Sum of Sequence of Squares

Theorem

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

Proof by Induction

Proof by induction:

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

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

Base Case

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

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:

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

Then we need to show:

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

Induction Step

This is our induction step:

Using the properties of summation, we have:

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

We can now apply our induction hypothesis, obtaining:

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

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

Therefore:

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

$\blacksquare$

Proof by Telescoping Sum

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

By taking the sum we'll get a telescoping one on the RHS and the conclusion follows.

$\blacksquare$

Direct Proof

We can observe from the above diagram that:

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

Therefore we have:

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

$\blacksquare$

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): $\S 1.1$: Exercise $1$
• Gary Chartrand: Introductory Graph Theory (1977): Appendix $\text{A}.6$: Problem $36$