Let's solve these summations step by step:

1. $\sum_{j=1}^{8}(2j-3)$

Expand the summation: $\sum_{j=1}^{8}(2j-3) = (2(1)-3) + (2(2)-3) + \cdots + (2(8)-3).$ Simplify and calculate each term: $\sum_{j=1}^{8}(2j-3) = -1 + 1 + 3 + 5 + 7 + 9 + 11 + 13 = 48.$

2. $\sum_{j=0}^{10}(j^2 - 4j)$

Separate the terms: $\sum_{j=0}^{10}(j^2 - 4j) = \sum_{j=0}^{10}j^2 - 4\sum_{j=0}^{10}j.$ Use formulas:

• Sum of squares: $\sum_{j=0}^{n} j^2 = \frac{n(n+1)(2n+1)}{6}$.
• Sum of integers: $\sum_{j=0}^{n} j = \frac{n(n+1)}{2}$.

For $n = 10$: $\sum_{j=0}^{10} j^2 = \frac{10(10+1)(2(10)+1)}{6} = \frac{10(11)(21)}{6} = 385.$ $\sum_{j=0}^{10} j = \frac{10(10+1)}{2} = 55.$ Thus: $\sum_{j=0}^{10}(j^2 - 4j) = 385 - 4(55) = 385 - 220 = 165.$

3. $\sum_{j=0}^{5} \frac{j+1}{j+2}$

Break it down: $\sum_{j=0}^{5} \frac{j+1}{j+2} = \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5} + \frac{5}{6} + \frac{6}{7}.$ Approximate numerically or compute fractions as: $\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5} + \frac{5}{6} + \frac{6}{7} \approx 4.736.$

4. $\sum_{j=1}^{6} 2^j$

This is a geometric series: $\sum_{j=1}^{6} 2^j = 2 + 4 + 8 + 16 + 32 + 64.$ $\sum_{j=1}^{6} 2^j = 126.$

5. $\sum_{j=1}^{6} \left(\sqrt{j+1} - \sqrt{j}\right)$

This is a telescoping sum: $\sum_{j=1}^{6} \left(\sqrt{j+1} - \sqrt{j}\right) = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{7} - \sqrt{6}).$ The terms cancel, leaving: $\sqrt{7} - \sqrt{1} = \sqrt{7} - 1.$

6. $\sum_{j=1}^{9} 2$

This is a constant sum: $\sum_{j=1}^{9} 2 = 2 + 2 + \cdots + 2 = 9 \cdot 2 = 18.$

7. $\sum_{j=0}^{4} (-2)^{j-1}$

Expand the terms: $\sum_{j=0}^{4} (-2)^{j-1} = (-2)^{-1} + (-2)^{0} + (-2)^{1} + (-2)^{2} + (-2)^{3}.$ Simplify: $\frac{-1}{2} + 1 - 2 + 4 - 8 = -\frac{5}{2}.$

8. $\sum_{j=4}^{10} 3$

This is a constant sum: $\sum_{j=4}^{10} 3 = 3 + 3 + 3 + 3 + 3 + 3 + 3 = 7 \cdot 3 = 21.$

9. $\sum_{j=1}^{n} n$

This uses the formula for the sum of integers: $\sum_{j=1}^{n} j = \frac{n(n+1)}{2}.$

10. $\sum_{j=1}^{n} j$

Same as above: $\sum_{j=1}^{n} j = \frac{n(n+1)}{2}.$

Do you want detailed steps for any specific summation? Here are some related questions:

1. What is a telescoping series, and why do terms cancel?
2. How do geometric series differ from arithmetic ones?
3. Can you explain formulas for sums of squares and integers in detail?
4. How can approximations be improved for numerical fractions like in question 3?
5. What are common mistakes when dealing with summations involving negative exponents?

Tip: When solving summations, always check if it's telescoping, geometric, or arithmetic to simplify calculations!