What is the number of term in the expansion of `[(2x + y)^8- (2x - y)^8] ` ?

Answer

To find the number of terms in the expression \([(2x + y)^8 - (2x - y)^8]\), we can follow these steps:

### Step 1: Expand both expressions
We will first expand \((2x + y)^8\) and \((2x - y)^8\) using the Binomial Theorem.

The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

For \((2x + y)^8\):
\[ (2x + y)^8 = \sum_{k=0}^{8} \binom{8}{k} (2x)^{8-k} y^k = \sum_{k=0}^{8} \binom{8}{k} 2^{8-k} x^{8-k} y^k \]

For \((2x - y)^8\):
\[ (2x - y)^8 = \sum_{k=0}^{8} \binom{8}{k} (2x)^{8-k} (-y)^k = \sum_{k=0}^{8} \binom{8}{k} 2^{8-k} x^{8-k} (-1)^k y^k \]

### Step 2: Combine the expansions
Now, we subtract the two expansions:
\[ (2x + y)^8 - (2x - y)^8 = \sum_{k=0}^{8} \binom{8}{k} 2^{8-k} x^{8-k} y^k - \sum_{k=0}^{8} \binom{8}{k} 2^{8-k} x^{8-k} (-1)^k y^k \]

### Step 3: Simplify the expression
Combining the two sums, we can factor out the common terms:
\[ = \sum_{k=0}^{8} \binom{8}{k} 2^{8-k} x^{8-k} \left( y^k - (-1)^k y^k \right) \]
\[ = \sum_{k=0}^{8} \binom{8}{k} 2^{8-k} x^{8-k} y^k (1 - (-1)^k) \]

### Step 4: Determine which terms remain
The term \(1 - (-1)^k\) is zero for even \(k\) and two for odd \(k\). Therefore, only the odd \(k\) terms will remain in the final expression.

The odd values of \(k\) from 0 to 8 are \(1, 3, 5, 7\).

### Step 5: Count the number of terms
The odd values of \(k\) are:
- \(k = 1\)
- \(k = 3\)
- \(k = 5\)
- \(k = 7\)

Thus, there are **4 terms** remaining in the final expression.

### Final Answer:
The number of terms in the expansion of \([(2x + y)^8 - (2x - y)^8]\) is **4**.

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