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Let's go through this step-by-step.

The problem is centered around the identity for the sum of odd integers up to a given number. You are tasked with developing this identity, and several sub-parts guide you toward that goal. Let's break it down and explain each part.

### Part (a)
We are asked to find \( \sum_{i=1}^{n} 1 \).

This is just summing 1 over \( n \) terms, so:
\[
\sum_{i=1}^{n} 1 = n.
\]

### Part (b)
We are asked to calculate \( \sum_{i=1}^{n} i = 1 + 2 + 3 + \cdots + n \), which is the sum of the first \( n \) integers.

Using the well-known formula for the sum of the first \( n \) integers:
\[
\sum_{i=1}^{n} i = \frac{n(n+1)}{2}.
\]

### Part (c)
We are tasked with calculating \( \sum_{i=1}^{n} i^2 \), but first, we need to show by direct expansion as a telescoping sum:
\[
\sum_{i=1}^{n} (1 + i)^3 - i^3 = (1 + n)^3 - 1.
\]

#### Expansion of \( (1 + i)^3 - i^3 \):
We can expand \( (1 + i)^3 \) using the binomial expansion:
\[
(1 + i)^3 = 1 + 3i + 3i^2 + i^3.
\]
Thus,
\[
(1 + i)^3 - i^3 = 1 + 3i + 3i^2.
\]

Summing this over \( i \) from 1 to \( n \):
\[
\sum_{i=1}^{n} \left[ (1 + i)^3 - i^3 \right] = \sum_{i=1}^{n} 1 + 3 \sum_{i=1}^{n} i + 3 \sum_{i=1}^{n} i^2.
\]
This gives us:
\[
(1 + n)^3 - 1 = n + 3 \sum_{i=1}^{n} i + 3 \sum_{i=1}^{n} i^2.
\]

Now, using the formula from part (b) for \( \sum_{i=1}^{n} i \), we can substitute:
\[
(1 + n)^3 - 1 = n + 3 \times \frac{n(n+1)}{2} + 3 \sum_{i=1}^{n} i^2.
\]

From here, solving for \( \sum_{i=1}^{n} i^2 \) gives:
\[
\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}.
\]

### Part (d)
We are asked to show that:
\[
\sum_{i=1}^{n} (2i - 1)^2 = \frac{n(2n-1)(2n+1)}{3}.
\]

Using the expansion:
\[
(2i - 1)^2 = 4i^2 - 4i + 1,
\]
we can write:
\[
\sum_{i=1}^{n} (2i - 1)^2 = 4 \sum_{i=1}^{n} i^2 - 4 \sum_{i=1}^{n} i + \sum_{i=1}^{n} 1.
\]

Substitute the known values for \( \sum_{i=1}^{n} i^2 \), \( \sum_{i=1}^{n} i \), and \( \sum_{i=1}^{n} 1 \):
\[
= 4 \times \frac{n(n+1)(2n+1)}{6} - 4 \times \frac{n(n+1)}{2} + n.
\]

Simplifying this expression gives:
\[
= \frac{n(2n-1)(2n+1)}{3}.
\]

This confirms the given identity.

### Do you want more details or have any questions? 

Here are some related questions you could explore further:

1. How would you derive the formula for \( \sum_{i=1}^{n} i^3 \)?
2. What happens when you sum only even integers? Can a similar identity be derived?
3. What is the geometrical interpretation of summing squares of integers?
4. How does this identity relate to other common summation formulas, such as for cubes?
5. Can this method of telescoping sums be generalized to higher powers?

**Tip:** Always remember that when handling summations, look for patterns like telescoping sums or use known formulas to simplify the work!

In this course we will need an identity for the sum of odd integers up to a given number. Consider the sum 1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 and show that it can be represented using a formula. The problem is divided into four parts: (a) calculating the sum of 1 over n terms, (b) calculating the sum of the first n integers, (c) calculating the sum of squares of the first n integers using a telescoping series, and (d) showing a specific identity for the sum of squares of odd integers up to (2n - 1)^2.

Learn how to derive the identity for the sum of squares of odd integers up to (2n - 1)^2 using formulas for arithmetic and telescoping series. Includes a detailed explanation of the sum of first n integers and squares, suitable for high school students.

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