If `1+x^(4)+x^(5)=sum_(i=0)^(5)a_(i)(1+x)^(i)` for all `x""inR`, then `a_(4)` is equal to ________

Explanation

To solve the equation \(1 + x^4 + x^5 = \sum_{i=0}^{5} a_i (1 + x)^i\) for \(a_4\), we will follow these steps:

### Step 1: Expand the right-hand side
The right-hand side can be expanded using the binomial theorem:
\[ (1 + x)^i = \sum_{k=0}^{i} \binom{i}{k} x^k \]
Thus, we can write:
\[ \sum_{i=0}^{5} a_i (1 + x)^i = a_0 + a_1(1 + x) + a_2(1 + x)^2 + a_3(1 + x)^3 + a_4(1 + x)^4 + a_5(1 + x)^5 \]

### Step 2: Write out the terms
Now, we can write out the terms for \(i = 0\) to \(5\):
- For \(i = 0\): \(a_0\)
- For \(i = 1\): \(a_1(1 + x) = a_1 + a_1 x\)
- For \(i = 2\): \(a_2(1 + x)^2 = a_2(1 + 2x + x^2)\)
- For \(i = 3\): \(a_3(1 + x)^3 = a_3(1 + 3x + 3x^2 + x^3)\)
- For \(i = 4\): \(a_4(1 + x)^4 = a_4(1 + 4x + 6x^2 + 4x^3 + x^4)\)
- For \(i = 5\): \(a_5(1 + x)^5 = a_5(1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5)\)

### Step 3: Combine all terms
Combining all these, we have:
\[ a_0 + (a_1 + a_2 + a_3 + a_4 + a_5) + (a_1 + 2a_2 + 3a_3 + 4a_4 + 5a_5)x + (a_2 + 3a_3 + 6a_4 + 10a_5)x^2 + (a_3 + 4a_4 + 10a_5)x^3 + (a_4 + 5a_5)x^4 + a_5 x^5 \]

### Step 4: Set coefficients equal
Now, we equate the coefficients of \(1 + x^4 + x^5\) on both sides:
- Constant term: \(a_0 = 1\)
- Coefficient of \(x\): \(a_1 + 2a_2 + 3a_3 + 4a_4 + 5a_5 = 0\)
- Coefficient of \(x^2\): \(a_2 + 3a_3 + 6a_4 + 10a_5 = 0\)
- Coefficient of \(x^3\): \(a_3 + 4a_4 + 10a_5 = 0\)
- Coefficient of \(x^4\): \(a_4 + 5a_5 = 1\)
- Coefficient of \(x^5\): \(a_5 = 1\)

### Step 5: Solve for \(a_5\)
From the coefficient of \(x^5\), we have:
\[ a_5 = 1 \]

### Step 6: Substitute \(a_5\) into the equations
Now substitute \(a_5 = 1\) into the equation for \(x^4\):
\[ a_4 + 5(1) = 1 \implies a_4 + 5 = 1 \implies a_4 = 1 - 5 = -4 \]

### Conclusion
Thus, the value of \(a_4\) is:
\[ \boxed{-4} \]