The value of the expression `Sigma_(k=0)^(27)k.^(27)C_(k)((1)/(3))^(k)((2)/(3))^(27-k)` is equal to

Answer

To solve the expression

\[ \sum_{k=0}^{27} k \cdot \binom{27}{k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{27-k} \]

we can follow these steps:

### Step 1: Rewrite the summation
Notice that when \( k = 0 \), the term becomes zero. Therefore, we can start the summation from \( k = 1 \):

\[ \sum_{k=1}^{27} k \cdot \binom{27}{k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{27-k} \]

### Step 2: Use the identity for \( k \cdot \binom{n}{k} \)
We can use the identity \( k \cdot \binom{n}{k} = n \cdot \binom{n-1}{k-1} \). In our case, \( n = 27 \):

\[ k \cdot \binom{27}{k} = 27 \cdot \binom{26}{k-1} \]

Substituting this into our summation gives:

\[ \sum_{k=1}^{27} 27 \cdot \binom{26}{k-1} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{27-k} \]

### Step 3: Factor out the constant
Now we can factor out the constant \( 27 \):

\[ 27 \sum_{k=1}^{27} \binom{26}{k-1} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{27-k} \]

### Step 4: Change the index of summation
Let \( j = k - 1 \). Then when \( k = 1 \), \( j = 0 \) and when \( k = 27 \), \( j = 26 \). Thus, we can rewrite the summation:

\[ 27 \sum_{j=0}^{26} \binom{26}{j} \left(\frac{1}{3}\right)^{j+1} \left(\frac{2}{3}\right)^{26-j} \]

### Step 5: Simplify the expression
This can be simplified as:

\[ 27 \cdot \frac{1}{3} \sum_{j=0}^{26} \binom{26}{j} \left(\frac{1}{3}\right)^{j} \left(\frac{2}{3}\right)^{26-j} \]

### Step 6: Recognize the binomial expansion
The summation

\[ \sum_{j=0}^{26} \binom{26}{j} \left(\frac{1}{3}\right)^{j} \left(\frac{2}{3}\right)^{26-j} \]

is the binomial expansion of \( \left(\frac{1}{3} + \frac{2}{3}\right)^{26} = 1^{26} = 1 \).

### Step 7: Combine the results
Thus, we have:

\[ 27 \cdot \frac{1}{3} \cdot 1 = 9 \]

### Final Result
The value of the expression is

\[ \boxed{9} \]