If `f(x)=(4+x)^(n),"n" epsilonN` and `f^(r)(0)` represents the `r^(th)` derivative of `f(x)` at `x=0`, then the value of `sum_(r=0)^(oo)((f^(r)(0)))/(r!)` is equal to

Options

A. `2 ^(n)`
B. `3 ^(n)`
C. `5 ^(n)`
D. `4 ^(n)`

Answer

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

\[ \sum_{r=0}^{\infty} \frac{f^{(r)}(0)}{r!} \]

where \( f(x) = (4 + x)^n \) and \( n \in \mathbb{N} \).

### Step 1: Find the derivatives of \( f(x) \)

The function \( f(x) = (4 + x)^n \) can be differentiated using the formula for the \( r \)-th derivative of a polynomial. The \( r \)-th derivative of \( f(x) \) at \( x = 0 \) can be expressed as:

\[ f^{(r)}(0) = n(n-1)(n-2)\cdots(n-r+1)(4)^{n-r} \]

for \( r \leq n \). For \( r > n \), \( f^{(r)}(0) = 0 \).

### Step 2: Substitute into the summation

Now, we can substitute this expression into the summation:

\[ \sum_{r=0}^{\infty} \frac{f^{(r)}(0)}{r!} = \sum_{r=0}^{n} \frac{n(n-1)(n-2)\cdots(n-r+1)(4)^{n-r}}{r!} \]

### Step 3: Recognize the binomial coefficient

The term \( n(n-1)(n-2)\cdots(n-r+1) \) can be rewritten as \( \frac{n!}{(n-r)!} \), so we have:

\[ \sum_{r=0}^{n} \frac{n!}{(n-r)! r!} (4)^{n-r} \]

This is the binomial expansion of \( (x+y)^n \) where \( x = 4 \) and \( y = 1 \):

\[ \sum_{r=0}^{n} \binom{n}{r} 4^{n-r} 1^r = (4 + 1)^n = 5^n \]

### Step 4: Conclusion

Thus, we conclude that:

\[ \sum_{r=0}^{\infty} \frac{f^{(r)}(0)}{r!} = 5^n \]

### Final Answer

The value of the sum is:

\[ \boxed{5^n} \]