Sum of Sequence of Squares/Proof by Induction
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Theorem
- $\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
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:
\(\ds \sum_{i \mathop = 1}^{k + 1} i^2\) | \(=\) | \(\ds \frac {k \paren {k + 1} \paren {2 k + 1} } 6 + \paren {k + 1}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k \paren {k + 1} \paren {2 k + 1} + 6 \paren {k + 1}^2} 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {k + 1} \paren {k \paren {2 k + 1} + 6 \paren {k + 1} } } 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {k + 1} \paren {2 k^2 + 7 k + 6} } 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 1} } 6\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$
$\blacksquare$
Sources
- 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: $\text {(i)}$