If `Sigma_(r=1)^(n) T_r=n/8(n+1)(n+2)(n+3)` then find `Sigma_(r=1)^(n) 1/T_r`

Explanation

To solve the problem, we need to find the summation \( \Sigma_{r=1}^{n} \frac{1}{T_r} \) given that \( \Sigma_{r=1}^{n} T_r = \frac{n}{8}(n+1)(n+2)(n+3) \).

### Step-by-step Solution:

1. **Find \( T_n \)**:
We know that \( T_n = S_n - S_{n-1} \), where \( S_n = \frac{n}{8}(n+1)(n+2)(n+3) \).
- First, we calculate \( S_{n-1} \):
\[ S_{n-1} = \frac{n-1}{8}(n)(n+1)(n+2) \]
- Now, we can find \( T_n \):
\[ T_n = S_n - S_{n-1} = \frac{n}{8}(n+1)(n+2)(n+3) - \frac{n-1}{8}(n)(n+1)(n+2) \]
- Simplifying this expression:
\[ T_n = \frac{1}{8} \left[ n(n+1)(n+2)(n+3) - (n-1)(n)(n+1)(n+2) \right] \]
- Factor out common terms:
\[ T_n = \frac{1}{8} n(n+1)(n+2) \left[ (n+3) - (n-1) \right] \]
- This simplifies to:
\[ T_n = \frac{1}{8} n(n+1)(n+2)(4) = \frac{n(n+1)(n+2)}{2} \]

2. **Find \( \frac{1}{T_n} \)**:
\[ \frac{1}{T_n} = \frac{2}{n(n+1)(n+2)} \]

3. **Calculate \( \Sigma_{r=1}^{n} \frac{1}{T_r} \)**:
\[ \Sigma_{r=1}^{n} \frac{1}{T_r} = \Sigma_{r=1}^{n} \frac{2}{r(r+1)(r+2)} \]
- We can use partial fractions to simplify \( \frac{2}{r(r+1)(r+2)} \):
\[ \frac{2}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2} \]
- Solving for \( A, B, C \):
\[ 2 = A(r+1)(r+2) + B(r)(r+2) + C(r)(r+1) \]
- By substituting convenient values for \( r \), we can find \( A, B, C \):
- Let \( r = 0 \): \( 2 = A(1)(2) \Rightarrow A = 1 \)
- Let \( r = -1 \): \( 2 = B(-1)(1) \Rightarrow B = -2 \)
- Let \( r = -2 \): \( 2 = C(-2)(-1) \Rightarrow C = 1 \)
- Thus, we have:
\[ \frac{2}{r(r+1)(r+2)} = \frac{1}{r} - \frac{2}{r+1} + \frac{1}{r+2} \]

4. **Sum the series**:
\[ \Sigma_{r=1}^{n} \left( \frac{1}{r} - \frac{2}{r+1} + \frac{1}{r+2} \right) \]
- This can be separated into three sums:
\[ \Sigma_{r=1}^{n} \frac{1}{r} - 2 \Sigma_{r=1}^{n} \frac{1}{r+1} + \Sigma_{r=1}^{n} \frac{1}{r+2} \]
- The first sum is \( H_n \) (the \( n \)-th harmonic number), the second sum is \( H_{n+1} \), and the third sum is \( H_{n+2} \):
\[ H_n - 2H_{n+1} + H_{n+2} \]

5. **Final Result**:
\[ \Sigma_{r=1}^{n} \frac{1}{T_r} = H_n - 2H_{n+1} + H_{n+2} \]