Sum of the series
`S=sum_(j=0)^(8)1/((j+1)(j+2))(""^(8)C_(j))` is

Options

A. `1003/90`
B. `1013/90`
C. `1023/90`
D. `1033/90`

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{j=0}^{8} \frac{1}{(j+1)(j+2)} \binom{8}{j} \]

### Step 1: Rewrite the term \(\frac{1}{(j+1)(j+2)}\)

We can rewrite \(\frac{1}{(j+1)(j+2)}\) using partial fractions:

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

### Step 2: Substitute the partial fraction into the sum

Now, substitute this back into the sum:

\[ S = \sum_{j=0}^{8} \left( \frac{1}{j+1} - \frac{1}{j+2} \right) \binom{8}{j} \]

### Step 3: Split the summation

This gives us two separate sums:

\[ S = \sum_{j=0}^{8} \frac{1}{j+1} \binom{8}{j} - \sum_{j=0}^{8} \frac{1}{j+2} \binom{8}{j} \]

### Step 4: Simplify the first sum

The first sum can be simplified using the identity:

\[ \sum_{j=0}^{n} \frac{1}{j+1} \binom{n}{j} = \frac{1}{n+1} \sum_{j=0}^{n} \binom{n+1}{j+1} = \frac{2^n}{n+1} \]

For \(n=8\):

\[ \sum_{j=0}^{8} \frac{1}{j+1} \binom{8}{j} = \frac{2^8}{9} = \frac{256}{9} \]

### Step 5: Simplify the second sum

The second sum can be simplified similarly:

\[ \sum_{j=0}^{8} \frac{1}{j+2} \binom{8}{j} = \frac{1}{10} \sum_{j=0}^{8} \binom{9}{j+2} = \frac{1}{10} \cdot 2^9 = \frac{512}{10} = \frac{256}{5} \]

### Step 6: Combine the results

Now, substitute these results back into the expression for \(S\):

\[ S = \frac{256}{9} - \frac{256}{5} \]

### Step 7: Find a common denominator and simplify

The common denominator of 9 and 5 is 45:

\[ S = \frac{256 \cdot 5}{45} - \frac{256 \cdot 9}{45} = \frac{1280 - 2304}{45} = \frac{-1024}{45} \]

### Step 8: Final Result

Thus, the final value of the sum \(S\) is:

\[ S = \frac{-1024}{45} \]