To evaluate the sum \(\sum_{r=1}^n (2r-1)^2\), we first expand the term inside the summation: \((2r-1)^2 = 4r^2 - 4r + 1\). The entire expression becomes \(\sum_{r=1}^n (4r^2 - 4r + 1)\). By the linearity of summation, this is equal to \(4 \sum_{r=1}^n r^2 - 4 \sum_{r=1}^n r + \sum_{r=1}^n 1\).
We use the standard power sum formulas: \(\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}\), \(\sum_{r=1}^n r = \frac{n(n+1)}{2}\), and \(\sum_{r=1}^n 1 = n\). Substituting these into our expression, we get:
\[4 \left( \frac{n(n+1)(2n+1)}{6} \right) - 4 \left( \frac{n(n+1)}{2} \right) + n\]
Simplifying this, we have:
\[\frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n = \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3}\]
Factoring out \(n\):
\[= \frac{n}{3} [2(n+1)(2n+1) - 6(n+1) + 3]\]
\[= \frac{n}{3} [2(2n^2 + 3n + 1) - 6n - 6 + 3]\]
\[= \frac{n}{3} [4n^2 + 6n + 2 - 6n - 3] = \frac{n}{3} [4n^2 - 1]\]
\[= \frac{1}{3} n(2n+1)(2n-1)\]
Thus, we have shown \(\sum_{r=1}^n (2r-1)^2 = \frac{1}{3} n(2n+1)(2n-1)\), where \(a=2\) and \(b=1\).
Now, we compute \(\sum_{r=1}^{2n} (2r-1)^2\) by substituting \(2n\) for \(n\) in the formula obtained in part (a).
\[\sum_{r=1}^{2n} (2r-1)^2 = \frac{1}{3} (2n) (2(2n)+1) (2(2n)-1)\]
\[= \frac{2n}{3} (4n+1)(4n-1)\]
\[= \frac{2n(16n^2-1)}{3} = \frac{32n^3 - 2n}{3}\]
Thus, the sum \(\sum_{r=1}^{2n} (2r-1)^2\) is \(\frac{2n(4n+1)(4n-1)}{3}\).
This method of calculating sums of squared terms is foundational in fields like statistics and signal processing, where you often need to calculate the variance of a sequence of numbers or determine the energy of discrete signals. For example, if you were analyzing a sequence of alternating voltage levels at set time intervals, you might need to calculate the total energy, which is proportional to the sum of the squares of these levels. Using these summation formulas makes these calculations much faster and more accurate than manual addition.