Smaller of `(19^10 + 20^10)` and `21^10` is

Options

A. a) `19^10 + 20^10`
B. b) `21^10`
C. c) Both are equal
D. d) can not be decided

Answer

To find the smaller of \(19^{10} + 20^{10}\) and \(21^{10}\), we can use the Binomial Theorem for expansion and comparison. Here’s a step-by-step solution:

### Step 1: Rewrite \(19^{10}\)
We can express \(19\) as \(20 - 1\):
\[ 19^{10} = (20 - 1)^{10} \]

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we can expand \((20 - 1)^{10}\):
\[ (20 - 1)^{10} = \sum_{k=0}^{10} \binom{10}{k} (20)^{10-k} (-1)^k \]
This gives us:
\[ 19^{10} = 20^{10} - \binom{10}{1} 20^9 + \binom{10}{2} 20^8 - \binom{10}{3} 20^7 + \ldots + (-1)^{10} \]

### Step 3: Combine with \(20^{10}\)
Now, we add \(20^{10}\) to \(19^{10}\):
\[ 19^{10} + 20^{10} = \left(20^{10} - \binom{10}{1} 20^9 + \binom{10}{2} 20^8 - \binom{10}{3} 20^7 + \ldots + 1\right) + 20^{10} \]
This simplifies to:
\[ 19^{10} + 20^{10} = 2 \cdot 20^{10} - \binom{10}{1} 20^9 + \binom{10}{2} 20^8 - \binom{10}{3} 20^7 + \ldots + 1 \]

### Step 4: Expand \(21^{10}\)
Next, we expand \(21^{10}\) using the Binomial Theorem:
\[ 21^{10} = (20 + 1)^{10} = \sum_{k=0}^{10} \binom{10}{k} 20^{10-k} (1)^k \]
This gives us:
\[ 21^{10} = 20^{10} + \binom{10}{1} 20^9 + \binom{10}{2} 20^8 + \binom{10}{3} 20^7 + \ldots + 1 \]

### Step 5: Compare \(19^{10} + 20^{10}\) and \(21^{10}\)
Now we need to compare:
\[ 19^{10} + 20^{10} \quad \text{and} \quad 21^{10} \]
From our expansions:
\[ 19^{10} + 20^{10} = 2 \cdot 20^{10} - \text{(terms with alternating signs)} \]
\[ 21^{10} = 20^{10} + \text{(terms with all positive signs)} \]

### Step 6: Analyze the difference
To find which is smaller, we can analyze:
\[ 21^{10} - (19^{10} + 20^{10}) \]
This results in:
\[ 21^{10} - (19^{10} + 20^{10}) = (20^{10} + \text{positive terms}) - (2 \cdot 20^{10} - \text{negative terms}) \]
This simplifies to:
\[ = -20^{10} + \text{(positive terms + negative terms)} \]
Since the positive terms are greater than the negative terms, we conclude:
\[ 21^{10} > 19^{10} + 20^{10} \]

### Conclusion
Thus, the smaller of \(19^{10} + 20^{10}\) and \(21^{10}\) is:
\[ \boxed{19^{10} + 20^{10}} \]