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Let's break this down step by step.

### Part (i)

We are asked to use the result:

\[
\frac{3}{r} - \frac{2}{r+1} - \frac{1}{r+2} = \frac{4r+6}{r(r+1)(r+2)}
\]

To find a constant \( C \) such that:

\[
\sum_{r=1}^{n} \frac{2r+3}{r(r+1)(r+2)} = C - \frac{3}{2(n+1)} - \frac{1}{2(n+2)}
\]

#### Solution Approach:

The first step is to recognize that the left-hand side of the equation is a summation of a term that is closely related to the given result. Specifically, we can rewrite:

\[
\frac{2r+3}{r(r+1)(r+2)} = \frac{3}{r} - \frac{2}{r+1} - \frac{1}{r+2}
\]

Thus, the sum becomes:

\[
\sum_{r=1}^{n} \frac{2r+3}{r(r+1)(r+2)} = \sum_{r=1}^{n} \left( \frac{3}{r} - \frac{2}{r+1} - \frac{1}{r+2} \right)
\]

Now, by summing the right-hand side, many terms will cancel out due to the telescoping nature of the series.

- The term \( \frac{3}{r} \) gives the sum \( 3 \sum_{r=1}^{n} \frac{1}{r} \).
- The term \( \frac{2}{r+1} \) gives the sum \( 2 \sum_{r=1}^{n} \frac{1}{r+1} \).
- The term \( \frac{1}{r+2} \) gives the sum \( \sum_{r=1}^{n} \frac{1}{r+2} \).

#### Simplifying:

- The first sum \( \sum_{r=1}^{n} \frac{1}{r} \) results in a harmonic sum \( H_n \).
- The second sum \( \sum_{r=1}^{n} \frac{1}{r+1} \) is just a shifted harmonic sum \( H_{n+1} - 1 \).
- The third sum \( \sum_{r=1}^{n} \frac{1}{r+2} \) is \( H_{n+2} - H_2 \), which simplifies further.

The constant \( C \) can be determined by comparing the result of the summation and simplifying the remaining terms on the right-hand side.

### Part (ii)

We are asked to find the sum:

\[
\frac{7}{2 \times 3 \times 4} + \frac{9}{3 \times 4 \times 5} + \frac{11}{4 \times 5 \times 6} + \cdots + \frac{2n+3}{n(n+1)(n+2)}
\]

This is a telescoping series similar to the previous problem. To solve this, we need to express the general term:

\[
\frac{2r+3}{r(r+1)(r+2)}
\]

and sum it over \( r = 1 \) to \( n \). Using the same result from part (i), we can express the sum in a simplified form, recognizing the pattern that results in cancellation of intermediate terms.

Would you like further details for each part? Here are some questions to extend your understanding:

1. How does a telescoping series work, and why is it useful here?
2. What are harmonic numbers, and how do they relate to summations of fractions?
3. Can you generalize the result for any term \( \frac{2r+3}{r(r+1)(r+2)} \) to see how the series evolves?
4. What other techniques could be used to sum such series if telescoping isn't possible?
5. How can you verify your final constant \( C \) by testing specific values of \( n \)?

**Tip:** Telescoping sums often involve breaking down fractions into simpler parts to observe the cancellation of terms across the summation boundary.

Using the result: 3/r - 2/(r+1) - 1/(r+2) = (4r+6) / (r(r+1)(r+2)), find the constant C such that Σ (from r=1 to n) (2r+3) / (r(r+1)(r+2)) = C - 3/(2(n+1)) - 1/(2(n+2)) and also find the sum: 7/(2×3×4) + 9/(3×4×5) + 11/(4×5×6) + ... + (2n+3)/(n(n+1)(n+2)).

This detailed solution explores how to solve a telescoping series involving summations of fractional terms. Learn to find the constant C and compute complex series sums. Suitable for advanced high school or university students.

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