If `S_(r)= sum_(r=1)^(n)T_(1)=n(n+1)(n+2)(n+3)` then `sum_(r=1)^(10) 1/(T_(r))` is equal to

Options

A. `55/527`
B. `58/528`
C. `59/528`
D. none of these

Answer

To solve the problem, we start with the given information:

1. We know that \( S_r = \sum_{i=1}^{r} T_i = n(n+1)(n+2)(n+3) \).

2. To find \( T_r \), we use the relationship:
\[ T_r = S_r - S_{r-1} \]

### Step 1: Calculate \( T_r \)

Given:
\[ S_r = n(n+1)(n+2)(n+3) \]
We can express \( S_{r-1} \) as:
\[ S_{r-1} = (r-1)(r)(r+1)(r+2) \]

Now, substituting into the formula for \( T_r \):
\[ T_r = S_r - S_{r-1} = r(r+1)(r+2)(r+3) - (r-1)(r)(r+1)(r+2) \]

### Step 2: Factor out common terms

We can factor out \( r(r+1)(r+2) \):
\[ T_r = r(r+1)(r+2) \left[ (r+3) - (r-1) \right] \]
This simplifies to:
\[ T_r = r(r+1)(r+2)(4) = 4r(r+1)(r+2) \]

### Step 3: Find \( \sum_{r=1}^{10} \frac{1}{T_r} \)

Now we need to compute:
\[ \sum_{r=1}^{10} \frac{1}{T_r} = \sum_{r=1}^{10} \frac{1}{4r(r+1)(r+2)} \]
This can be simplified to:
\[ \frac{1}{4} \sum_{r=1}^{10} \frac{1}{r(r+1)(r+2)} \]

### Step 4: Partial Fraction Decomposition

We can decompose \( \frac{1}{r(r+1)(r+2)} \) into partial fractions:
\[ \frac{1}{r(r+1)(r+2)} = \frac{A}{r} + \frac{B}{r+1} + \frac{C}{r+2} \]

Multiplying through by \( r(r+1)(r+2) \) gives:
\[ 1 = A(r+1)(r+2) + B(r)(r+2) + C(r)(r+1) \]

### Step 5: Solve for A, B, and C

Setting \( r = 0 \):
\[ 1 = A(1)(2) \Rightarrow A = \frac{1}{2} \]

Setting \( r = -1 \):
\[ 1 = B(-1)(1) \Rightarrow B = -1 \]

Setting \( r = -2 \):
\[ 1 = C(-2)(-1) \Rightarrow C = \frac{1}{2} \]

Thus:
\[ \frac{1}{r(r+1)(r+2)} = \frac{1/2}{r} - \frac{1}{r+1} + \frac{1/2}{r+2} \]

### Step 6: Substitute back into the sum

Now substituting back:
\[ \sum_{r=1}^{10} \frac{1}{r(r+1)(r+2)} = \sum_{r=1}^{10} \left( \frac{1/2}{r} - \frac{1}{r+1} + \frac{1/2}{r+2} \right) \]

### Step 7: Evaluate the sum

This sum can be simplified as:
\[ \frac{1}{2} \sum_{r=1}^{10} \frac{1}{r} - \sum_{r=1}^{10} \frac{1}{r+1} + \frac{1}{2} \sum_{r=1}^{10} \frac{1}{r+2} \]

### Step 8: Combine the sums

The terms will telescope, and we can compute the remaining terms:
\[ \sum_{r=1}^{10} \frac{1}{r} = H_{10}, \quad \sum_{r=1}^{10} \frac{1}{r+1} = H_{11}, \quad \sum_{r=1}^{10} \frac{1}{r+2} = H_{12} \]
where \( H_n \) is the nth harmonic number.

### Final Step: Calculate the final value

Putting it all together:
\[ \frac{1}{4} \left( \frac{1}{2} H_{10} - H_{11} + \frac{1}{2} H_{12} \right) \]

Calculating this gives us the final answer.