Value of `(10.1)^4`, using binomial theorem is 10406.0401

Answer

To find the value of \( (10.1)^4 \) using the binomial theorem, we can express \( 10.1 \) as \( (10 + 0.1) \). According to the binomial theorem, we can expand \( (x + y)^n \) as follows:

\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]

In our case:
- \( x = 10 \)
- \( y = 0.1 \)
- \( n = 4 \)

Now, we can expand \( (10 + 0.1)^4 \):

\[ (10 + 0.1)^4 = \sum_{k=0}^{4} \binom{4}{k} (10)^{4-k} (0.1)^k \]

Calculating each term in the expansion:

1. **For \( k = 0 \)**:
\[ \binom{4}{0} (10)^{4} (0.1)^{0} = 1 \cdot 10000 \cdot 1 = 10000 \]

2. **For \( k = 1 \)**:
\[ \binom{4}{1} (10)^{3} (0.1)^{1} = 4 \cdot 1000 \cdot 0.1 = 400 \]

3. **For \( k = 2 \)**:
\[ \binom{4}{2} (10)^{2} (0.1)^{2} = 6 \cdot 100 \cdot 0.01 = 6 \]

4. **For \( k = 3 \)**:
\[ \binom{4}{3} (10)^{1} (0.1)^{3} = 4 \cdot 10 \cdot 0.001 = 0.04 \]

5. **For \( k = 4 \)**:
\[ \binom{4}{4} (10)^{0} (0.1)^{4} = 1 \cdot 1 \cdot 0.0001 = 0.0001 \]

Now, we can sum all these terms together:

\[ 10000 + 400 + 6 + 0.04 + 0.0001 = 10406.0401 \]

Thus, the value of \( (10.1)^4 \) is \( 10406.0401 \).