Use Binomial theorem to find `(1.02)^6` , , correct to five decimal places.

Answer

To find \( (1.02)^6 \) using the Binomial Theorem and rounding to five decimal places, we can follow these steps:

### Step 1: Identify the Binomial Expansion
The Binomial Theorem states that:
\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
In our case, we can express \( 1.02 \) as \( 1 + 0.02 \), where \( x = 1 \), \( y = 0.02 \), and \( n = 6 \).

### Step 2: Write the Expansion
Using the Binomial Theorem, we have:
\[ (1 + 0.02)^6 = \sum_{k=0}^{6} \binom{6}{k} (1)^{6-k} (0.02)^k \]

### Step 3: Calculate Each Term
Now we will calculate each term in the expansion up to \( k = 5 \) (since we want to round to five decimal places):

1. For \( k = 0 \):
\[ \binom{6}{0} (1)^{6} (0.02)^0 = 1 \]

2. For \( k = 1 \):
\[ \binom{6}{1} (1)^{5} (0.02)^1 = 6 \cdot 0.02 = 0.12 \]

3. For \( k = 2 \):
\[ \binom{6}{2} (1)^{4} (0.02)^2 = 15 \cdot (0.02)^2 = 15 \cdot 0.0004 = 0.006 \]

4. For \( k = 3 \):
\[ \binom{6}{3} (1)^{3} (0.02)^3 = 20 \cdot (0.02)^3 = 20 \cdot 0.000008 = 0.00016 \]

5. For \( k = 4 \):
\[ \binom{6}{4} (1)^{2} (0.02)^4 = 15 \cdot (0.02)^4 = 15 \cdot 0.00000016 = 0.0000024 \]

6. For \( k = 5 \):
\[ \binom{6}{5} (1)^{1} (0.02)^5 = 6 \cdot (0.02)^5 = 6 \cdot 0.00000032 = 0.00000192 \]

### Step 4: Sum the Terms
Now, we sum the significant terms calculated:
\[ 1 + 0.12 + 0.006 + 0.00016 + 0.0000024 + 0.00000192 \]
Calculating this step-by-step:
- First, sum the first three terms:
\[ 1 + 0.12 + 0.006 = 1.126 \]
- Now add the next term:
\[ 1.126 + 0.00016 = 1.12616 \]
- Next, add the next term:
\[ 1.12616 + 0.0000024 \approx 1.1261624 \]
- Finally, add the last term:
\[ 1.1261624 + 0.00000192 \approx 1.12616432 \]

### Step 5: Round to Five Decimal Places
Now we round \( 1.12616432 \) to five decimal places:
\[ 1.12616 \]

### Final Result
Thus, the value of \( (1.02)^6 \) correct to five decimal places is:
\[ \boxed{1.12616} \]