How would you write the series 3+12+48+192%C2%A0using sigma notation?(1 point)%0AResponses%0A%0A∑4n=1(3)(4)n−1%0Asigma-summation underscript n equals 1 overscript 4 End Scripts left parenthesis 3 right parenthesis left parenthesis 4 right parenthesis superscript n minus 1 baseline%0A%0A∑3n=1(3)(4)n−1%0Asigma-summation underscript n equals 1 overscript 3 End Scripts left parenthesis 3 right parenthesis left parenthesis 4 right parenthesis superscript n minus 1 baseline%0A%0A∑4n=1(4)(3)n−1%0Asigma-summation underscript n equals 1 overscript 4 End Scripts left parenthesis 4 right parenthesis left parenthesis 3 right parenthesis superscript n minus 1 baseline%0A%0A∑4n=1(3)(−4)n−1

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8 months ago

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1 answer

The correct way to write the series 3+12+48+192 using sigma notation is:

∑n=1^4 3(4)^n-1

answered by GPT 3.5

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How would you write the series 3+12+48+192%C2%A0using sigma notation?(1 point)%0AResponses%0A%0A∑4n=1(3)(4)n−1%0Asigma-summation underscript n equals 1 overscript 4 End Scripts left parenthesis 3 right parenthesis left parenthesis 4 right parenthesis superscript n minus 1 baseline%0A%0A∑3n=1(3)(4)n−1%0Asigma-summation underscript n equals 1 overscript 3 End Scripts left parenthesis 3 right parenthesis left parenthesis 4 right parenthesis superscript n minus 1 baseline%0A%0A∑4n=1(4)(3)n−1%0Asigma-summation underscript n equals 1 overscript 4 End Scripts left parenthesis 4 right parenthesis left parenthesis 3 right parenthesis superscript n minus 1 baseline%0A%0A∑4n=1(3)(−4)n−1

1 answer

The correct way to write the series 3+12+48+192 using sigma notation is:

∑n=1^4 3(4)^n-1

GPT 3.5 · Answer

The correct way to write the series 3+12+48+192 using sigma notation is: ∑n=1^4 3(4)^n-1