Find the number of terms in the following expansions.
(v) `(3x+7)^(8) + (3x-7)^(8)`

Answer

To find the number of terms in the expansion of \((3x + 7)^8 + (3x - 7)^8\), we can follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \((a + b)^n\) is given by:
\[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, we will apply this to both \((3x + 7)^8\) and \((3x - 7)^8\).

### Step 2: Expand \((3x + 7)^8\)
Using the binomial theorem:
\[ (3x + 7)^8 = \sum_{k=0}^{8} \binom{8}{k} (3x)^{8-k} (7)^k \]
This will yield terms of the form \((3x)^{8-k} \cdot 7^k\).

### Step 3: Expand \((3x - 7)^8\)
Similarly, for \((3x - 7)^8\):
\[ (3x - 7)^8 = \sum_{k=0}^{8} \binom{8}{k} (3x)^{8-k} (-7)^k \]
This will yield terms of the form \((3x)^{8-k} \cdot (-7)^k\).

### Step 4: Combine the Two Expansions
Now, we add the two expansions:
\[ (3x + 7)^8 + (3x - 7)^8 \]
When we combine these, we notice that terms with odd powers of \(7\) will cancel out because they will have opposite signs.

### Step 5: Identify the Remaining Terms
The remaining terms will be those with even powers of \(7\). The powers of \(x\) in the expansion will be:
- From \((3x + 7)^8\) and \((3x - 7)^8\), the powers of \(x\) are \(8, 6, 4, 2, 0\) (even powers).

### Step 6: Count the Unique Terms
The unique powers of \(x\) that remain after cancellation are:
- \(x^8\)
- \(x^6\)
- \(x^4\)
- \(x^2\)
- Constant term (which is \(7^8 + 7^8\))

Thus, the total number of unique terms is \(5\).

### Final Answer
The number of terms in the expansion of \((3x + 7)^8 + (3x - 7)^8\) is **5**.

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