`(x+y)^(n) + (x-y)^(n)` is equal to

Options

A. `2[.^(n)C_(1)x^(n)y^(0)+.^(n)C_(3)x^(n-2)y^(2)+"….."]`
B. `2[.^(n)C_(0)x^(n)y^(0)+.^(n)C_(2)x^(n-2)y^(2)+"….."]`
C. `2[.^(n)C_(1)x^(n-1)y^(1)+.^(n)C_(3)x^(n-3)y^(3)+"….."]`
D. `3[.^(n)C_(0)x^(n)y^(0)+.^(n)C_(2)x^(n-2)y^(2)+"……"]`

Answer

To solve the expression \((x+y)^{n} + (x-y)^{n}\), we can use the Binomial Theorem, which states that:

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

### Step 1: Expand \((x+y)^{n}\)

Using the Binomial Theorem, we can expand \((x+y)^{n}\):

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

This gives us:

\[ (x+y)^{n} = \binom{n}{0} x^{n} + \binom{n}{1} x^{n-1} y + \binom{n}{2} x^{n-2} y^{2} + \ldots + \binom{n}{n} y^{n} \]

### Step 2: Expand \((x-y)^{n}\)

Now, we expand \((x-y)^{n}\) using the same theorem:

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

This gives us:

\[ (x-y)^{n} = \binom{n}{0} x^{n} + \binom{n}{1} x^{n-1} (-y) + \binom{n}{2} x^{n-2} (-y)^{2} + \ldots + \binom{n}{n} (-y)^{n} \]

### Step 3: Combine the two expansions

Now, we add the two expansions together:

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

### Step 4: Simplify the expression

When we combine the two expansions, we notice that terms with odd powers of \(y\) will cancel out because they will have opposite signs. Therefore, we only keep the terms with even powers of \(y\):

\[ = \sum_{k \text{ even}} \binom{n}{k} x^{n-k} y^{k} + \sum_{k \text{ even}} \binom{n}{k} x^{n-k} y^{k} \]

This can be simplified to:

\[ = 2 \sum_{k \text{ even}} \binom{n}{k} x^{n-k} y^{k} \]

### Step 5: Final Result

Thus, the final result is:

\[ (x+y)^{n} + (x-y)^{n} = 2 \sum_{k \text{ even}} \binom{n}{k} x^{n-k} y^{k} \]