Respuesta :

Sum of the geometric series: S=?
S=(-2)(-3)^(1-1)+(-2)(-3)^(2-1)+(-2)(-3)^(3-1)+(-2)(-3)^(4-1)
S=(-2)[(-3)^0+(-3)^1+(-3)^2+(-3)^3]
S=(-2)[1+(-3)+9+(-27)]
S=(-2)(1-3+9-27)
S=(-2)(-20)
S=40

Answer: Third option 40

Answer:

The sum is 40.

Step-by-step explanation:

Given,

[tex]\sum_{n=1}^{4} (-2)(-3)^{n-1}[/tex]

We know that,

[tex]\sum_{n=1}^{4} (-2)(-3)^{n-1}=(-2)(-3)^{1-1}+(-2)(-3)^{2-1}+(-2)(-3)^{3-1}+(-2)(-3)^{4-1}[/tex]

[tex]=(-2)(-3)^{0}+(-2)(-3)^1+(-2)(-3)^2+(-2)(-3)^3[/tex]

[tex]=-2\times 1 - 2\times - 3 - 2\times 9-2\times -27[/tex]

[tex]=-2+6-18+54[/tex]

[tex]=40[/tex]

Hence,

[tex]\sum_{n=1}^{4} (-2)(-3)^{n-1}=40[/tex]

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