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Sagot :
The summation of the considered expression in terms of n from n = 1 to 14 is given by: Option D: 343
How to find the sum of consecutive integers?
[tex]1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}[/tex]
What are the properties of summation?
[tex]\sum_{i=r}^s (a \times f(i) + b) = a \times [\: \sum_{i=r}^s f(i)] + (s-r)b[/tex]
where a, b, r, and s are constants, f(i) is function of i, i ranging from r to s (integral assuming).
For the given case, the considered summation can be written symbolically as:
[tex]\sum_{n=1}^{14} (3n + 2)[/tex]
It is evaluated as;
[tex]\sum_{n=1}^{14} (3n + 2) = 3 \times [ \: \sum_{n=1}^{14} n ] + \sum_{n=1}^{14} 2\\\\\sum_{n=1}^{14} (3n + 2) = 3 \times \dfrac{(14)(14 + 1)}{2} + (2 + 2 + .. + 2(\text{14 times}))\\\\\sum_{n=1}^{14} (3n + 2) = 3 \times 105 + 28 = 343\\[/tex]
Thus, the summation of the considered expression in terms of n from n = 1 to 14 is given by: Option D: 343
Learn more about summation here:
https://brainly.com/question/14322177
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