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Did I do this correctly?

(basis) $\sum_{i=1}^{1} i^2=\frac{1(1+1)(2(1)+1)}{6}$ thus both sides are equal to 1.

(induction) Fixed $n$ and $k$ to be natural numbers. Assume $n=k$, then

$\sum_{i=1}^{k+1}i^2= \frac{k(k+1)(2k+1)}{6} + (k+1)^2$ by induction hypothesis.

$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$

$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$

$=\frac{(k+1)(2k^2+7k+6))}{6}$

$=\frac{(k+1)(2k+3)(k+2))}{6}$

Eroge, the statement is true for the natural number n whenever it is true for any natural number less than n.

Seth
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George
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2 Answers2

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I would write your last sum in terms of $k+1$. What i mean is, if you assume when $n=k$ that

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

then your last sum, which you have written as

$$\sum_{i=1}^{k+1}i^2=\frac{(k+1)(2k+3)(k+2)}{6}$$

I would write as

$$\sum_{i=1}^{k+1}i^2=\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}$$

Because you are trying to show that the formula is valid for $n=k+1$, which is much more obvious at first glance. But your proof looks correct. Just my two cents...

Eleven-Eleven
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HINT

For the induction step start from here

$$\sum_{i=1}^{k+1}i^2= \sum_{i=1}^{k}(i^2)+(k+1)^2=\ldots$$

and then by the induction hypothesis

$$\ldots=\frac{k(k+1)(2k+1)}{6} +(k+1)^2=\ldots$$

and we need to show that it is equal to

$$\ldots=\frac{(k+1)(k+2)(2k+3)}{6}$$

user
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