Respuesta :

lcannady2001
the correct answer is C)216 / Let f(n) = 2n + 5. Then we must simply calculate (f1), f(2), ..., (f12). And we have that f(1) = 2*1 + 5 <=> f(1) = 7; f(2) = 2*2 + 5 <=> f(2) = 9; f(3) = 2*3 + 5 <=> f(3) = 11. And so on. It's easy to see that the rest of the results will be 13, 15, 17, 19, 21 and so on. So the summation will be S = 7 + 9 + 11 +... + 29 <=> S = 216

Answer:

Option: C is the correct answer.

                          C.  216

Step-by-step explanation:

We are asked to evaluate the sum of the following summation:

[tex]\sum_{n=1}^{12} (2n+5)[/tex]

Now, this summation could also be written as:

[tex]\sum_{n=1}^{12} (2n+5)=\sum_{n=1}^{12} 2n+\sum_{n=1}^{12} 5[/tex]

i.e.

[tex]\sum_{n=1}^{12} (2n+5)=2\sum_{n=1}^{12} n+\sum_{n=1}^{12} 5[/tex]

Now, we know that:

[tex]\sum_{n=1}^{12} n=1+2+3+4+....+12\\\\i.e.\\\\\sum_{n=1}^{12} n=\dfrac{12(12+1)}{2}[/tex]

Since,

[tex]\sum_{i=1}^{n} i=\dfrac{n(n+1)}{2}[/tex]

i.e.

[tex]\sum_{n=1}^{12} n=78[/tex]

and

[tex]\sum_{n=1}^{12} 5=5+5+....+5\ (\text{twelve\ times})\\\\i.e.\\\\\sum_{n=1}^{12} 12=12\times 5\\\\i.e.\\\\\sum_{n=1}^{12} n=60[/tex]

Hence, we have:

[tex]\sum_{n=1}^{12} (2n+5)=2\times 78+60[/tex]

i.e.

[tex]\sum_{n=1}^{12} (2n+5)=156+60[/tex]

i.e.

[tex]\sum_{n=1}^{12} (2n+5)=216[/tex]

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