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I need to demonstrate that:

(a)

$$1^2+2^2+3^2+\dots+n^2 = \frac{n(n+1)(2n+1)}6$$

using "mathematical induction". If I make some interpretation of the Peano's postulates then the way to demonstrate this is proving that (a) statement is true for any value of $n$ and for doing that we need to implicate that must be true for any number $k=n+1$.

If I make a geometric interpretation of the problem. Plotting the right side of (a). $f(x)= \frac{n(n+1)(2n+1)}6$

graphic interpretation of f(x) on (a)

We can see that (a) is a CONTINUOUS FUNCTION.

My question:

Is it valid to say that if a function is a "continuos function" is the same as making the math induction of the validity of (a) for any number k=n+1?

Michael Hardy
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Haizum Skallah
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1 Answers1

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No, simply saying that this is a continuous function does not work as an induction step.

The function $f(n) = n$ is continuous, but it is certainly not true that

$$\sum\limits_{i = 1}^n i^2 = n.$$

Edit: To actually do the induction, (I'm assuming that you have already established a base case):

First, assume that the result ($\sum\limits_{i = 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$) holds for some $k$.

Then, this gives that

$$\sum\limits_{i = 1}^k i^2 = \frac{k(k + 1)(2k + 1)}{6},$$ so

$$\sum\limits_{i = 1}^{k + 1} i^2 = \frac{k(k + 1)(2k + 1)}{6} + (k + 1)^2.$$

Now, multiplying out the numerator of this fraction and rewriting $(k + 1)^2$ as $\frac{6(k+1)^2}{6}$ gives that

$$\sum\limits_{i = 1}^{k + 1} i^2 = \frac{2k^3 + 2k^2 + k^2 + k + 6(k^2 + 2k + 1)}{6} = \frac{2k^3 + 9k^2 + 13k + 6}{6}.$$

Finally, factoring this cubic gives that

$$\sum\limits_{i = 1}^{k + 1} i^2 = \frac{(k + 1)(k + 2)(2k + 3)}{6} = \frac{[k + 1]([k + 1] + 1)(2[k + 1] + 1)}{6}.$$

Thus, since the result holds for $n = 1$, and since if the result holds for some $n = k$, then it also holds for $n = k + 1$, by the principle of mathematical induction, the result holds for all $n \in \mathbb{N} \setminus \{0\}$ (Or $\mathbb{N}$ if zero isn't a natural number for you.)

Edit again: Also, this works with the lower bounds changed to zero, which might be a bit neater as a result.

qaphla
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