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 \(T_k\) in the expansion of \((a + b)^n\) is given by:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
In our case, \(a = 2\), \(b = 5x\), and \(n = 21\). Therefore, the general term becomes:
\[ T_k = \binom{21}{k} (2)^{21-k} (5x)^k \]

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

### Step 3: Find the Maximum Term
To find the numerically greatest term, we need to determine the value of \(k\) that maximizes \(T_k\). The maximum term occurs around the middle of the distribution of terms. Since \(n = 21\) (which is odd), the middle terms are \(T_{10}\) and \(T_{11}\).

### Step 4: Calculate \(T_{10}\) and \(T_{11}\)
Calculating \(T_{10}\):
\[ T_{10} = \binom{21}{10} (2)^{21} \]
Calculating \(T_{11}\):
\[ T_{11} = \binom{21}{11} (2)^{21} \]
Since \(\binom{21}{10} = \binom{21}{11}\), both terms are equal:
\[ T_{10} = T_{11} \]

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