Find the coefficient of `t^8` in the expansion of `(1+2t^2-t^3)^9`.

Explanation

To find the coefficient of \( t^8 \) in the expansion of \( (1 + 2t^2 - t^3)^9 \), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b + c)^n = \sum_{i=0}^{n} \sum_{j=0}^{n-i} \frac{n!}{i! j! (n-i-j)!} a^i b^j c^{n-i-j} \]

In our case, we have:

- \( a = 1 \)
- \( b = 2t^2 \)
- \( c = -t^3 \)
- \( n = 9 \)

We need to find the terms where the total power of \( t \) is 8.

### Step 1: Identify the powers of \( t \)

The general term in the expansion can be expressed as:

\[ T_{i,j} = \frac{9!}{i! j! (9-i-j)!} (1)^i (2t^2)^j (-t^3)^{9-i-j} \]

This simplifies to:

\[ T_{i,j} = \frac{9!}{i! j! (9-i-j)!} 2^j (-1)^{9-i-j} t^{2j + 3(9-i-j)} \]

### Step 2: Set up the equation for the power of \( t \)

We want the exponent of \( t \) to equal 8:

\[ 2j + 3(9-i-j) = 8 \]

Expanding this gives:

\[ 2j + 27 - 3i - 3j = 8 \]

Simplifying further:

\[ -3i - j + 27 = 8 \]
\[ -3i - j = -19 \]
\[ 3i + j = 19 \quad \text{(Equation 1)} \]

### Step 3: Determine the constraints for \( i \) and \( j \)

Since \( i + j + (9-i-j) = 9 \), we have:

\[ 9 - i - j \geq 0 \quad \Rightarrow \quad i + j \leq 9 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations

From Equation 1, we can express \( j \) in terms of \( i \):

\[ j = 19 - 3i \]

Substituting into Equation 2:

\[ i + (19 - 3i) \leq 9 \]
\[ 19 - 2i \leq 9 \]
\[ -2i \leq -10 \]
\[ i \geq 5 \]

Now substituting \( i = 5 \) into \( j = 19 - 3i \):

\[ j = 19 - 15 = 4 \]

### Step 5: Calculate \( k \)

Now, we need to find \( k \):

\[ k = 9 - i - j = 9 - 5 - 4 = 0 \]

### Step 6: Calculate the coefficient

The coefficient for \( i = 5 \), \( j = 4 \), and \( k = 0 \) is:

\[ \frac{9!}{5! 4! 0!} \cdot 2^4 \cdot (-1)^0 \]

Calculating this gives:

\[ \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \cdot 16 = 126 \cdot 16 = 2016 \]

### Step 7: Check for other combinations

We also check for \( i = 6 \):

\[ j = 19 - 3 \cdot 6 = 1 \quad \Rightarrow \quad k = 9 - 6 - 1 = 2 \]

Calculating the coefficient for \( i = 6 \), \( j = 1 \), and \( k = 2 \):

\[ \frac{9!}{6! 1! 2!} \cdot 2^1 \cdot (-1)^2 \]

Calculating this gives:

\[ \frac{9 \times 8 \times 7}{2 \times 1} \cdot 2 = 252 \cdot 2 = 504 \]

### Final Coefficient Calculation

Adding both contributions:

\[ 2016 + 504 = 2520 \]

Thus, the coefficient of \( t^8 \) in the expansion of \( (1 + 2t^2 - t^3)^9 \) is **2520**.