Find the numerically greatest term in the expansion of `(2+5x)^(21)` when `x = 2/5`.

Answer

To find the numerically greatest term in the expansion of \((2 + 5x)^{21}\) when \(x = \frac{2}{5}\), we can follow these steps:

### Step 1: Identify the General Term
The general term \(T_r\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = 2\), \(b = 5x\), and \(n = 21\). Thus, the general term becomes:
\[ T_r = \binom{21}{r} (2)^{21-r} (5x)^r \]

### Step 2: Substitute \(x\)
Substituting \(x = \frac{2}{5}\) into the general term:
\[ T_r = \binom{21}{r} (2)^{21-r} \left(5 \cdot \frac{2}{5}\right)^r \]
This simplifies to:
\[ T_r = \binom{21}{r} (2)^{21-r} (2)^r = \binom{21}{r} (2)^{21} \]

### Step 3: Simplify the General Term
Thus, we can rewrite the general term as:
\[ T_r = \binom{21}{r} (2)^{21} \]
This shows that the term depends on \(\binom{21}{r}\) since \((2)^{21}\) is a constant factor.

### Step 4: Find the Maximum Value of \(\binom{21}{r}\)
The binomial coefficient \(\binom{21}{r}\) is maximized when \(r\) is around \(\frac{n}{2}\). Here, \(n = 21\), so we check \(r = 10\) and \(r = 11\).

### Step 5: Calculate the Terms for \(r = 10\) and \(r = 11\)
1. For \(r = 10\):
\[ T_{10} = \binom{21}{10} (2)^{21} \]
2. For \(r = 11\):
\[ T_{11} = \binom{21}{11} (2)^{21} \]

### Step 6: Compare \(\binom{21}{10}\) and \(\binom{21}{11}\)
Since \(\binom{21}{10} = \binom{21}{11}\), both terms are equal.

### Step 7: Conclusion
The numerically greatest terms in the expansion are \(T_{10}\) and \(T_{11}\).

### Final Answer
The numerically greatest term in the expansion of \((2 + 5x)^{21}\) when \(x = \frac{2}{5}\) is:
\[ T_{10} = T_{11} = \binom{21}{10} (2)^{21} \]