The value of `4{ C_1 +4. C_2 +4^2 C_3 + ... +4^(n-1)}` is

Options

A. 0
B. `5^(n)+1`
C. `5^(n)`
D. `5^(n)-1`

Answer

To solve the expression \( 4C_1 + 4C_2 + 4^2C_3 + \ldots + 4^{n-1} \), we will follow these steps:

### Step 1: Rewrite the expression
We can rewrite the expression as:
\[ \sum_{k=1}^{n} 4^{k-1} C_k \]
where \( C_k \) represents the binomial coefficient \( nCk \).

### Step 2: Recognize the binomial expansion
We know from the Binomial Theorem that:
\[ (a + b)^n = \sum_{k=0}^{n} C_k a^{n-k} b^k \]
For our case, let \( a = 4 \) and \( b = 1 \). Thus, we can write:
\[ (4 + 1)^n = 5^n = \sum_{k=0}^{n} C_k 4^{n-k} 1^k \]

### Step 3: Separate the terms
From the expansion, we can separate the \( k=0 \) term:
\[ 5^n = C_0 4^n + \sum_{k=1}^{n} C_k 4^{n-k} \]
Here, \( C_0 = 1 \), so we have:
\[ 5^n = 4^n + \sum_{k=1}^{n} C_k 4^{n-k} \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ \sum_{k=1}^{n} C_k 4^{n-k} = 5^n - 4^n \]

### Step 5: Change the index of summation
Now, we can change the index of summation to match our original expression:
\[ \sum_{k=1}^{n} C_k 4^{n-k} = \sum_{j=0}^{n-1} C_{j+1} 4^{n-(j+1)} = \sum_{j=0}^{n-1} C_{j+1} 4^{n-1-j} \]
This is equivalent to:
\[ \sum_{j=0}^{n-1} C_{j+1} 4^{j} = 4C_1 + 4^2C_2 + \ldots + 4^{n-1}C_n \]

### Step 6: Final expression
Thus, we can conclude that:
\[ 4C_1 + 4^2C_2 + \ldots + 4^{n-1}C_n = 5^n - 4^n \]

### Step 7: Conclusion
The final value of the expression \( 4C_1 + 4C_2 + 4^2C_3 + \ldots + 4^{n-1} \) is:
\[ 5^n - 4^n \]