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Calculate:
[tex]\[ \sum_{n=1}^{10}(10n + 15)(2n - 3) \][/tex]

A. 4,450
B. 5,225
C. 7,250
D. 6,750
E. 6,400

Sagot :

GinnyAnswer
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.
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