Sum of Sequence of Squares

Theorem

$\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6$

Proof 1

Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \sum_{i \mathop = 1}^n i^2 = \frac{n \left({n+1}\right)\left({2n+1}\right)} 6$

When $n = 0$, we see from the definition of vacuous sum that:

$0 = \displaystyle \sum_{i \mathop = 1}^0 i^2 = \frac{0 \left({1}\right)\left({1}\right)} 6 = 0$

and so $P(0)$ holds.

Base Case

When $n=1$, we have $\displaystyle \sum_{i \mathop = 1}^1 i^2 = 1^2 = 1$.

Now, we have:

$\displaystyle \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6 = \frac {1 \left({1 + 1}\right) \left({2 \cdot 1 + 1}\right)} 6 = \frac 6 6 = 1$

So $P(1)$ is true. This is our base case.

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:

$\displaystyle \sum_{i \mathop = 1}^k i^2 = \frac {k \left({k + 1}\right) \left({2 k + 1}\right)} 6$

Then we need to show:

$\displaystyle \sum_{i \mathop = 1}^{k+1} i^2 = \frac{\left({k+1}\right) \left({k+2}\right) \left({2 \left({k+1}\right) + 1}\right)} 6$

Induction Step

This is our induction step:

Using the properties of summation, we have:

$\displaystyle \sum_{i \mathop = 1}^{k+1} i^2 = \sum_{i \mathop = 1}^k i^2 + \left({k+1}\right)^2$

We can now apply our induction hypothesis, obtaining:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^{k+1} i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{k \left({k + 1}\right) \left({2 k + 1}\right)} 6 + \left({k + 1}\right)^2\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{k \left({k + 1}\right) \left({2 k + 1}\right) + 6 \left({k + 1}\right)^2} 6\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{\left({k + 1}\right) \left({k \left({2 k + 1}\right) + 6 \left({k + 1}\right)}\right)} 6\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{\left({k + 1}\right) \left({2 k^2 + 7 k + 6}\right)} 6\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{\left({k + 1}\right) \left({k + 2}\right) \left({2 \left({k + 1}\right) + 1}\right)} 6\)	\(\displaystyle \)	\(\displaystyle \)

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6$

$\blacksquare$

Proof 2

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n 3 i \left({i + 1}\right)\)	\(=\)	\(\displaystyle \)	\(\displaystyle n \left({n + 1}\right) \left({n + 2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	Sum of Sequence of Products of Consecutive Integers
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n 3 i^2 + \sum_{i \mathop = 1}^n 3 i\)	\(=\)	\(\displaystyle \)	\(\displaystyle n \left({n + 1}\right) \left({n + 2}\right)\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n 3 i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle n \left({n + 1}\right) \left({n + 2}\right) - 3 \frac {n \left({n + 1}\right) } 2\)	\(\displaystyle \)	\(\displaystyle \)	Closed Form for Triangular Numbers
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {n \left({n + 1}\right) \left({n + 2}\right)} 3 - \frac {n \left({n + 1}\right) } 2\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {2 n \left({n + 1}\right) \left({n + 2}\right) - 3 n \left({n + 1}\right) } 6\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{n \left({n + 1}\right) \left({2n + 1}\right)} 6\)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof 3

Observe that:

$3 i \left({i + 1}\right) = i \left({i + 1}\right) \left({i + 2}\right) - i \left({i + 1}\right) \left({i - 1}\right)$

That is:

$(1): \quad 6 T_i = \left({i + 1}\right) \left({\left({i + 1}\right) + 1}\right) \left({\left({i + 1}\right) - 1}\right) - i \left({i + 1}\right) \left({i - 1}\right)$

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle n^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {n^2 + n + n^2 - n} 2\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {n \left({n+1}\right) + n \left({n-1}\right)} 2\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {n \left({n+1}\right)} 2 + \frac{n \left({n-1}\right)} 2\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle T_n + T_{n-1}\)	\(\displaystyle \)	\(\displaystyle \)

where $T_n$ is the $n$th Triangular number.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle 1 + \left({T_1 + T_2}\right) + \left({T_2 + T_3}\right) + \left({T_3 + T_4}\right) + \cdots + \left({T_{n-1} + T_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle 1 + 2T_2 + 2T_3 + 2T_4 + \cdots + 2T_{n-1} + T_n\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle 1 - T_1 - T_n + 2 \left({T_1 + T_2 + T_3 + T_4 + \cdots + T_n}\right)\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle 2 \left({\sum_{i \mathop = 1}^n T_i} \right) - \frac{n \left({n+1}\right)} 2\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle 2 \left({\frac{n \left({n+1}\right)\left({n+2}\right)} 6}\right) - \frac{n \left({n+1}\right)} 2\)	\(\displaystyle \)	\(\displaystyle \)	Telescoping Series from $(1)$ above
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{2n \left({n^2 + 3n+2}\right) - \left({3n^2 + 3n}\right)} 6\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{2n^3 + 3n^2 + n} 6\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac{n \left({n + 1}\right) \left({2n + 1}\right)} 6\)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof 4

We can observe from the above diagram that:

$\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \sum_{i \mathop = 1}^n \left({\sum_{j \mathop = i}^n j}\right)$

Therefore we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n \left({\sum_{j \mathop = i}^n j}\right)\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n \left( {\sum_{j \mathop = 1}^n j - \sum_{j \mathop = 1}^{i-1} j} \right)\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n \left({\frac {n \left({n+1}\right)} 2 - \frac {i \left({i-1}\right)} 2}\right)\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 2 \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle n^2 \left({n+1}\right) - \sum_{i \mathop = 1}^n i^2 + \sum_{i \mathop = 1}^n i\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 3 \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle n^2 \left({n+1}\right) + \sum_{i \mathop = 1}^n i\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 3 \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle n^2 \left({n+1}\right) + \frac {n \left({n+1}\right)} 2\)	\(\displaystyle \)	\(\displaystyle \)	Closed Form for Triangular Numbers
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 6 \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle 2 n^2 \left({n+1}\right) + n \left({n+1}\right)\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle n \left({n+1}\right) \left({2n+1}\right)\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \sum_{i \mathop = 1}^n i^2\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {n \left({n+1}\right) \left({2n+1}\right)} 6\)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Historical Note

This result was documented by Āryabhaṭa in his work Āryabhaṭīya of 499 CE.

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $18.10 \ \text{(b)}$
• George E. Andrews: Number Theory (1971)... (previous)... (next): $\S 1.1$: Exercise $1$
• Gary Chartrand: Introductory Graph Theory (1977): Appendix $\text{A}.6$: Problem $36$