Find the middle term (s) in the expansion of ` ((3)/(p^3) + 5p^4)^20`

Answer

To find the middle term in the expansion of \(\left(\frac{3}{p^3} + 5p^4\right)^{20}\), we will follow these steps:

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, \(a = \frac{3}{p^3}\), \(b = 5p^4\), and \(n = 20\).

### Step 2: Write the General Term
Substituting the values into the general term formula, we have:
\[ T_{r+1} = \binom{20}{r} \left(\frac{3}{p^3}\right)^{20-r} (5p^4)^r \]
This simplifies to:
\[ T_{r+1} = \binom{20}{r} \cdot \frac{3^{20-r}}{(p^3)^{20-r}} \cdot 5^r \cdot (p^4)^r \]
\[ = \binom{20}{r} \cdot 3^{20-r} \cdot 5^r \cdot \frac{p^{4r}}{p^{3(20-r)}} \]
\[ = \binom{20}{r} \cdot 3^{20-r} \cdot 5^r \cdot p^{4r - 60 + 3r} \]
\[ = \binom{20}{r} \cdot 3^{20-r} \cdot 5^r \cdot p^{7r - 60} \]

### Step 3: Determine the Middle Term
Since \(n = 20\) is even, the middle term will be the \(\frac{n}{2} + 1\)th term, which is the 11th term (\(T_{11}\)).

To find \(T_{11}\), we set \(r = 10\) (since \(T_{r+1}\) corresponds to \(r\)):
\[ T_{11} = \binom{20}{10} \cdot 3^{20-10} \cdot 5^{10} \cdot p^{7(10) - 60} \]
\[ = \binom{20}{10} \cdot 3^{10} \cdot 5^{10} \cdot p^{70 - 60} \]
\[ = \binom{20}{10} \cdot 3^{10} \cdot 5^{10} \cdot p^{10} \]

### Step 4: Final Expression for the Middle Term
Thus, the middle term in the expansion of \(\left(\frac{3}{p^3} + 5p^4\right)^{20}\) is:
\[ T_{11} = \binom{20}{10} \cdot 3^{10} \cdot 5^{10} \cdot p^{10} \]