If `1+x^(4)-x^(5)=sum_(i=0)^(5)a_(i)(1+x)^(i)` for all `x""inR`, then `a_(2)` is equal to ________

Explanation

To solve the equation \(1 + x^4 - x^5 = \sum_{i=0}^{5} a_i (1+x)^i\) for \(a_2\), we can follow these steps:

### Step 1: Expand the Right-Hand Side
We start by expanding the right-hand side of the equation:
\[ \sum_{i=0}^{5} a_i (1+x)^i = a_0 (1+x)^0 + a_1 (1+x)^1 + a_2 (1+x)^2 + a_3 (1+x)^3 + a_4 (1+x)^4 + a_5 (1+x)^5 \]
This gives us:
\[ = a_0 + a_1 (1+x) + a_2 (1+x)^2 + a_3 (1+x)^3 + a_4 (1+x)^4 + a_5 (1+x)^5 \]

### Step 2: Differentiate Both Sides
Next, we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(1 + x^4 - x^5) = \frac{d}{dx}\left(\sum_{i=0}^{5} a_i (1+x)^i\right) \]
The left-hand side becomes:
\[ 0 + 4x^3 - 5x^4 = 4x^3 - 5x^4 \]
The right-hand side, using the chain rule, becomes:
\[ a_1 + 2a_2(1+x) + 3a_3(1+x)^2 + 4a_4(1+x)^3 + 5a_5(1+x)^4 \]

### Step 3: Set \(x = -1\)
To find \(a_2\), we can set \(x = -1\):
\[ 4(-1)^3 - 5(-1)^4 = a_1 + 2a_2(1-1) + 3a_3(1-1)^2 + 4a_4(1-1)^3 + 5a_5(1-1)^4 \]
Calculating the left-hand side:
\[ 4(-1) - 5(1) = -4 - 5 = -9 \]
The right-hand side simplifies to:
\[ a_1 + 0 + 0 + 0 + 0 = a_1 \]
Thus, we have:
\[ -9 = a_1 \]

### Step 4: Differentiate Again
We differentiate again to find a relation involving \(a_2\):
\[ \frac{d}{dx}(4x^3 - 5x^4) = 12x^2 - 20x^3 \]
The right-hand side becomes:
\[ 0 + 2a_2 + 6a_3(1+x) + 12a_4(1+x)^2 + 20a_5(1+x)^3 \]

### Step 5: Set \(x = -1\) Again
Setting \(x = -1\) again gives us:
\[ 12(-1)^2 - 20(-1)^3 = 2a_2 + 0 + 0 + 0 + 0 \]
Calculating the left-hand side:
\[ 12 + 20 = 32 \]
Thus, we have:
\[ 32 = 2a_2 \]

### Step 6: Solve for \(a_2\)
Now we can solve for \(a_2\):
\[ a_2 = \frac{32}{2} = 16 \]

### Final Answer
Thus, the value of \(a_2\) is:
\[ \boxed{16} \]