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Find the sum of the geometric series

1-3+3^(2)-3^(3)+dots-3^(29)
Choose 1 answer:
(A) 
-1.03*10^(14)
(B) 
-6.86*10^(13)
(c) 
-5.15*10^(13)
(D) 
1.72*10^(13)

Find the sum of the geometric series\newline13+3233+329 1-3+3^{2}-3^{3}+\ldots-3^{29} \newlineChoose 11 answer:\newline(A) 1.031014 -1.03 \cdot 10^{14} \newline(B) 6.861013 -6.86 \cdot 10^{13} \newline(C) 5.151013 -5.15 \cdot 10^{13} \newline(D) 1.721013 1.72 \cdot 10^{13}

Full solution

Q. Find the sum of the geometric series\newline13+3233+329 1-3+3^{2}-3^{3}+\ldots-3^{29} \newlineChoose 11 answer:\newline(A) 1.031014 -1.03 \cdot 10^{14} \newline(B) 6.861013 -6.86 \cdot 10^{13} \newline(C) 5.151013 -5.15 \cdot 10^{13} \newline(D) 1.721013 1.72 \cdot 10^{13}
  1. Identify Geometric Series: The given series is a geometric series with the first term a=1a = 1 and the common ratio r=3r = -3. The sum of a finite geometric series can be calculated using the formula Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r}, where nn is the number of terms.
  2. Determine Number of Terms: First, we need to determine the number of terms in the series. Since the series starts at 303^0 (which is 11) and goes up to 3293^{29}, and the powers of 33 increase by 11 each time, there are 3030 terms in total.
  3. Apply Sum Formula: Now we can apply the formula for the sum of a geometric series: Sn=a(1rn)/(1r)S_n = a(1 - r^n) / (1 - r). Here, a=1a = 1, r=3r = -3, and n=30n = 30.
  4. Calculate Numerator: Plugging the values into the formula, we get S30=1(1(3)30)1(3)S_{30} = \frac{1(1 - (-3)^{30})}{1 - (-3)}.
  5. Calculate Denominator: Calculating the numerator: 1(3)30=13301 - (-3)^{30} = 1 - 3^{30}. Since 3303^{30} is a very large number, we can leave it in the exponential form for now.
  6. Substitute Values: Calculating the denominator: 1(3)=1+3=41 - (-3) = 1 + 3 = 4.
  7. Calculate Exponential Value: Now we have S30=13304S_{30} = \frac{1 - 3^{30}}{4}. To find the exact value of 3303^{30}, we can use a calculator or a computer.
  8. Perform Subtraction: Using a calculator, we find that 3303^{30} is approximately 2.05×10142.05 \times 10^{14}.
  9. Divide by 44: Substituting this value into our sum expression, we get S30=(12.05×1014)4S_{30} = \frac{(1 - 2.05 \times 10^{14})}{4}.
  10. Express in Scientific Notation: Now we perform the subtraction in the numerator: 12.05×1014=2.05×1014+11 - 2.05 \times 10^{14} = -2.05 \times 10^{14} + 1. Since 11 is negligible compared to 2.05×10142.05 \times 10^{14}, we can approximate this as 2.05×1014-2.05 \times 10^{14}.
  11. Express in Scientific Notation: Now we perform the subtraction in the numerator: 12.05×1014=2.05×1014+11 - 2.05 \times 10^{14} = -2.05 \times 10^{14} + 1. Since 11 is negligible compared to 2.05×10142.05 \times 10^{14}, we can approximate this as 2.05×1014-2.05 \times 10^{14}.Finally, we divide by 44: S30=2.05×1014/4=0.5125×1014S_{30} = -2.05 \times 10^{14} / 4 = -0.5125 \times 10^{14}.
  12. Express in Scientific Notation: Now we perform the subtraction in the numerator: 12.05×1014=2.05×1014+11 - 2.05 \times 10^{14} = -2.05 \times 10^{14} + 1. Since 11 is negligible compared to 2.05×10142.05 \times 10^{14}, we can approximate this as 2.05×1014-2.05 \times 10^{14}.Finally, we divide by 44: S30=2.05×1014/4=0.5125×1014S_{30} = -2.05 \times 10^{14} / 4 = -0.5125 \times 10^{14}.We can write 0.5125×1014-0.5125 \times 10^{14} in scientific notation as 5.125×1013-5.125 \times 10^{13}.

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