The sum of the series `(1)/(1xx2)""^(25)C_(0)+(1)/(2xx3)""^(25)C_(1)+(1)/(3xx4)""^(25)C_(2)+…+(1)/(26xx27)""^(25)C_(25)` is

Options

A. `(2^(27)-1)/(26xx27)`
B. `(2^(27)-28)/(26xx27)`
C. `(2^(26)-1)/(52)`
D. `(1)/(2)((2^(26)+1)/(26xx27))`

Answer

To find the sum of the series

\[ S = \sum_{k=0}^{25} \frac{1}{(k+1)(k+2)} \binom{25}{k} \]

we can start by rewriting the term \(\frac{1}{(k+1)(k+2)}\) in a more manageable form. Notice that:

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

This allows us to rewrite the series \(S\) as:

\[ S = \sum_{k=0}^{25} \left( \frac{1}{k+1} - \frac{1}{k+2} \right) \binom{25}{k} \]

Now, we can split the sum into two separate sums:

\[ S = \sum_{k=0}^{25} \frac{1}{k+1} \binom{25}{k} - \sum_{k=0}^{25} \frac{1}{k+2} \binom{25}{k} \]

Next, we can simplify each of these sums. The first sum can be simplified using the identity:

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

For \(n = 25\):

\[ \sum_{k=0}^{25} \frac{1}{k+1} \binom{25}{k} = \frac{1}{26} \cdot 2^{26} \]

The second sum can be simplified similarly:

\[ \sum_{k=0}^{25} \frac{1}{k+2} \binom{25}{k} = \frac{1}{27} \cdot 2^{26} \]

Now substituting these results back into our expression for \(S\):

\[ S = \frac{1}{26} \cdot 2^{26} - \frac{1}{27} \cdot 2^{26} \]

Factoring out \(2^{26}\):

\[ S = 2^{26} \left( \frac{1}{26} - \frac{1}{27} \right) \]

Calculating the expression inside the parentheses:

\[ \frac{1}{26} - \frac{1}{27} = \frac{27 - 26}{26 \cdot 27} = \frac{1}{26 \cdot 27} \]

Thus, we have:

\[ S = 2^{26} \cdot \frac{1}{26 \cdot 27} = \frac{2^{26}}{26 \cdot 27} \]

Finally, we can express the final answer:

\[ S = \frac{2^{26}}{702} \]