Find the middle term (s) in the expansion of `(4x^2 + 5x^3)^17`

Answer

To find the middle terms in the expansion of \((4x^2 + 5x^3)^{17}\), we will follow these steps:

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, \(n = 17\), \(a = 4x^2\), and \(b = 5x^3\). Thus, the general term becomes:
\[ T_{r+1} = \binom{17}{r} (4x^2)^{17-r} (5x^3)^r \]

### Step 2: Simplify the General Term
Now, we simplify the general term:
\[ T_{r+1} = \binom{17}{r} (4^{17-r} (x^2)^{17-r}) (5^r (x^3)^r) \]
This can be rewritten as:
\[ T_{r+1} = \binom{17}{r} 4^{17-r} 5^r x^{2(17-r) + 3r} \]
\[ = \binom{17}{r} 4^{17-r} 5^r x^{34 - 2r + 3r} \]
\[ = \binom{17}{r} 4^{17-r} 5^r x^{34 + r} \]

### Step 3: Determine the Middle Terms
Since \(n = 17\) is odd, there are two middle terms. The middle terms are given by:
- \(T_{\frac{n+1}{2}}\) and \(T_{\frac{n+3}{2}}\)
Calculating these:
- \(\frac{n+1}{2} = \frac{17+1}{2} = 9\)
- \(\frac{n+3}{2} = \frac{17+3}{2} = 10\)

Thus, we need to find \(T_9\) and \(T_{10}\).

### Step 4: Calculate the 9th Term
For \(T_9\) (where \(r = 8\)):
\[ T_9 = \binom{17}{8} 4^{17-8} 5^8 x^{34 + 8} \]
\[ = \binom{17}{8} 4^9 5^8 x^{42} \]

### Step 5: Calculate the 10th Term
For \(T_{10}\) (where \(r = 9\)):
\[ T_{10} = \binom{17}{9} 4^{17-9} 5^9 x^{34 + 9} \]
\[ = \binom{17}{9} 4^8 5^9 x^{43} \]

### Final Answer
The middle terms in the expansion of \((4x^2 + 5x^3)^{17}\) are:
- \(T_9 = \binom{17}{8} 4^9 5^8 x^{42}\)
- \(T_{10} = \binom{17}{9} 4^8 5^9 x^{43}\)