
=== Formulation 2 ===
== Also see ==
* Sum of Cosines of Arithmetic Sequence of Angles
Category:Sine Function
Sum of Sines of Arithmetic Sequence of Angles/Formulation 1

== Theorem ==
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
== Proof ==
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 1:
:$\ds \sum_{k \mathop = 0}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} } \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} }$
It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the is not defined.
From Euler's Formula, this can be expressed as:
:$\ds \sum_{k \mathop = 0}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \paren {\map \cos {\theta + \frac {n \alpha} 2} + i \map \sin {\theta + \frac {n \alpha} 2} }$
Equating imaginary parts:
:$\ds \sum_{k \mathop = 0}^n \map \sin {\theta + k \alpha} = \frac {\map \sin {\alpha \paren {n + 1} / 2} } {\map \sin {\alpha / 2} } \map \sin {\theta + \frac {n \alpha} 2}$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $143 \ \text{(b)}$
Category:Sine Function
Sum of Sines of Arithmetic Sequence of Angles/Formulation 2

== Theorem ==
Let $\alpha \in \R$ be a real number such that $\alpha \ne 2 \pi k$ for $k \in \Z$.
Then:
== Proof ==
From Sum of Complex Exponentials of i times Arithmetic Sequence of Angles: Formulation 2:
:$\ds \sum_{k \mathop = 1}^n e^{i \paren {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
It is noted that, from Sine of Multiple of Pi, when $\alpha = 2 \pi k$ for $k \in \Z$, $\map \sin {\alpha / 2} = 0$ and the is not defined.
From Euler's Formula, this can be expressed as:
:$\ds \sum_{k \mathop = 1}^n \paren {\map \cos {\theta + k \alpha} + i \map \sin {\theta + k \alpha} } = \paren {\map \cos {\theta + \frac {n + 1} 2 \alpha} + i \map \sin {\theta + \frac {n + 1} 2 \alpha} } \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
Equating imaginary parts:
:$\ds \sum_{k \mathop = 1}^n \map \sin {\theta + k \alpha} = \map \sin {\theta + \frac {n + 1} 2 \alpha} \frac {\map \sin {n \alpha / 2} } {\map \sin {\alpha / 2} }$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $1$
Category:Sine Function
Sum of Sines of Fractions of Pi

== Theorem ==
Let $n \in \Z$ such that $n > 1$.
Then:
:$\ds \sum_{k \mathop = 1}^{n - 1} \sin \frac {2 k \pi} n = 0$
== Proof ==
Consider the equation:
:$z^n - 1 = 0$
whose solutions are the complex roots of unity:
:$1, e^{2 \pi i / n}, e^{4 \pi i / n}, e^{6 \pi i / n}, \ldots, e^{2 \paren {n - 1} \pi i / n}$
By Sum of Roots of Polynomial:
:$1 + e^{2 \pi i / n} + e^{4 \pi i / n} + e^{6 \pi i / n} + \cdots + e^{2 \paren {n - 1} \pi i / n} = 0$
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
from which comes:
:$\paren {1 + \cos \dfrac {2 \pi} n + \cos \dfrac {4 \pi} n + \cdots + \cos \dfrac {2 \paren {n - 1} \pi} n} + i \paren {\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n} = 0$
Equating imaginary parts:
:$\sin \dfrac {2 \pi} n + \sin \dfrac {4 \pi} n + \cdots + \sin \dfrac {2 \paren {n - 1} \pi} n = 0$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: The $n$th Roots of Unity: $38 \ \text{(b)}$
Category:Sine Function
Sum of Sines of Twice Angles of Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\sin 2 A + \sin 2 B + \sin 2 C = 4 \sin A \sin B \sin C$
== Proof ==
First we note that:
That is, $C$ is the supplement of $A + B$.
Then:
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $18$.
Category:Triangles
Category:Sine Function
Sum of Squared Deviations from Mean

== Theorem ==
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a set of real numbers.
Let $\overline x$ denote the arithmetic mean of $S$.
Then: 
:$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n \paren { {x_i}^2 - {\overline x}^2}$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
Category:Arithmetic Mean
Category:Sum of Squared Deviations from Mean
Sum of Squared Deviations from Mean/Corollary 1

== Theorem ==
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a set of real numbers.
Let $\overline x$ denote the arithmetic mean of $S$.
Then: 
:$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n x_i^2 - n \overline x^2$
 
== Proof ==
Category:Sum of Squared Deviations from Mean
Sum of Squared Deviations from Mean/Corollary 2

== Theorem ==
Let $S = \set {x_1, x_2, \ldots, x_n}$ be a set of real numbers.
Let $\overline x$ denote the arithmetic mean of $S$.
Then: 
:$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n x_i^2 - \frac 1 n \paren {\sum_{i \mathop = 1}^n x_i}^2$
 
== Proof ==
For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.
Then:
Category:Sum of Squared Deviations from Mean
Sum of Squared Deviations from Mean/Proof 2

== Theorem ==
== Proof ==
In this context, $x_1, x_2, \ldots, x_n$ are instances of a discrete random variable.
Hence the result Variance as Expectation of Square minus Square of Expectation can be applied:
:$\var X = \expect {X^2} - \paren {\expect X}^2$
which means the same as this but in the language of probability theory.
Category:Sum of Squared Deviations from Mean
Sum of Squares of Binomial Coefficients

== Theorem ==
:$\ds \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$
where $\dbinom n i$ denotes a binomial coefficient.
== Combinatorial Proof ==
== Inductive Proof ==
== Algebraic Proof ==
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.12$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.12.$
Category:Sum of Squares of Binomial Coefficients
Category:Binomial Coefficients
Sum of Squares of Binomial Coefficients/Algebraic Proof

== Theorem ==
== Proof ==
Consider the Binomial Theorem:
Let $m$ be the coefficient of $x^n$ in the expansion of $\paren {1 + x}^{2 n} = \paren {1 + x}^n \times \paren {1 + x}^n$.
For all $k \in \set {0, 1, \ldots, n}$, by matching up the coefficient of $x^k$ with that of $x^{n - k}$ in $(1)$, we have:
:$m = \ds \sum_{k \mathop = 0}^n \dbinom n k^2$
But from the Binomial Theorem, this given by:
:$m = \dbinom {2 n} n$
Hence the result.
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Relations between coeffficients
Category:Sum of Squares of Binomial Coefficients
Sum of Squares of Binomial Coefficients/Combinatorial Proof

== Theorem ==
== Proof ==
Consider the number of paths in the integer lattice from $\tuple {0, 0}$ to $\tuple {n, n}$ using only single steps of the form:
:$\tuple {i, j} \to \tuple {i + 1, j}$
:$\tuple {i, j} \to \tuple {i, j + 1}$
that is, either to the right or up.
This process takes $2 n$ steps, of which $n$ are steps to the right.
Thus the total number of paths through the graph is equal to $\dbinom {2 n} n$.
Now let us count the paths through the grid by first counting the paths:
:$(1): \quad$ from $\tuple {0, 0}$ to $\tuple {k, n - k}$
and then the paths:
:$(2): \quad$ from $\tuple {k, n - k}$ to $\tuple {n, n}$.
Note that each of these paths is of length $n$.
Since each path is $n$ steps long, every endpoint will be of the form $\tuple {k, n - k}$ for some $k \in \set {1, 2, \ldots, n}$, representing $k$ steps right and $n-k$ steps up.
Note that the number of paths through $\tuple {k, n - k}$ is equal to $\dbinom n k$, since we are free to choose the $k$ steps right in any order.
We can also count the number of $n$-step paths from the point $\tuple {k, n - k}$ to $\tuple {n, n}$.
These paths will be composed of $n - k$ steps to the right and $k$ steps up.
Therefore the number of these paths is equal to $\dbinom n {n - k} = \dbinom n k$.
Thus the total number of paths from $\tuple {0, 0}$ to $\tuple {n, n}$ that pass through $\tuple {k, n - k}$ is equal to the product of:
:the number of possible paths from $\tuple {0, 0}$ to $\tuple {k, n - k}$, which equals $\dbinom n k$
and:
:the number of possible paths from $\tuple {k, n - k}$ to $\tuple {n, n}$, which equals $\dbinom n k$.
So the total number of paths through $\tuple {k, n - k}$ is equal to $\dbinom n k^2$.
Summing over all possible values of $k \in 0, \ldots, n$ gives the total number of paths.
Thus we get:
:$\ds \sum_{k \mathop = 0}^n \binom n k^2 = \binom {2 n} n$
Category:Sum of Squares of Binomial Coefficients
Sum of Squares of Binomial Coefficients/Inductive Proof

== Theorem ==
== Proof ==
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$
$\map P 0$ is true, as this just says:
:$\dbinom 0 0^2 = 1 = \dbinom {2 \times 0} 0$
This holds by definition.
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$\dbinom 1 0^2 + \dbinom 1 1^2 = 1^2 + 1^2 = 2 = \dbinom 2 1$
This also holds by definition.
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:
:$\ds \sum_{i \mathop = 0}^k \binom k i^2 = \binom {2 k} k$
Then we need to show:
:$\ds \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i^2 = \binom {2 \paren {k + 1} } {k + 1}$
=== Induction Step ===
This is our induction step:
Now we look at $\ds 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}$.
Using the Chu-Vandermonde Identity:
:$\ds \sum_i \binom r i \binom s {n - i} = \binom {r + s} n$
From the Symmetry Rule for Binomial Coefficients, this can be written:
:$\ds \sum_i \binom r i \binom s {s - n + i} = \binom {r + s} n$
Putting $r = k, s = k, s - n = -1$ from whence $n = k + 1$:
:$\ds \sum_i \binom k i \binom k {i - 1} = \binom {2 k} {k + 1}$
So:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$
Category:Sum of Squares of Binomial Coefficients
Sum of Squares of Complex Moduli of Sum and Differences of Complex Numbers

== Theorem ==
Let $\alpha, \beta \in \C$ be complex numbers.
Then:
:$\cmod {\alpha + \beta}^2 + \cmod {\alpha - \beta}^2 = 2 \cmod \alpha^2 + 2 \cmod \beta^2$
== Proof ==
Let:
:$\alpha = x_1 + i y_1$
:$\beta = x_2 + i y_2$
Then:
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $9$
Category:Complex Modulus
Sum of Squares of Consecutive Fibonacci Numbers

== Theorem ==
:${F_n}^2 + {F_{n + 1} }^2 = F_{2 n + 1}$
where $F_n$ denotes the $n$th Fibonacci number.
== Proof ==
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $18$
Category:Fibonacci Numbers
Sum of Squares of Divisors of 24 and 26 are Equal

== Theorem ==
The sum of the squares of the divisors of $24$ equals the sum of the squares of the divisors of $26$:
:$\map {\sigma_2} {24} = \map {\sigma_2} {26}$
where $\sigma_\alpha$ denotes the divisor function.
== Proof ==
The divisors of $24$ are:
:$1, 2, 3, 4, 6, 8, 12, 24$
The divisors of $26$ are:
:$1, 2, 13, 26$
Then we have:
== Sources ==
* 
* : $24$
Category:Divisors
Category:Square Numbers
Category:24
Category:26