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Find the 
3^(rd) partial sum of the series 
sum_(n=1)(-1)^(n)(4)/(2n^((4)/(3))+6). Then use the alternating series error bound to find an upper and lower bound for the sum of the series. Round to three decimal places.

Find the 3rd 3^{\mathrm{rd}} partial sum of the series n=1(1)n42n43+6 \sum_{n=1}(-1)^{n} \frac{4}{2 n^{\frac{4}{3}}+6} . Then use the alternating series error bound to find an upper and lower bound for the sum of the series. Round to three decimal places.

Full solution

Q. Find the 3rd 3^{\mathrm{rd}} partial sum of the series n=1(1)n42n43+6 \sum_{n=1}(-1)^{n} \frac{4}{2 n^{\frac{4}{3}}+6} . Then use the alternating series error bound to find an upper and lower bound for the sum of the series. Round to three decimal places.
  1. Calculate total tape needed: Step 11: Determine the total amount of tape needed and the amount per roll.\newlineTotal tape needed = 8,000cm8,000 \, \text{cm}, Tape per roll = 2,000cm2,000 \, \text{cm}.\newlineWe need to divide the total tape needed by the tape per roll to find out how many rolls are required.\newlineCalculation: 8,000÷2,000=48,000 \div 2,000 = 4.
  2. Calculate number of rolls: Step 22: Conclusion based on the calculation.\newlineSince 8,0008,000 divided by 2,0002,000 equals 44, the electrician needs to order 44 rolls of tape.

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