hcju/proofwikiqa huggingface dataset:
First we prove the theorem for the 2-color case, by induction on $r + s$. It is clear from the definition that :$\forall n \in \N: \map R {n, 1} = \map R {1, n} = 1$ because the complete graph on one node has no edges. This is the base case. We prove that $R \left({r, s}\right)$ exists by finding an explicit bound for it. By the inductive hypothesis, $\map R {r - 1, s}$ and $\map R {r, s - 1}$ exist. === Proof for Two Colors === We will show that: :$\map R {r, s} \le \map R {r - 1, s} + \map R {r, s - 1}$ Consider a complete graph on $\map R {r - 1, s} + \map R {r, s - 1}$ vertices. Pick a vertex $v$ from the graph, and partition the remaining vertices into two sets $M$ and $N$, such that for every vertex $w$: :$w \in M$ if $\tuple {v, w}$ is blue :$w \in N$ if $\tuple {v, w}$ is red. Because the graph has $\map R {r - 1, s} + \map R {r, s - 1} = \size M + \size N + 1$ vertices, it follows that either $\size M \ge \map R {r - 1, s}$ or $\size N \ge \map R {r, s - 1}$. In the former case, if $M$ has a red $K_s$ then so does the original graph and we are finished. Otherwise $M$ has a blue $K_{r− 1}$, and so $M \cup \set v$ has blue $K_r$ by definition of $M$. The latter case is analogous. Thus the claim is true and we have completed the proof for 2 colours. We now prove the result for the general case of $c$ colors. The proof is again by induction, this time on the number of colors $c$. We have the result for $c = 1$ (trivially) and for $c = 2$ (above). Now let $c > 2$. === Proof for More than Two Colors === We will show that: :$\map R {n_1, \ldots, n_c} \le \map R {n_1, \ldots, n_{c - 2}, \map R {n_{c - 1}, n_c} }$ We note that the {{RHS}} only contains only Ramsey numbers for $c - 1$ colors and 2 colors, and therefore exists. Thus it is the finite number $t$, by the inductive hypothesis. So proving this will prove the theorem.  Consider a graph on $t$ vertices and color its edges with $c$ colors. Now "go color-blind" and pretend that $c - 1$ and $c$ are the same color. Thus the graph is now $\paren {c - 1}$-colored. By the inductive hypothesis, it contains either: :a complete monochromatic graph $K_{n_i}$ with color $i$ for some $1 \le i \le \paren {c - 2}$, or :a complete monochromatic graph $K_{\map R {n_{c - 1}, n_c}}$-colored in the "blurred color". In the former case we are finished. In the latter case, we recover our sight again and see from the definition of $\map R {n_{c - 1}, n_c}$ we must have either: :a complete $\paren {c - 1}$-monochromatic graph $K_{n_{c - 1} }$, or :a complete $c$-monochromatic graph $K_{n_c}$. In either case the proof is complete. {{qed}} {{Namedfor|Frank Plumpton Ramsey|cat = Ramsey}}
Let: :$n = \map R {3, \ldots, 3}$ where $\map R {3, \ldots, 3}$ denotes the Ramsey number on $r$ colors. Take $S$ to be $n$. {{refactor|Extract the below process of "coloring" a partition into its own page}} partition the integers $\set {1, \ldots, n}$ into $r$ parts, which we denote by '''colors'''. That is: :the integers in the first part are said to be '''colored''' $c_1$ :the integers in the second part are said to be colored $c_2$  and so on till color $c_r$. Thus $\set {1, \ldots, S}$ has been '''$r$-colored'''. (This terminology is common in Ramsey theory.) Now consider the complete graph $K_n$. Now color the edges of $K_n$ as follows: :An edge $xy$ is given color $c$ if $\size {x - y}$ was colored $c$ in the partitioning. {{explain|When the page defining a "coloring" of a partition is written, make sure that the links are assigned appropriately from the two difference senses of "coloring" in the above.}} From the definition of $\map R {3, \ldots, 3}$ and Ramsey's Theorem, $K_n$ will definitely contain a monochromatic triangle, say built out of the vertices $i > j > k$. Let the triangle be colored $c_m$. Now $i - j$, $i - k$ and $j - k$ will also be colored $c_m$. That is, $i - j$, $i - k$ and $j - k$ will belong to the same part in the partition. It only remains to take $x = i - j$, $y = j - k$ and $z = i - k$ to complete the proof. {{qed}}
:300px Let $ABC$ be a circle and let $D$ be a point outside the circle. Let $M$ be the center of the circle and let $E$, $F$, and $C$ be points on the concave circumference. Draw straight lines $DM$, $DC$, $DE$, and $DF$, then extend $DM$ to $A$ on the circle. Place $H$, $L$, $K$, and $G$ on the intersections of the circle and $DC$, $DF$, $DE$, and $DA$ respectively. Then of the straight lines falling on the concave circumference $AEFC$, the line $DA$ through the center is the greatest, with $DE > DF > DC$, and of the straight lines falling on the convex circumference $HLKG$, the line $DG$ is the least, with $DK < DL < DH$. The proof is as follows: Join $ME$, $MF$, $MC$, $MH$, $ML$, and $MK$. Since $AM = EM$, add $DM$ to each, so $AD = EM + MD$. But we know that $EM + MD > ED$, so it follows that $AD > ED$.  Since this is true for any $E$, $AD$ is the greatest line from $D$ to the concave circumference. Since $EM = FM$, $DM$ is common, and $\angle EMD > \angle FMD$, it follows that $ED > FD$. Similarly we can show that $FD > CD$.  Thus $DA > DE > DF > DC$. We know that $MK + KD > MD$, and since $MG = MK$, we have $KD > GD$.  Since this is true for any $K$, $GD$ is the least line from $D$ to the convex circumference. Since $K$ is inside $\triangle MLD$, it follows that $ML + LD > MK + KD$.  Then since $MK = ML$, we have $DK < DL$. Similarly we can show that $DL < DH$.  Thus $DG < DK < DL < DH$. ---- 250px Also, from the point $D$ only two equal straight lines fall on the circle $ABC$, one on each side of the line $DA$. Construct the angle $\angle DMB = \angle KMD$ on the straight line $MD$ at the point $M$.  Join $DB$. Then since $MK = MB$ and $MD$ is common, from Triangle Side-Angle-Side Equality $DK = DB$. Another straight line equal to $DK$ will not fall on the circle from $D$. For if this is possible, let $DN$ be this straight line. Then $DN = DB$, but from what we proved above either $DN > DB$ (or $DN < DB$ depending on where $N$ falls on the circle), a contradiction. Therefore the point $N$ cannot exist as described, and no more than two equal straight lines fall on the circle from $D$, one on each side of $DA$. {{qed}} {{Euclid Note|8|III}}
Let $ABCD$ be a cyclic quadrilateral with sides $a, b, c, d$. :300px Area of $ABCD$ = Area of $\triangle ABC$ + Area of $\triangle ADC$ From Area of Triangle in Terms of Two Sides and Angle: {{begin-eqn}} {{eqn | l = \triangle ABC       | r = \frac 1 2 a b \sin \angle ABC       | c =  }} {{eqn | l = \triangle ADC       | r = \frac 1 2 c d \sin \angle ADC       | c =  }} {{end-eqn}} From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\angle ABC + \angle ADC$ equals two right angles, that is, are supplementary. Hence we have: {{begin-eqn}} {{eqn | l = \sin \angle ABC       | r = \sin \angle ADC       | c = Sine and Cosine of Supplementary Angles }} {{eqn | l = \cos \angle ABC       | r = -\cos \angle ADC       | c = Sine and Cosine of Supplementary Angles }} {{end-eqn}} This leads to: {{begin-eqn}} {{eqn | l = \Area       | r = \frac 1 2 a b \sin \angle ABC + \frac 1 2 c d \sin \angle ABC       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\Area}^2       | r = \frac 1 4 \paren {a b + c d} \sin^2 \angle ABC       | c =  }} {{eqn | ll= \leadsto       | l = 4 \paren {\Area}^2       | r = \paren {a b + c d} \paren {1 - \cos^2 \angle ABC}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \paren {a b + c d} - \cos^2 \angle ABC \paren {a b + c d}       | c =  }} {{end-eqn}} Applying the Law of Cosines for $\triangle ABC$ and $\triangle ADC$ and equating the expressions for side $AC$: :$a^2 + b^2 - 2 a b \cos \angle ABC = c^2 + d^2 - 2 c d \cos \angle ADC$ From the above: :$\cos \angle ABC = -\cos \angle ADC$ Hence: :$2 \cos \angle ABC \paren {a b + c d} = a^2 + b^2 - c^2 - d^2$ Substituting this in the above equation for the area: {{begin-eqn}} {{eqn | l = 4 \paren {\Area}^2       | r = \paren {a b + c d}^2 - \frac 1 4 \paren {a^2 + b^2 - c^2 - d^2}^2       | c =  }} {{eqn | ll= \leadsto       | l = 16 \paren {\Area}^2       | r = 4 \paren {a b + c d}^2 - \paren {a^2 + b^2 - c^2 - d^2}^2       | c =  }} {{end-eqn}} This is of the form $x^2 - y^2$. Hence, by Difference of Two Squares, it can be written in the form $\paren {x + y} \paren {x - y}$ as: {{begin-eqn}} {{eqn | o =       | r = \paren {2 \paren {a b + c d} + a^2 + b^2 - c^2 - d^2} \paren {2 \paren {a b + c d} - a^2 - b^2 + c^2 + d^2}       | c =  }} {{eqn | r = \paren {\paren {a + b}^2 - \paren {c - d}^2} \paren {\paren {c + d}^2 - \paren {a - b}^2}       | c =  }} {{eqn | r = \paren {a + b + c - d} \paren {a + b + d - c} \paren {a + c + d - b} \paren {b + c + d - a}       | c =  }} {{end-eqn}} When we introduce the expression for the semiperimeter: :$\displaystyle s = \frac {a + b + c + d} 2$ the above converts to: :$16 \paren {\Area}^2 = 16 \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d}$ Taking the square root: :$\Area = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$ {{qed}}
Let $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$. First by the lemma to Cassini's Identity:  :$(1): \quad \forall n \in \Z_{>1}: A^n = \begin{bmatrix} F_{n + 1} & F_n \\ F_n       & F_{n - 1} \end{bmatrix}$ Next is is demonstrated that $A$ has the eigenvalues $\phi$ and $\hat \phi$, where $\hat \phi = 1 - \phi$. Now we have that: {{begin-eqn}} {{eqn | n = 2       | l = A \begin{pmatrix} \phi \\ 1 \end{pmatrix}       | r = \begin{pmatrix} \phi + 1 \\ \phi \end{pmatrix} }} {{eqn | r = \phi \begin{pmatrix} \phi \\ 1 \end{pmatrix}       | c = since $\phi$ is a solution of $x^2 - x - 1 = 0$ }} {{eqn | l = A \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}       | r = \begin{pmatrix} \hat \phi + 1 \\ \hat \phi \end{pmatrix} }} {{eqn | r = \hat \phi \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}       | c = since $\hat \phi$ is a solution of $x^2 - x - 1 = 0$ }} {{end-eqn}} This shows that: :$\displaystyle \begin{pmatrix} \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$ and: :$\displaystyle \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\hat \phi$. Thus: :$\displaystyle \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$ and: :$\displaystyle \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\hat \phi$. By Eigenvalue of Matrix Powers we get for a positive integer $n$: {{begin-eqn}} {{eqn | l = \phi^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}       | r = A^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} }} {{eqn | l = {\hat \phi}^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}       | r = A^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} }} {{end-eqn}} From $(1)$ we get: :$A^n \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}$ Substituting: :$\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} - \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ we get: {{begin-eqn}} {{eqn | l = \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}       | r = \frac 1 {\sqrt 5} A^n \left({\begin{pmatrix} \phi \\ 1 \end{pmatrix} - \begin{pmatrix} \hat \phi \\ 1 \end{pmatrix} }\right) }} {{eqn | r = A^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} - A^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} }} {{eqn | r = \phi^n \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} - \hat \phi^n \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} }} {{eqn | r = \frac 1 {\sqrt 5} \begin{pmatrix} \phi^n \cdot \phi - \hat \phi^n \cdot \hat \phi \\ \phi^n - \hat \phi^n \end{pmatrix} }} {{end-eqn}} Hence the result. {{Qed}}
Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition: :$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$ === Basis for the Induction === $\map P 0$ is true, as this just says: :$\dfrac {\phi^0 - \hat \phi^0} {\sqrt 5} = \dfrac {1 - 1} {\sqrt 5} = 0 = F_0$ $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \frac {\phi - \hat \phi} {\sqrt 5}       | r = \frac {\paren {\frac {1 + \sqrt 5} 2} - \paren {\frac {1 - \sqrt 5} 2} } {\sqrt 5}       | c =  }} {{eqn | r = \frac {\paren {1 - 1} + \paren {\sqrt 5 + \sqrt 5} } {2 \sqrt 5}       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = F_1       | c =  }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P j$ is true for all $0 \le j \le k + 1$, then it logically follows that $\map P {k + 2}$ is true. So this is our induction hypothesis: :$\forall 0 \le j \le k + 1: F_j = \dfrac {\phi^j - \hat \phi^j} {\sqrt 5}$ Then we need to show: :$F_{k + 2} = \dfrac {\phi^{k + 2} - \hat \phi^{k + 2} } {\sqrt 5}$ === Induction Step === This is our induction step: We observe that we have the following two identities: :$\phi^2 = \paren {\dfrac {1 + \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 + 2 \sqrt 5} = \dfrac {3 + \sqrt 5} 2 = 1 + \phi$ :$\hat \phi^2 = \paren {\dfrac {1 - \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 - 2 \sqrt 5} = \dfrac {3 - \sqrt 5} 2 = 1 + \hat \phi$ This can also be deduced from the definition of the golden mean: the fact that $\phi$ and $\hat \phi$ are the solutions to the quadratic equation $x^2 = x + 1$. Thus: {{begin-eqn}} {{eqn | l = \phi^{k + 2} - \hat \phi^{k + 2}       | r = \paren {1 + \phi} \phi^k - \paren {1 + \hat \phi} \hat \phi^k }} {{eqn | r = \paren {\phi^k - \hat \phi^k} + \paren {\phi^{k + 1} - \hat \phi^{k + 1} } }} {{eqn | r = \sqrt 5 \paren {F_k + F_{k + 1} }       | c = Induction Hypothesis }} {{eqn | r = \sqrt 5 F_{k + 2}       | c = {{Defof|Fibonacci Numbers}} }} {{end-eqn}} The result follows by the Second Principle of Mathematical Induction. Therefore: :$\forall n \in \N: F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$ {{qed}}
This follows as a direct application of the first Binet form: :$U_n = m U_{n - 1} + U_{n - 2}$ where: {{begin-eqn}} {{eqn | l = U_0       | r = 0 }} {{eqn | l = U_1       | r = 1 }} {{end-eqn}} has the closed-form solution: :$U_n = \dfrac {\alpha^n - \beta^n} {\Delta}$ where: {{begin-eqn}} {{eqn | l = \Delta       | r = \sqrt {m^2 + 4} }} {{eqn | l = \alpha       | r = \frac {m + \Delta} 2 }} {{eqn | l = \beta       | r = \frac {m - \Delta} 2 }} {{end-eqn}} where $m = 1$. {{qed}}
From Generating Function for Fibonacci Numbers, a generating function for the Fibonacci numbers is: :$\map G z = \dfrac z {1 - z - z^2}$ Hence: {{begin-eqn}} {{eqn | l = \map G z       | r = \dfrac z {1 - z - z^2}       | c =  }} {{eqn | r = \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} }       | c = Partial Fraction Expansion }} {{end-eqn}} where: :$\phi = \dfrac {1 + \sqrt 5} 2$ :$\hat \phi = \dfrac {1 - \sqrt 5} 2$ By Sum of Infinite Geometric Sequence: :$\dfrac 1 {1 - \phi z} = 1 + \phi z + \phi^2 z^2 + \cdots$ and so: :$\map G z = \dfrac 1 {\sqrt 5} \paren {1 + \phi z + \phi^2 z^2 + \cdots - 1 - \hat \phi z - \hat \phi^2 z^2 - \cdots}$ By definition, the coefficient of $z^n$ in $\map G z$ is exactly the $n$th Fibonacci number. That is: :$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$ {{qed}}
{{WLOG}}, assume that $\mathbf x$ and $\mathbf y$ are non-zero. Define: :$\mathbf u = \sequence {u_n} = \dfrac {\mathbf x} {\norm {\mathbf x}_p}$ :$\mathbf v = \sequence {v_n} = \dfrac {\mathbf y} {\norm {\mathbf y}_q}$ Then: :$\displaystyle \norm {\mathbf u}_p = \dfrac 1 {\norm {\mathbf x}_p} \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^p}^{1/p} = 1$ Similarly: :$\norm {\mathbf v}_q = 1$ By Young's Inequality for Products: :$(1): \quad \forall n \in \N: \size {u_n v_n} \le \dfrac 1 p \size {u_n}^p + \dfrac 1 q \size {v_n}^q$ By the comparison test, it follows that: :$\mathbf u \mathbf v = \sequence {u_n v_n} \in \ell^1$ :$\mathbf x \mathbf y = \norm {\mathbf x}_p \norm {\mathbf y}_q \mathbf u \mathbf v \in \ell^1$ From $(1)$, it follows that: :$\norm {\mathbf u \mathbf v}_1 \le \dfrac 1 p \norm {\mathbf u}_p + \dfrac 1 q \norm {\mathbf v}_q = 1$ Therefore: :$\norm {\mathbf x \mathbf y}_1 = \norm {\mathbf x}_p \norm {\mathbf y}_q \norm {\mathbf u \mathbf v}_1 \le \norm {\mathbf x}_p \norm {\mathbf y}_q$ as desired. {{qed}}
Proof by induction: Let $\map P n$ be the proposition: :$U_{n - 1} + U_{n + 1} = V_n$ === Basis for the Induction === We have: {{begin-eqn}} {{eqn | l = U_0 + U_2       | r = 0 + m U_1 + U_0 }} {{eqn | r = m \times 1 + 0 }} {{eqn | r = m }} {{eqn | r = V_1       | c = From definition }} {{eqn | l = U_1 + U_3       | r = 1 + m U_2 + U_1 }} {{eqn | r = m \times m + 2 }} {{eqn | r = m \times V_1 + V_0 }} {{eqn | r = V_2       | c = From definition }} {{end-eqn}} Therefore $\map P 1$ and $\map P 2$ are true. This is the basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :For some $k \in \N_{> 0}$, both $\map P k$ and $\map P {k + 1}$ are true. That is: :$U_{k - 1} + U_{k + 1} = V_k$ :$U_k + U_{k + 2} = V_{k + 1}$ Now we need to show true for $n = k + 2$: :$\map P {k + 2}$ is true. That is: :$U_{k + 1} + U_{k + 3} = V_{k + 2}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = V_{k + 2}       | r = m V_{k + 1} + V_k       | c = Definition of the recursive sequence }} {{eqn | r = m \paren {U_k + U_{k + 2} } + \paren {U_{k - 1} + U_{k + 1} }       | c = Induction Hypothesis }} {{eqn | r = \paren {m U_k + U_{k - 1} } + \paren {m U_{k + 2} + U_{k + 1} } }} {{eqn | r = U_{k + 1} + U_{k + 3} }} {{end-eqn}} This show that $\map P {k + 2}$ is true. By principle of mathematical induction, $\map P n$ is true for all $n \in \N _{> 0}$. {{qed}} {{Namedfor|Jacques Philippe Marie Binet|cat = Binet}}
This is a special case of the Cauchy-Binet Formula: :$\displaystyle \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \, \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$ where: :$\mathbf A$ is an $m \times n$ matrix :$\mathbf B$ is an $n \times m$ matrix. :For $1 \le j_1, j_2, \ldots, j_m \le n$: ::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$. ::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$. In this case $m = 2$, giving: :$\displaystyle \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop \le n} \map \det {\mathbf A_{j_1 j_2} } \, \map \det {\mathbf B_{j_1 j_2} }$ {{finish}}
Proof by induction: For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: : $n^3 = \left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 + n - 1}\right)$ === Basis for the Induction === $P(1)$ is true, as this just says $1^3 = 1$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$k^3 = \left({k^2 - k + 1}\right) + \left({k^2 - k + 3}\right) + \ldots + \left({k^2 + k - 1}\right)$ Then we need to show: :$\left({k + 1}\right)^3 = \left({\left({k + 1}\right)^2 - \left({k + 1}\right) + 1}\right) + \left({\left({k + 1}\right)^2 - \left({k + 1}\right) + 3}\right) + \ldots + \left({\left({k + 1}\right)^2 + \left({k + 1}\right) - 1}\right)$ === Induction Step === Let $T_k = \left({k^2 - k + 1}\right) + \left({k^2 - k + 3}\right) + \ldots + \left({k^2 + k - 1}\right)$. We can express this as: :$T_k = \left({k^2 - k + 1}\right) + \left({k^2 - k + 3}\right) + \ldots + \left({k^2 - k + 2k - 1}\right)$ We see that there are $k$ terms in $T_k$. Let us consider the general term $\left({\left({k + 1}\right)^2 - \left({k + 1}\right) + j}\right)$ in $T_{k+1}$: {{begin-eqn}} {{eqn | l = \left({k + 1}\right)^2 - \left({k + 1}\right) + j       | r = k^2 + 2 k + 1 - \left({k + 1}\right) + j       | c =  }} {{eqn | r = k^2 - k + j + 2 k       | c =  }} {{end-eqn}} So, in $T_{k+1}$, each of the terms is $2k$ larger than the corresponding term for $T_k$. So: {{begin-eqn}} {{eqn | l = T_{k+1}       | r = T_k + k \left({2k}\right) + \left({k + 1}\right)^2 + \left({k + 1}\right) - 1       | c =  }} {{eqn | r = k^3 + k \left({2k}\right) + \left({k + 1}\right)^2 + \left({k + 1}\right) - 1       | c = Induction Hypothesis }} {{eqn | r = k^3 + 2k^2 + k^2 + 2k + 1 + k + 1 - 1       | c =  }} {{eqn | r = k^3 + 3k^2 + 3k + 1       | c =  }} {{eqn | r = \left({k + 1}\right)^3       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>0}: n^3 = \left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 + n - 1}\right)$   Finally, note that the first term in the expansion for $\left({n + 1}\right)^3$ is $n^2 - n + 1 + 2 n = n^2 + n + 1$. This is indeed two more than the last term in the expansion for $n^3$ . {{qed}}
From the definition: :$\left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 + n - 1}\right)$ can be written: :$\left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 - n + 2 n - 1}\right)$ Writing this in sum notation: {{begin-eqn}} {{eqn | o =        | r = \left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 - n + 2 n - 1}\right)       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n \left({n^2 - n + 2 k - 1}\right)       | c =  }} {{eqn | r = n \left({n^2 - n}\right) + \sum_{k \mathop = 1}^n \left({2 k - 1}\right)       | c =  }} {{eqn | r = n^3 - n^2 + n^2       | c = Odd Number Theorem }} {{eqn | r = n^3       | c =  }} {{end-eqn}} {{Qed}}