Sum of Sequence of Squares/Proof by Induction

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

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:

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

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 20 \beta$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 3.11 \ (1) \ \text{(i)}$