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Find the sum of the following series. Round to the nearest hundredth if necessary.

5+15+45+dots+7971615
Sum of a finite geometric series:

S_(n)=(a_(1)-a_(1)r^(n))/(1-r)

Find the sum of the following series. Round to the nearest hundredth if necessary.\newline5+15+45++7971615 5+15+45+\ldots+7971615 \newlineSum of a finite geometric series:\newlineSn=a1a1rn1r S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r}

Full solution

Q. Find the sum of the following series. Round to the nearest hundredth if necessary.\newline5+15+45++7971615 5+15+45+\ldots+7971615 \newlineSum of a finite geometric series:\newlineSn=a1a1rn1r S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r}
  1. Find Number of Terms: Next, we need to find the number of terms nn in the series. We can do this by using the formula for the nth term of a geometric series, which is an=a1r(n1)a_n = a_1 \cdot r^{(n-1)}, where ana_n is the nth term, a1a_1 is the first term, and rr is the common ratio. We will solve for nn using the last term given, which is 79716157971615.\newline7971615=53(n1)7971615 = 5 \cdot 3^{(n-1)}
  2. Isolate nn in Equation: To solve for nn, we need to isolate nn on one side of the equation. We start by dividing both sides by 55.\newline79716155=3(n1)\frac{7971615}{5} = 3^{(n-1)}\newline1594323=3(n1)1594323 = 3^{(n-1)}
  3. Take Logarithm: Now, we take the logarithm of both sides to solve for nn. We can use the natural logarithm or the common logarithm; here, we will use the natural logarithm for convenience.\newlineln(1594323)=ln(3(n1))\ln(1594323) = \ln(3^{(n-1)})
  4. Rewrite Equation: Using the property of logarithms that ln(ab)=b×ln(a)\ln(a^b) = b \times \ln(a), we can rewrite the equation as:\newlineln(1594323)=(n1)×ln(3)\ln(1594323) = (n-1) \times \ln(3)
  5. Solve for n1n-1: Now, we divide both sides by ln(3)\ln(3) to solve for n1n-1.\newline(n1)=ln(1594323)ln(3)(n-1) = \frac{\ln(1594323)}{\ln(3)}
  6. Calculate n1n-1: We calculate the value of n1n-1 using a calculator.\newlinen1=ln(1594323)ln(3)14n-1 = \frac{\ln(1594323)}{\ln(3)} \approx 14
  7. Find nn: To find nn, we add 11 to the result.\newlinen=14+1n = 14 + 1\newlinen=15n = 15
  8. Use Sum Formula: Now that we have the number of terms, we can use the formula for the sum of a finite geometric series to find the sum.\newlineSn=a1a1rn1rS_n = \frac{a_1 - a_1 \cdot r^n}{1 - r}
  9. Substitute Values: We substitute the values we know into the formula.\newlineS15=55×31513S_{15} = \frac{5 - 5 \times 3^{15}}{1 - 3}
  10. Calculate Numerator and Denominator: We calculate the numerator and the denominator separately.\newlineNumerator: 55×3155 - 5 \times 3^{15}\newlineDenominator: 131 - 3
  11. Divide Numerator by Denominator: We calculate the numerator and the denominator using a calculator.\newlineNumerator: 55×31555×143489075717445355 - 5 \times 3^{15} \approx 5 - 5 \times 14348907 \approx 5 - 71744535\newlineDenominator: 13=21 - 3 = -2
  12. Calculate Final Sum: Now we divide the numerator by the denominator to find the sum.\newlineS15=5717445352S_{15} = \frac{5 - 71744535}{-2}
  13. Calculate Final Sum: Now we divide the numerator by the denominator to find the sum.\newlineS15=5717445352S_{15} = \frac{5 - 71744535}{-2} We calculate the final sum using a calculator.\newlineS15=717445302S_{15} = \frac{-71744530}{-2}\newlineS15=35872265S_{15} = 35872265

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