Certainly! Let's evaluate the summation [tex]\(\sum_{n=2}^7 \frac{(-1)^n(2n+1)}{3n-1}\)[/tex] step-by-step.
1. Identify the range of summation:
- The summation runs from [tex]\( n = 2 \)[/tex] to [tex]\( n = 7 \)[/tex].
2. Evaluate each term in the series:
- For [tex]\( n = 2 \)[/tex]:
[tex]\[
\frac{(-1)^2 (2 \cdot 2 + 1)}{3 \cdot 2 - 1} = \frac{1 \cdot 5}{6 - 1} = \frac{5}{5} = 1.0
\][/tex]
- For [tex]\( n = 3 \)[/tex]:
[tex]\[
\frac{(-1)^3 (2 \cdot 3 + 1)}{3 \cdot 3 - 1} = \frac{-1 \cdot 7}{9 - 1} = \frac{-7}{8} = -0.875
\][/tex]
- For [tex]\( n = 4 \)[/tex]:
[tex]\[
\frac{(-1)^4 (2 \cdot 4 + 1)}{3 \cdot 4 - 1} = \frac{1 \cdot 9}{12 - 1} = \frac{9}{11} \approx 0.8181818181818182
\][/tex]
- For [tex]\( n = 5 \)[/tex]:
[tex]\[
\frac{(-1)^5 (2 \cdot 5 + 1)}{3 \cdot 5 - 1} = \frac{-1 \cdot 11}{15 - 1} = \frac{-11}{14} \approx -0.7857142857142857
\][/tex]
- For [tex]\( n = 6 \)[/tex]:
[tex]\[
\frac{(-1)^6 (2 \cdot 6 + 1)}{3 \cdot 6 - 1} = \frac{1 \cdot 13}{18 - 1} = \frac{13}{17} \approx 0.7647058823529411
\][/tex]
- For [tex]\( n = 7 \)[/tex]:
[tex]\[
\frac{(-1)^7 (2 \cdot 7 + 1)}{3 \cdot 7 - 1} = \frac{-1 \cdot 15}{21 - 1} = \frac{-15}{20} = -0.75
\][/tex]
3. Sum the evaluated terms:
- The terms computed in the series are: [tex]\( 1.0, -0.875, 0.8181818181818182, -0.7857142857142857, 0.7647058823529411, -0.75 \)[/tex].
- Adding these together:
[tex]\[
1.0 + (-0.875) + 0.8181818181818182 + (-0.7857142857142857) + 0.7647058823529411 + (-0.75) \approx 0.17217341482047366
\][/tex]
Therefore, the sum of the series [tex]\(\sum_{n=2}^7 \frac{(-1)^n(2n+1)}{3n-1}\)[/tex] is approximately [tex]\(0.17217341482047366\)[/tex].
