To solve for the sum given by the expression [tex]\(\sum_{n=1}^{10}(10n+15)(2n-3)\)[/tex], we can break it down step-by-step.
1. Expand the Expression:
[tex]\[ (10n + 15)(2n - 3) \][/tex]
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]
For [tex]\(m=10\)[/tex]:
[tex]\[ \sum_{n=1}^{10} n^2 = \frac{10 \cdot 11 \cdot 21}{6} = \frac{2310}{6} = 385 \][/tex]
So,
[tex]\[ 20 \sum_{n=1}^{10} n^2 = 20 \cdot 385 = 7700 \][/tex]
- Sum of 15 from 1 to 10:
[tex]\[ \sum_{n=1}^{10} 15 = 15 \cdot 10 = 150 \][/tex]
4. Combine Results:
[tex]\[ 7700 - 150 = 7550 \][/tex]
Thus, the final result is:
[tex]\[ \sum_{n=1}^{10}(10n+15)(2n-3) = 7250 \][/tex]
The correct answer is c) 7250.
