[5 marks] Using the result
3
r
− 2
r + 1
− 1
r + 2
= 4r + 6
r(r + 1)(r + 2),
find the constant C such that
nΣ
r=1
2r + 3
r(r + 1)(r + 2)
= C − 3
2(n + 1)
− 1
2(n + 2).

We are tasked with finding the constant \( C \) such that

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

### Step 1: Start with the given result

We know from the problem that:

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

Now, we rewrite the expression on the left-hand side in a similar form. We can decompose \( \frac{2r + 3}{r(r+1)(r+2)} \) using partial fractions. Let’s express it as:

\[
\frac{2r + 3}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2}.
\]

### Step 2: Solving for partial fractions

To find the values of \( A \), \( B \), and \( C \), we equate the numerator \( 2r + 3 \) to the expression obtained by combining the right-hand side:

\[
2r + 3 = A(r+1)(r+2) + B(r)(r+2) + C(r)(r+1).
\]

Expand both sides and match the coefficients of powers of \( r \). After solving for \( A \), \( B \), and \( C \), we find:

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

### Step 3: Substituting into the sum

Now, substitute this result into the sum:

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

This sum can be broken down as:

\[
\sum_{r=1}^n \frac{1}{r} - \sum_{r=1}^n \frac{1}{r+1} - \sum_{r=1}^n \frac{1}{r+2}.
\]

### Step 4: Simplifying the sum

The first two sums are telescoping series, and simplify to:

\[
\sum_{r=1}^n \frac{1}{r} - \sum_{r=1}^n \frac{1}{r+1} = 1 - \frac{1}{n+1}.
\]

For the third sum:

\[
\sum_{r=1}^n \frac{1}{r+2} = \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{n+2}.
\]

Thus, the total sum becomes:

\[
1 - \frac{1}{n+1} - \frac{1}{n+2}.
\]

### Step 5: Comparing with the given expression

We are given that this sum equals \( C - \frac{3}{2(n+1)} - \frac{1}{2(n+2)} \). Comparing the two expressions:

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

Equating the coefficients, we find that \( C = 1 \).

Thus, the constant \( C \) is \( \boxed{1} \).

---

Do you want more details or have any questions? 

Here are 5 questions that relate to this problem:
1. How do telescoping series simplify sums?
2. What are partial fractions, and how are they used in this context?
3. Why do we decompose the fraction \( \frac{2r+3}{r(r+1)(r+2)} \)?
4. What is the significance of equating coefficients in partial fraction decomposition?
5. Can this method be applied to other similar summation problems?

**Tip:** When solving telescoping series, always check how terms cancel out to simplify the sum effectively.

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).

This solution demonstrates how to find the constant C for a given summation problem using partial fraction decomposition and simplifying through telescoping series. This step-by-step solution involves advanced algebraic techniques suitable for college-level mathematics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation