Sum of Sequence of Squares/Proof by Induction

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:

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.

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.

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:

$\ds \sum_{i \mathop = 1}^{k + 1} i^2 = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 1} } 6$

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:

$\ds \sum_{i \mathop = 1}^{k + 1} i^2$

$=$

$\ds \frac {k \paren {k + 1} \paren {2 k + 1} } 6 + \paren {k + 1}^2$

$\ds $

$=$

$\ds \frac {k \paren {k + 1} \paren {2 k + 1} + 6 \paren {k + 1}^2} 6$

$\ds $

$=$

$\ds \frac {\paren {k + 1} \paren {k \paren {2 k + 1} + 6 \paren {k + 1} } } 6$

$\ds $

$=$

$\ds \frac {\paren {k + 1} \paren {2 k^2 + 7 k + 6} } 6$

$\ds $

$=$

$\ds \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 1} } 6$

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

Therefore:

$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Method of Induction: Example $1$
1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Properties of the Natural Numbers: $\S 20 \beta$
1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 3$: Natural Numbers: Exercise $\S 3.11 \ (1) \ \text{(i)}$
1979: John E. Hopcroft and Jeffrey D. Ullman: Introduction to Automata Theory, Languages, and Computation ... (previous) ... (next): Chapter $1$: Preliminaries: $1.3$ Inductive Proofs: Example $1.1$
1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction
2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): mathematical induction: $\text {(i)}$