Use Binomial theorem to find : `(51)^6`

Answer

To find \( 51^6 \) using the Binomial Theorem, we can express \( 51 \) as \( 50 + 1 \). Then, we can apply the Binomial Theorem to expand \( (50 + 1)^6 \).

### Step-by-Step Solution:

1. **Identify the Binomial Expansion**:
The Binomial Theorem states that:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
Here, we will let \( x = 50 \), \( y = 1 \), and \( n = 6 \).

2. **Set Up the Expansion**:
Using the theorem, we have:
\[ (50 + 1)^6 = \sum_{k=0}^{6} \binom{6}{k} 50^{6-k} \cdot 1^k \]

3. **Calculate Each Term**:
We will calculate each term in the expansion:
- For \( k = 0 \):
\[ \binom{6}{0} 50^6 = 1 \cdot 50^6 \]
- For \( k = 1 \):
\[ \binom{6}{1} 50^5 \cdot 1 = 6 \cdot 50^5 \]
- For \( k = 2 \):
\[ \binom{6}{2} 50^4 \cdot 1^2 = 15 \cdot 50^4 \]
- For \( k = 3 \):
\[ \binom{6}{3} 50^3 \cdot 1^3 = 20 \cdot 50^3 \]
- For \( k = 4 \):
\[ \binom{6}{4} 50^2 \cdot 1^4 = 15 \cdot 50^2 \]
- For \( k = 5 \):
\[ \binom{6}{5} 50^1 \cdot 1^5 = 6 \cdot 50 \]
- For \( k = 6 \):
\[ \binom{6}{6} 50^0 \cdot 1^6 = 1 \]

4. **Substituting Values**:
Now, we substitute the powers of \( 50 \):
\[ 50^6 = 15625000000, \quad 50^5 = 312500000, \quad 50^4 = 6250000, \quad 50^3 = 125000, \quad 50^2 = 2500, \quad 50^1 = 50 \]

5. **Calculating Each Term**:
- \( 1 \cdot 15625000000 = 15625000000 \)
- \( 6 \cdot 312500000 = 1875000000 \)
- \( 15 \cdot 6250000 = 93750 \)
- \( 20 \cdot 125000 = 2500000 \)
- \( 15 \cdot 2500 = 37500 \)
- \( 6 \cdot 50 = 300 \)
- \( 1 \cdot 1 = 1 \)

6. **Summing All Terms**:
Now, we sum all the terms:
\[ 15625000000 + 1875000000 + 93750 + 2500000 + 37500 + 300 + 1 \]

7. **Final Calculation**:
Performing the addition:
\[ 15625000000 + 1875000000 = 17500000000 \]
\[ 17500000000 + 93750 = 17500093750 \]
\[ 17500093750 + 2500000 = 17502543750 \]
\[ 17502543750 + 37500 = 17502581250 \]
\[ 17502581250 + 300 = 17502581550 \]
\[ 17502581550 + 1 = 17502581551 \]

Thus, the final result is:
\[ \boxed{17592151000} \]