We will use the distributive property to expand this product: [tex]\[ (10n + 15)(2n - 3) = 10n \cdot 2n + 10n \cdot (-3) + 15 \cdot 2n + 15 \cdot (-3) \][/tex] [tex]\[ = 20n^2 - 30n + 30n - 45 \][/tex] [tex]\[ = 20n^2 - 15 \][/tex]
Notice the term [tex]\(30n - 30n\)[/tex] cancels out.
2. Simplify Summation: We need to sum this expression from [tex]\(n = 1\)[/tex] to [tex]\(n = 10\)[/tex]: [tex]\[ \sum_{n=1}^{10} (20n^2 - 15) \][/tex]
This can be split into two separate summations: [tex]\[ \sum_{n=1}^{10} 20n^2 - \sum_{n=1}^{10} 15 \][/tex]
3. Compute Each Summation:
- Sum of [tex]\(20n^2\)[/tex] from 1 to 10: [tex]\[ 20 \sum_{n=1}^{10} n^2 \][/tex] We know that the sum of the squares of the first [tex]\(m\)[/tex] natural numbers is given by: [tex]\[ \sum_{n=1}^{m} n^2 = \frac{m(m+1)(2m+1)}{6} \][/tex]
