The value of `sum_(r=1)^n(sum_(p=0)^(r-1) ^nC_r ^rC_p 2^p)` is equal to (a) `4^(n)-3^(n)+1` (b) `4^(n)-3^(n)-1` (c) `4^(n)-3^(n)+2` (d) `4^(n)-3^(n)`

Options

A. `4^(n)-3^(n)+1`
B. `4^(n)-3^(n)-1`
C. `4^(n)-3^(n)+2`
D. `4^(n)-3^(n)`

Explanation

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

\[ \sum_{r=1}^{n} \sum_{p=0}^{r-1} \binom{n}{r} \binom{r}{p} 2^p \]

### Step 1: Rewrite the Inner Summation
We start with the inner summation:

\[ \sum_{p=0}^{r-1} \binom{r}{p} 2^p \]

This can be recognized as a binomial expansion. The sum can be rewritten using the binomial theorem:

\[ \sum_{p=0}^{r} \binom{r}{p} 2^p = (1 + 2)^r = 3^r \]

However, since we are summing up to \(r-1\), we need to subtract the last term:

\[ \sum_{p=0}^{r-1} \binom{r}{p} 2^p = 3^r - \binom{r}{r} 2^r = 3^r - 2^r \]

### Step 2: Substitute Back into the Outer Summation
Now, we substitute this back into the outer summation:

\[ \sum_{r=1}^{n} \binom{n}{r} (3^r - 2^r) \]

This can be split into two separate summations:

\[ \sum_{r=1}^{n} \binom{n}{r} 3^r - \sum_{r=1}^{n} \binom{n}{r} 2^r \]

### Step 3: Evaluate Each Summation
Using the binomial theorem, we know:

\[ \sum_{r=0}^{n} \binom{n}{r} x^r = (1+x)^n \]

Thus, we can evaluate our summations:

1. For \(x = 3\):

\[ \sum_{r=0}^{n} \binom{n}{r} 3^r = (1 + 3)^n = 4^n \]

2. For \(x = 2\):

\[ \sum_{r=0}^{n} \binom{n}{r} 2^r = (1 + 2)^n = 3^n \]

Now, we need to subtract the \(r=0\) term from both summations:

- For \(x = 3\):

\[ \sum_{r=1}^{n} \binom{n}{r} 3^r = 4^n - \binom{n}{0} 3^0 = 4^n - 1 \]

- For \(x = 2\):

\[ \sum_{r=1}^{n} \binom{n}{r} 2^r = 3^n - \binom{n}{0} 2^0 = 3^n - 1 \]

### Step 4: Combine the Results
Now we can combine the results from both summations:

\[ \sum_{r=1}^{n} \binom{n}{r} 3^r - \sum_{r=1}^{n} \binom{n}{r} 2^r = (4^n - 1) - (3^n - 1) \]

This simplifies to:

\[ 4^n - 3^n \]

### Final Answer
Thus, the final value of the original expression is:

\[ \boxed{4^n - 3^n} \]