nyuuzyou/wikis huggingface dataset:
Sum of Sequence of Product of Lucas Numbers with Powers of 2

== Theorem ==
Let $L_k$ be the $k$th Lucas number.
Let $F_k$ be the $k$th Fibonacci number.
Then:
:$\ds \forall n \in \N_{>0}: \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$
That is:
:$2^0 L_1 + 2^1 L_2 + 2^2 L_3 + \cdots + 2^{n - 1} L^n = 2^n F_{n + 1} - 1$
== Proof ==
Proof by induction:
For all $\forall n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $L_1 = 1 = 2^1 \times F_1 - 1$ which follows directly from the definitions of Fibonacci numbers and Lucas numbers.
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:
:$\ds \sum_{j \mathop = 1}^k 2^{j - 1} L_j = 2^k F_{k + 1} - 1$
Then we need to show:
:$\ds \sum_{j \mathop = 1}^{k + 1} 2^{j - 1} L_j = 2^{k + 1} F_{k + 2} - 1$
=== 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:
:$\ds \forall n \in \N_{>0}: \sum_{j \mathop = 1}^n 2^{j - 1} L_j = 2^n F_{n + 1} - 1$
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $16$
Category:Lucas Numbers
Category:Fibonacci Numbers
Sum of Sequence of Products of 3 Consecutive Integers

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2} = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to be true.
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:
:$\ds \sum_{j \mathop = 1}^k j \paren {j + 1} \paren {j + 2} = \dfrac {k \paren {k + 1} \paren {k + 2} \paren {k + 3} } 4$
Then we need to show:
:$\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1} \paren {j + 2} = \dfrac {\paren {k + 1} \paren {k + 2} \paren {k + 3} \paren {k + 4} } 4$
=== 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:
:$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2} = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4$
== Sources ==
* : $\text I$. Algebra: The Method of Induction: Exercises $\text {II}$: $5$
Category:Sums of Sequences
Category:Proofs by Induction
Sum of Sequence of Products of 3 Consecutive Reciprocals

== Theorem ==
:$\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} } = \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }$
=== Corollary ===
== Proof ==
We observe that:
Hence:
Category:Sums of Sequences
Category:Reciprocals
Category:Sum of Sequence of Products of 3 Consecutive Reciprocals
Sum of Sequence of Products of 3 Consecutive Reciprocals/Corollary

== Theorem ==
:$\ds \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} \paren {j + 2} } = \frac 1 4$
== Proof ==
Category:Limits of Series
Category:Reciprocals
Category:Sum of Sequence of Products of 3 Consecutive Reciprocals
Sum of Sequence of Products of Consecutive Integers

== Theorem ==
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $4$
* : $\S 1.17$: Finite Induction and Well-Ordering for Positive Integers: Exercise $6$
Category:Sums of Sequences
Category:Sum of Sequence of Products of Consecutive Integers
Sum of Sequence of Products of Consecutive Integers/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} = \frac {n \paren {n + 1} \paren {n + 2} } 3$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $1 \times 2 = 2 = \dfrac {1 \times 2 \times 3} 3$.
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:
:$\ds \sum_{j \mathop = 1}^k j \paren {j + 1} = \frac {k \paren {k + 1} \paren {k + 2} } 3$
Then we need to show:
:$\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1} = \frac {\paren {k + 1} \paren {k + 2} \paren {k + 3} } 3$
=== 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:
:$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} = \frac {n \paren {n + 1} \paren {n + 2} } 3$
== Sources ==
* : $\text I$. Algebra: The Method of Induction: Exercises $\text {II}$: $2$
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $1 \ \text {(c)}$
Category:Sum of Sequence of Products of Consecutive Integers
Category:Proofs by Induction
Sum of Sequence of Products of Consecutive Integers/Proof 2

== Theorem ==
== Proof ==
Observe that:
Thus:
:$3 i \paren {i + 1} = b \paren {i + 1} - b \paren i$
where:
:$b \paren i = i \paren {i + 1} \paren {i - 1}$
This can therefore be used as the basis of a telescoping series, as follows:
Hence the result.
Category:Sum of Sequence of Products of Consecutive Integers
Category:Telescoping Series
Sum of Sequence of Products of Consecutive Odd Reciprocals

== Theorem ==
=== Corollary ===
== Proof ==
We observe that:
and that $\ds \sum_{j \mathop = 0}^n \paren {\frac 1 {2 j + 1} - \frac 1 {2 j + 3} }$ is a telescoping series. 
Therefore:
Category:Sum of Sequence of Products of Consecutive Odd Reciprocals
Category:Sums of Sequences
Category:Reciprocals
Sum of Sequence of Products of Consecutive Odd Reciprocals/Corollary

== Corollary to Sum of Sequence of Products of Consecutive Odd Reciprocals ==
== Proof ==
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.30$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.30.$
Category:Sum of Sequence of Products of Consecutive Odd Reciprocals
Sum of Sequence of Products of Consecutive Odd and Consecutive Even Numbers

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$
=== 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:
:$\ds \sum_{j \mathop = 1}^k j \paren {j + 2} = \frac {k \paren {k + 1} \paren {2 k + 7} } 6$
from which it is to be shown that:
:$\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2} = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6$
=== 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:
:$\ds \forall n \in \Z_{>0}: \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$
</onlyinclude>
== Sources ==
* : $\text I$. Algebra: The Method of Induction: Exercises $\text {II}$: $3$
* : Appendix $\text{A}.6$: Mathematical Induction: Problem Set $\text{A}.6$: $40$
Category:Sums of Sequences
Category:Proofs by Induction
Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals/Corollary

== Corollary to Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals ==
== Proof ==
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.31$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.31.$
Category:Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals
Sum of Sequence of Products of Consecutive Reciprocals

== Theorem ==
:$\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $6$
* : Appendix $\text{A}.6$: Mathematical Induction: Problem Set $\text{A}.6$: $39$
Category:Sum of Sequence of Products of Consecutive Reciprocals
Category:Sums of Sequences
Category:Reciprocals
Sum of Sequence of Products of Consecutive Reciprocals/Corollary

== Corollary to Sum of Sequence of Products of Consecutive Reciprocals ==
:$\ds \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = 1$
== Proof ==
From Sum of Sequence of Products of Consecutive Reciprocals:
:$\ds \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1} = 1 - \frac 1 {n + 1}$
We have that:
:$\dfrac 1 {n + 1} < \dfrac 1 n$
and that $\sequence {\dfrac 1 n}$ is a basic null sequence.
Thus by the Squeeze Theorem:
:$\ds \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = 1$
== Sources ==
* : $\S 6.3$
* : $\S 1.2$: Infinite Series of Constants: Example $1$
Category:Limits of Series
Category:Sum of Sequence of Products of Consecutive Reciprocals
Sum of Sequence of Products of Consecutive Reciprocals/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $\dfrac 1 2 = \dfrac 1 2$.
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:
:$\ds \sum_{j \mathop = 1}^k \frac 1 {j \paren {j + 1} } = \frac k {k + 1}$
Then we need to show:
:$\ds \sum_{j \mathop = 1}^{k + 1} \frac 1 {j \paren {j + 1} } = \frac {k + 1} {k + 2}$
=== 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:
:$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$
Category:Sum of Sequence of Products of Consecutive Reciprocals
Category:Proofs by Induction
Sum of Sequence of Products of Consecutive Reciprocals/Proof 2

== Theorem ==
== Proof ==
We can observe that:
:$\dfrac 1 {j \paren {j + 1} } = \dfrac 1 j - \dfrac 1 {j + 1}$
and that $\ds \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 1 {j + 1} }$ is a telescoping series. 
Therefore:
== Sources ==
* : $\S 6.3$
Category:Sum of Sequence of Products of Consecutive Reciprocals
Category:Telescoping Series
Sum of Sequence of Products of Squares of 3 Consecutive Reciprocals

== Theorem ==
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.33$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.33.$
Category:Sum of Sequence of Products of Squares of 3 Consecutive Reciprocals
Category:Limits of Series
Category:Reciprocals
Sum of Sequence of Products of Squares of 3 Consecutive Reciprocals/Proof 1

== Theorem ==
== Proof ==
We use partial fraction expansion to expand $\dfrac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2}$.
Let $\dfrac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2} = \dfrac A j + \dfrac B {j^2} + \dfrac C {j + 1} + \dfrac D {\paren {j + 1}^2} + \dfrac E {j + 2} + \dfrac F {\paren {j + 2}^2}$.
This has solutions $A = - \dfrac 3 4, B = \dfrac 1 4, C = 0, D = 1, E = \dfrac 3 4, F = \dfrac 1 4$.
Thus:
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.33$
Category:Limits of Series
Category:Reciprocals
Category:Sum of Sequence of Products of Squares of 3 Consecutive Reciprocals
Sum of Sequence of Products of Squares of 3 Consecutive Reciprocals/Proof 2

== Theorem ==
== Proof ==
Note that we have:
Thus:
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.33$
Category:Limits of Series
Category:Reciprocals
Category:Sum of Sequence of Products of Squares of 3 Consecutive Reciprocals
Sum of Sequence of Products of Squares of Consecutive Odd Reciprocals

== Theorem ==
== Proof ==
Note that we have: 
:$\dfrac 1 {2 j + 1} - \dfrac 1 {2 j + 3} = \dfrac 2 {\paren {2 j + 1} \paren {2 j + 3} }$
So:
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.32$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.32.$
Category:Limits of Series
Category:Reciprocals
Sum of Sequence of Reciprocals of 3 n + 1 Alternating in Sign

== Theorem ==
== Proof ==
From Primitive of Power and the Fundamental Theorem of Calculus, we have: 
:$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {3 n + 1} = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{3 n} \rd x$
We can rewrite: 
We now use Lebesgue's Dominated Convergence Theorem to swap integral and summation. 
Specifically, we will prove that: 
:$\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x$ 
From , we have: 
:$\ds \size {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \le 1$
for all $x \in \closedint 0 1$ and $N \in \N$. 
All functions involved are continuous, so the conditions for Lebesgue's Dominated Convergence Theorem are satisfied, and we have: 
:$\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x$
Then, by the definition of infinite series, we have: 
:$\ds \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x = \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-x^3}^n} \rd x$
Then we have: 
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.16$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.16.$
Category:Logarithms
Category:Reciprocals
Sum of Sequence of Reciprocals of 3 n + 2 Alternating in Sign

== Theorem ==
== Proof ==
From Primitive of Power and the Fundamental Theorem of Calculus, we have: 
:$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {3 n + 2} = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{3 n + 1} \rd x$
We can rewrite: 
We now use Lebesgue's Dominated Convergence Theorem to swap integral and summation. 
Specifically, we will prove that: 
:$\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n + 1} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n + 1} } \rd x$ 
From , we have: 
:$\ds \size {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n + 1} } \le x$
for all $x \in \closedint 0 1$ and $N \in \N$. 
So:
:$\ds \size {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n + 1} } \le 1$ 
for all $x \in \closedint 0 1$.
All functions involved are continuous, so the conditions for Lebesgue's Dominated Convergence Theorem are satisfied, and we have: 
:$\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n + 1} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n + 1} } \rd x$ 
Then we have: 
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.18$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.18.$
Category:Logarithms
Category:Reciprocals
Sum of Sequence of Reciprocals of 4 n + 1 Alternating in Sign

== Theorem ==
== Proof ==
From Primitive of Power and the Fundamental Theorem of Calculus, we have: 
:$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {4 n + 1} = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{4 n} \rd x$
We can rewrite: 
We now use Lebesgue's Dominated Convergence Theorem to swap integral and summation. 
Specifically, we will prove that: 
:$\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x$ 
From Alternating Series Test: Lemma, we have: 
:$\ds \size {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \le 1$
for all $x \in \closedint 0 1$ and $N \in \N$. 
All functions involved are continuous, so the conditions for Lebesgue's Dominated Convergence Theorem are satisfied, and we have: 
:$\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x$
Then, by the definition of infinite series, we have: 
:$\ds \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x = \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-x^4}^n} \rd x$
Then we have: 
We have: 
and: 
Putting these together gives: 
:$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {4 n + 1} = \int_0^1 \frac 1 {1 + x^4} \rd x = \frac {\pi \sqrt 2} 8 + \frac {\sqrt 2 \map \ln {1 + \sqrt 2} } 4$
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.17$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.17.$
Category:Logarithms
Category:Reciprocals
Sum of Sequence of Reciprocals of Triangular Numbers

== Theorem ==
:$\ds \sum_{k \mathop \ge 1} \dfrac 1 {T_k} = 2$
where $T_k$ denotes the $k$th triangular number.
== Proof ==
== Sources ==
* : $15$
* : $15$
Category:Triangular Numbers
Category:Sums of Sequences
Sum of Sequence of Seventh Powers

== Theorem ==
:$\ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== 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:
:$\ds \sum_{j \mathop = 0}^k j^7 = \dfrac {k^2 \paren {k + 1}^2 \paren {3 k^4 + 6 k^3 - k^2 - 4 k + 2} } {24}$
from which it is to be shown that:
:$\ds \sum_{j \mathop = 0}^{k + 1} j^7 = \dfrac {\paren {k + 1}^2 \paren{k + 2}^2 \paren {3 \paren {k + 1}^4 + 6 \paren {k + 1}^3 - \paren {k + 1}^2 - 4 \paren {k + 1} + 2} } {24}$
=== 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:
:$\forall n \in \Z_{\ge 0}: \ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$
== Also see ==
* Sum of Sequence of Fifth Powers
== Sources ==
* 
Category:Sums of Sequences
Category:Seventh Powers
Sum of Sequence of Squares

== Theorem ==
:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
== Proof by Induction ==
== Proof by Products of Consecutive Integers ==
== Proof by Telescoping Series ==
== Proof by Summation of Summations ==
== Proof by Sum of Differences of Cubes ==
== Proof by Binomial Coefficients ==
== Proof using Bernoulli Numbers ==
== Also presented as ==
== Also see ==
* Sum of Sequence of Cubes
== Historical Note ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Exercise $18.10 \ \text{(b)}$
* : $\text {1-1}$ Principle of Mathematical Induction: Exercise $1$
* : $\S 1.17$: Finite Induction and Well-Ordering for Positive Integers: Exercise $3$
* : Appendix $\text{A}.6$: Mathematical Induction: Problem Set $\text{A}.6$: $36$
* : Chapter $1$: Preliminaries: $1.3$ Inductive Proofs
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
Category:Sum of Sequence of Squares
Category:Square Numbers
Category:Sums of Sequences
Sum of Sequence of Squares/Also presented as

== Theorem ==
The Sum of Sequence of Squares can also be presented as:
:$\ds \forall n \in \N: \sum_{i \mathop = 0}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
This is seen to be equivalent to the given form by the fact that the first term evaluates to $\dfrac {0 \paren {0 + 1} \paren {2 \times 0 + 1} } 6$ which is zero.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients
Category:Sum of Sequence of Squares
Sum of Sequence of Squares/Historical Note

== Historical Note on Sum of Sequence of Squares ==
The closed-form expression for the Sum of Sequence of Squares was proved by during the course of his proofs of the volumes of various solids of revolution in his .
It was also documented by in his work of $499$ CE.
== Sources ==
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
Category:Historical Notes
Sum of Sequence of Squares/Proof by Binomial Coefficients

== Theorem ==
== 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:
Hence:
== Also see ==
* Sum of Sequence of Fourth Powers/Proof using Binomial Coefficients
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $4 \ \text {(b)}$
* : $\S 1.2.6$: Binomial Coefficients
Category:Sum of Sequence of Squares
Sum of Sequence of Squares/Proof by Induction

== Theorem ==
== 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:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
== Sources ==
* : $\text I$. Algebra: The Method of Induction: Example $1$
* : Chapter $1$: Properties of the Natural Numbers: $\S 20 \beta$
* : $\S 3$: Natural Numbers: Exercise $\S 3.11 \ (1) \ \text{(i)}$
* : Chapter $1$: Preliminaries: $1.3$ Inductive Proofs: Example $1.1$
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction
* : $\text {(i)}$
Category:Sum of Sequence of Squares
Category:Proofs by Induction