Coefficient of `x^15` in `(1+ x + x^3+ x^4)^n` is

Options

A. A. `sum_(r=0)^(5) ""^(n)C_(5-r) *""^(n)C_(3r)`
B. B. `sum_(r=0)^(5) ""^(n)C_(5r) `
C. C. `sum_(r=0)^(5) ""^(n)C_(2r) `
D. D. `sum_(r=0)^(5) ""^(n)C_(3-r) *""^(n)C_(5r)`

Answer

To find the coefficient of \( x^{15} \) in the expression \( (1 + x + x^3 + x^4)^n \), we can break down the problem step by step.

### Step 1: Rewrite the Expression
We can express \( (1 + x + x^3 + x^4)^n \) in a more manageable form. Notice that we can group terms:
\[ (1 + x + x^3 + x^4)^n = (1 + x + x^3 + x^4)^n = (1 + x)^n (1 + x^3)^n \]

### Step 2: Use the Binomial Theorem
Using the Binomial Theorem, we can expand both \( (1 + x)^n \) and \( (1 + x^3)^n \):
- The expansion of \( (1 + x)^n \) gives us:
\[ \sum_{r=0}^{n} \binom{n}{r} x^r \]
- The expansion of \( (1 + x^3)^n \) gives us:
\[ \sum_{s=0}^{n} \binom{n}{s} x^{3s} \]

### Step 3: Find the Coefficient of \( x^{15} \)
To find the coefficient of \( x^{15} \) in the product of these two expansions, we need to consider combinations of \( r \) and \( s \) such that:
\[ r + 3s = 15 \]
where \( r \) is from \( (1 + x)^n \) and \( s \) is from \( (1 + x^3)^n \).

### Step 4: Determine Valid Combinations
We can express \( r \) in terms of \( s \):
\[ r = 15 - 3s \]
Now, \( r \) must be non-negative, which gives us the constraint:
\[ 15 - 3s \geq 0 \implies s \leq 5 \]
Thus, \( s \) can take values from \( 0 \) to \( 5 \).

### Step 5: Calculate Coefficients for Each Valid \( s \)
For each valid \( s \), we can find the corresponding \( r \):
- If \( s = 0 \): \( r = 15 \) → Coefficient: \( \binom{n}{15} \)
- If \( s = 1 \): \( r = 12 \) → Coefficient: \( \binom{n}{12} \)
- If \( s = 2 \): \( r = 9 \) → Coefficient: \( \binom{n}{9} \)
- If \( s = 3 \): \( r = 6 \) → Coefficient: \( \binom{n}{6} \)
- If \( s = 4 \): \( r = 3 \) → Coefficient: \( \binom{n}{3} \)
- If \( s = 5 \): \( r = 0 \) → Coefficient: \( \binom{n}{0} \)

### Step 6: Sum the Coefficients
Now, we can sum the coefficients for each valid combination:
\[ \text{Coefficient of } x^{15} = \sum_{s=0}^{5} \binom{n}{15 - 3s} \binom{n}{s} \]
This can be expressed as:
\[ \sum_{s=0}^{5} \binom{n}{15 - 3s} \binom{n}{s} \]

### Final Answer
Thus, the coefficient of \( x^{15} \) in \( (1 + x + x^3 + x^4)^n \) is given by:
\[ \sum_{s=0}^{5} \binom{n}{15 - 3s} \binom{n}{s} \]