Respuesta :

caylus

Answer: C

Step-by-step explanation:

[tex]Method\ 1\\\\\displaystyle \sum _{i=1}^{12} (3i+1)=\dfrac{((3*12+1)+(3*1+1))*12}{2} =(37+4)*6=246\\\\Methode\ 2\\\\\displaystyle \sum _{i=1}^{12} (3i+1)=\sum _{i=1}^{12} (1)+3*\sum _{i=1}^{12} (i)=12+3*\dfrac{(12+1)*12}{2} =12+3*13*6=246\\[/tex]

Answer C

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