Given that \( \sum_{r=1}^{n} r^{2}=\frac{1}{6} n(n+1)(2 n+1) \), find an expression, in simplified form, for \( \sum_{r=n+1}^{2 n}(2 r-1)^{2} \). (b) [A level \( \mathbf{N} 2006 / \mathbf{P} 1 / \) Q11 modified] Given \( \sum_{r=1}^{n} r^{3}=\frac{1}{4} n^{2}(n+1)^{2} \), show that \( 2^{3}+4^{3}+6^{3}+\ldots \ldots+(2 n)^{3}=2 n^{2}(n+1)^{2} \) Hence or otherwise find \( \sum_{r=1}^{n}(2 r-1)^{3} \), simplifying your answer.

To solve the problem, we will break it down into two parts as requested. ### Part (a) We need to find an expression for \[ \sum_{r=n+1}^{2n} (2r-1)^{2}. \] First, we can rewrite the term \( (2r-1)^{2} \): \[ (2r-1)^{2} = 4r^{2} - 4r + 1. \] Now, we can express the sum as: \[ \sum_{r=n+1}^{2n} (2r-1)^{2} = \sum_{r=n+1}^{2n} (4r^{2} - 4r + 1). \] This can be separated into three sums: \[ = 4\sum_{r=n+1}^{2n} r^{2} - 4\sum_{r=n+1}^{2n} r + \sum_{r=n+1}^{2n} 1. \] Now, we will calculate each of these sums separately. 1. **Calculating \( \sum_{r=n+1}^{2n} r^{2} \)**: Using the formula for the sum of squares, we have: \[ \sum_{r=1}^{m} r^{2} = \frac{1}{6} m(m+1)(2m+1). \] Thus, \[ \sum_{r=n+1}^{2n} r^{2} = \sum_{r=1}^{2n} r^{2} - \sum_{r=1}^{n} r^{2}. \] Calculating these: \[ \sum_{r=1}^{2n} r^{2} = \frac{1}{6} (2n)(2n+1)(4n+1), \] \[ \sum_{r=1}^{n} r^{2} = \frac{1}{6} n(n+1)(2n+1). \] So, \[ \sum_{r=n+1}^{2n} r^{2} = \frac{1}{6} (2n)(2n+1)(4n+1) - \frac{1}{6} n(n+1)(2n+1). \] Factoring out \( \frac{1}{6}(2n+1) \): \[ = \frac{1}{6}(2n+1) \left( 2n(4n+1) - n(n+1) \right). \] Calculating the expression inside the parentheses: \[ 2n(4n+1) - n(n+1) = 8n^{2} + 2n - n^{2} - n = 7n^{2} + n. \] Thus, \[ \sum_{r=n+1}^{2n} r^{2} = \frac{1}{6}(2n+1)(7n^{2}+n). \] 2. **Calculating \( \sum_{r=n+1}^{2n} r \)**: This is a simple arithmetic series: \[ \sum_{r=n+1}^{2n} r = \sum_{r=1}^{2n} r - \sum_{r=1}^{n} r = \frac{1}{2}(2n)(2n+1) - \frac{1}{2}n(n+1). \] Calculating this gives: \[ = n(2n+1) - \frac{1}{2}n(n+1) = n(2n+1) - \frac{1}{2}(n^{2}+n) = n(2n+1) - \frac{1}{2}n^{2} - \frac{1}{2}n. \] Combining terms: \[ = \frac{3}{2}n^{2} + \frac{3}{2}n. \] 3. **Calculating \( \sum_{r=n+1}^{2n} 1 \)**: This is simply the count of terms from \( n+1 \) to \( 2n \), which is \( n \). Now substituting back into our expression: \[ \sum_{r=n+1}^{2n} (2r-1)^{2} = 4\left(\frac{1}{6}(2n+1)(7n^{2}+n)\right) - 4\left(\frac{3}{2}n^{2} + \frac{3}{2}n\right) + n. \] Calculating this gives: \[ = \frac{2}{3}(2n+1)(7n^{2}+n) - 6n^{2} - 6n + n. \] Now we can simplify this expression to get the final result. ### Part (b) We need to show that \[ 2^{3}+4^{3}+6^{3}+\ldots+(2n)^{3}=2n^{2}(n+1)^{2}. \] Using the formula for the sum of cubes: \[ \sum_{r=1}^{n} r^{3} = \left(\frac{n(n+1)}{2}\right)^{2}. \] Thus, \[ \sum_{r=1}^{n} (2r)^{3} = 8\sum_{r=1}^{n} r^{3} = 8\left(\frac{n(n+1)}{2}\right)^{2} = 2n^{2}(n+1)^{2}. \] This shows that \[ 2^{3}+4^{3}+6^{3}+\ldots+(2n)^{3}=2n^{2}(n+1)^{2}. \] Now, to find \( \sum_{r=1}^{n}(2r-1)^{3} \): \[ (2r-1)^{3} = 8r^{3} - 12r^{2} + 6r - 1. \] Thus, \[ \sum_{r=1}^{n} (2r-1)^{3} = \sum_{r=1}^{n} (8r^{3} - 12r^{2} + 6r - 1). \] Calculating each term: 1. \( 8\sum_{r=1}^{n} r^{3} = 2n^{2}(n+1)^{2} \). 2. \( -12\sum_{r=1}^{n} r^{2} = -12 \cdot \frac{1}{6} n(n+1)(2n+1) = -2n(n+1)(2n+1) \). 3. \( 6\sum_{r=1}^{n} r = 6 \cdot \frac{n(n+1)}{2} = 3n(n+