Using binomial theorem, evaluate each of the following:
(i)`(104)^(4)` (ii) `(98)^(4)` (iii)`(1.2)^(4)`

Explanation

To evaluate the expressions \(104^4\), \(98^4\), and \(1.2^4\) using the Binomial Theorem, we will follow these steps:

### Step 1: Evaluate \(104^4\)

We can express \(104\) as \(100 + 4\). According to the Binomial Theorem:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

For \(104^4\), we have \(a = 100\), \(b = 4\), and \(n = 4\):

\[ 104^4 = (100 + 4)^4 \]

Using the Binomial Theorem:

\[ = \sum_{r=0}^{4} \binom{4}{r} (100)^{4-r} (4)^r \]

Calculating each term:

- For \(r = 0\):
\[ \binom{4}{0} (100)^4 (4)^0 = 1 \cdot 100^4 \cdot 1 = 100000000 \]

- For \(r = 1\):
\[ \binom{4}{1} (100)^3 (4)^1 = 4 \cdot 100^3 \cdot 4 = 1600000 \]

- For \(r = 2\):
\[ \binom{4}{2} (100)^2 (4)^2 = 6 \cdot 100^2 \cdot 16 = 960000 \]

- For \(r = 3\):
\[ \binom{4}{3} (100)^1 (4)^3 = 4 \cdot 100 \cdot 64 = 25600 \]

- For \(r = 4\):
\[ \binom{4}{4} (100)^0 (4)^4 = 1 \cdot 1 \cdot 256 = 256 \]

Now, summing all these terms:

\[ 104^4 = 100000000 + 1600000 + 960000 + 25600 + 256 = 11652961 \]

### Step 2: Evaluate \(98^4\)

We can express \(98\) as \(100 - 2\). Using the Binomial Theorem again:

\[ 98^4 = (100 - 2)^4 \]

Using the Binomial Theorem:

\[ = \sum_{r=0}^{4} \binom{4}{r} (100)^{4-r} (-2)^r \]

Calculating each term:

- For \(r = 0\):
\[ \binom{4}{0} (100)^4 (-2)^0 = 1 \cdot 100^4 \cdot 1 = 100000000 \]

- For \(r = 1\):
\[ \binom{4}{1} (100)^3 (-2)^1 = 4 \cdot 100^3 \cdot (-2) = -8000000 \]

- For \(r = 2\):
\[ \binom{4}{2} (100)^2 (-2)^2 = 6 \cdot 100^2 \cdot 4 = 2400000 \]

- For \(r = 3\):
\[ \binom{4}{3} (100)^1 (-2)^3 = 4 \cdot 100 \cdot (-8) = -3200 \]

- For \(r = 4\):
\[ \binom{4}{4} (100)^0 (-2)^4 = 1 \cdot 1 \cdot 16 = 16 \]

Now, summing all these terms:

\[ 98^4 = 100000000 - 8000000 + 2400000 - 3200 + 16 = 92236816 \]

### Step 3: Evaluate \(1.2^4\)

We can express \(1.2\) as \(1 + 0.2\). Using the Binomial Theorem:

\[ 1.2^4 = (1 + 0.2)^4 \]

Using the Binomial Theorem:

\[ = \sum_{r=0}^{4} \binom{4}{r} (1)^{4-r} (0.2)^r \]

Calculating each term:

- For \(r = 0\):
\[ \binom{4}{0} (1)^4 (0.2)^0 = 1 \cdot 1 \cdot 1 = 1 \]

- For \(r = 1\):
\[ \binom{4}{1} (1)^3 (0.2)^1 = 4 \cdot 1 \cdot 0.2 = 0.8 \]

- For \(r = 2\):
\[ \binom{4}{2} (1)^2 (0.2)^2 = 6 \cdot 1 \cdot 0.04 = 0.24 \]

- For \(r = 3\):
\[ \binom{4}{3} (1)^1 (0.2)^3 = 4 \cdot 1 \cdot 0.008 = 0.032 \]

- For \(r = 4\):
\[ \binom{4}{4} (1)^0 (0.2)^4 = 1 \cdot 1 \cdot 0.0016 = 0.0016 \]

Now, summing all these terms:

\[ 1.2^4 = 1 + 0.8 + 0.24 + 0.032 + 0.0016 = 2.0736 \]

### Final Answers:
- \(104^4 = 11652961\)
- \(98^4 = 92236816\)
- \(1.2^4 = 2.0736\)