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
We can start by noting that when \( k = 0 \), the term becomes zero (since \( k \cdot \binom{27}{0} = 0 \)). Therefore, we can change the limits of the summation from \( k=0 \) to \( k=1 \):

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

### Step 2: Use the Identity for Binomial Coefficients
We can use the identity \( k \cdot \binom{n}{k} = n \cdot \binom{n-1}{k-1} \). Here, \( n = 27 \):

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

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: Change the Index of Summation
Now, we can change the index of summation by letting \( j = k - 1 \). When \( k = 1 \), \( j = 0 \) and when \( k = 27 \), \( j = 26 \):

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

### Step 4: Factor Out Constants
We can factor out \( \frac{1}{3} \):

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

### Step 5: Recognize the Binomial Expansion
The remaining summation is the binomial expansion of \( \left(\frac{1}{3} + \frac{2}{3}\right)^{26} \):

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

### Step 6: Final Calculation
Putting it all together, we have:

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

Thus, the value of the expression is

\[ \boxed{9}. \]