Solutions

(i) A natural approach to exercises that require finding a formula for a partial sum, is to evaluate it for certain values. Then, from these values, it might be possible to establish a conjecture and, from there, prove the conjecture. This is exactly what will be done for this exercise. We start by evaluating the partial sum for certain values of $n$.

n	Partial Sum
1	1
2	4
3	9

With only three evaluations, we should be able to deduce the conjecture that our partial sum is equal to $n^{2}$. That is,

$$\begin{aligned} \sum_{i = 1}^{n} (2i - 1) = n^{2} \end{aligned}$$

Now, to prove this conjecture, we can either use induction or a direct proof. In this solutions manual, a direct proof will always be favored over a proof by induction with the reason being that, as opposed to proofs by induction, a direct proof shows the way to obtain the formula.

Thus, how are we to obtain this formula with some simple algebraic manipulations? First, it definitely is a good idea to rewrite the sum to further observe it.

$$\begin{aligned} 1 + 3 + 5 + \dots + (2n -1) &= 1 + 3 + \dots + (2(n - 1) - 1) + (2n - 1) \\ &= 1 + 3 + \dots + (2n - 3) + (2n - 1) \end{aligned}$$

We realize that if we sum a number from the $i^{th}$ position with the $i^{th}$ number from the last, we obtain the term $2n$. For instance, the sum of the first and last term is:

$$\begin{aligned} 1 + (2n - 1) = 1 - 1 + 2n = 2n \end{aligned}$$

Thus, all we need to do now, is handle the case when $n$ is odd and the other where $n$ is even. If $n$ is even, then it is quite simple:

$$\begin{aligned} 1 + 3 + \dots + (2n - 3) + (2n - 1) &= 1 + (2n - 1) + 3 + (2n - 3) + \dots \\ &= \underbrace{2n + 2n + \dots}_{n/2} \\ &= 2n(n/2) \\ &= n^{2} \end{aligned}$$

Now, if $n$ is odd, we have a term "in the middle" that does not have a term to sum up with. However, it is compensated by the fact that there are not $n/2$ $2n$ terms but $\frac{n - 1}{2}$ of them. Furthermore, we know the middle term's index: $\frac{n + 1}{2}$. Thus, putting all the information together, we obtain,

$$\begin{aligned} 1 + & 3 + \dots + \underbrace{(2(\frac{n + 1}{2}) - 1)}_{\text{middle term}} + \dots + (2n - 3) + (2n - 1) \\ &= (1 + (2n - 1)) + (3 + (2n - 3)) + \dots + (n + 1 - 1) \\ &= \underbrace{2n + 2n + \dots}_{\frac{n - 1}{2}} + n \\ &= 2n (\frac{n - 1}{2}) + n \\ &= n^{2} - n + n \\ &= n^{2} \end{aligned}$$

Just like that, we have found and proven that the formula for the partial sum $\sum_{i = 1}^{n} (2i - 1) = 1 + 3 + 5 + \dots + (2n -1)$ is $n^{2}$.

NB. We have implicitly assumed that the natural numbers $\mathbb{N}$ are comprised of even and odd positive numbers.

(ii) When no answer seems in sight, it is best to simply scribble and hope for an opening.

Nonetheless, while solving exercises from Chapter 1, one might notice that those preceding one were usually used to help solve it. With that in mind, a small hint for this exercise would be to look at the formula of exercise 1.1.

$$\begin{aligned} 1^{2} + \dots + n^{2} = \frac{n(n+1)(2n+1)}{6} \end{aligned}$$

Since we know the formula for the partial sum of squared natural numbers, would it not be convenient to subtract from it the sum of squared even numbers leading up to an odd number? You might say: But we would have to find the formula for the sum of squared even numbers! That would make two formulas to find! 'Deed that does make it two formulas to find. Let's get to work!

$$\begin{aligned} \sum_{i = 1}^{n} (2i - 1)^{2} &= 1^{2} + 3^{2} + \dots + (2n -1)^{2} \\ &= 1^{2} + 2^{2} + \dots + {(2n - 2)}^{2} + {(2n - 1)}^{2} \\ & \qquad - (2^{2} + 4^{2} + \dots + {(2n - 4)}^{2} + {(2n - 2)}^{2}) \\ &= \frac{(2n - 1)(2n - 1 + 1)(2 (2n - 1) + 1)}{6} \\ & \qquad - ({(2 \cdot 1)}^{2} + {(2 \cdot 2)}^{2} + \dots + {(2 \cdot (n - 2) )}^{2} + {(2 \cdot (n - 1))}^{2} ) \\ &= \frac{(2n - 1)(2n)(4n - 2 + 1)}{6} \\ & \qquad - 2^{2} ({1}^{2} + {2}^{2} + \dots + {(n - 2)}^{2} + {(n - 1)}^{2} ) \\ &= \frac{n(2n - 1)(4n - 1)}{3} - 4 \frac{(n - 1) (n - 1 + 1) (2(n - 1) + 1)}{6} \\ &= \frac{n(2n - 1)(4n - 1) - 2n (n - 1) (2n - 1)}{3} \\ &= \frac{n(2n - 1) ((4n - 1) - 2 (n - 1) )}{3} \\ &= \frac{n(2n - 1) (4n - 1 - 2n + 2 )}{3} \\ &= \frac{n(2n - 1) (2n + 1)}{3} \\ &= \frac{n (4n^{2} - 1)}{3} \end{aligned}$$