What is the number of terms in the expansion of the following?
`(a+bx)^(17) - (a-bx)^(17)`

Answer

To find the number of terms in the expression \((a + bx)^{17} - (a - bx)^{17}\), we will follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \((a + b)^n\) is given by:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For \((a + bx)^{17}\) and \((a - bx)^{17}\), we can apply this formula.

### Step 2: Expand \((a + bx)^{17}\)
Using the binomial theorem:
\[ (a + bx)^{17} = \sum_{k=0}^{17} \binom{17}{k} a^{17-k} (bx)^k = \sum_{k=0}^{17} \binom{17}{k} a^{17-k} b^k x^k \]

### Step 3: Expand \((a - bx)^{17}\)
Similarly, for \((a - bx)^{17}\):
\[ (a - bx)^{17} = \sum_{k=0}^{17} \binom{17}{k} a^{17-k} (-bx)^k = \sum_{k=0}^{17} \binom{17}{k} a^{17-k} (-1)^k b^k x^k \]

### Step 4: Combine the Expansions
Now, we subtract the two expansions:
\[ (a + bx)^{17} - (a - bx)^{17} = \sum_{k=0}^{17} \binom{17}{k} a^{17-k} b^k x^k - \sum_{k=0}^{17} \binom{17}{k} a^{17-k} (-1)^k b^k x^k \]
This simplifies to:
\[ = \sum_{k=0}^{17} \binom{17}{k} a^{17-k} b^k x^k - \sum_{k=0}^{17} \binom{17}{k} a^{17-k} (-1)^k b^k x^k \]

### Step 5: Identify Terms that Remain
Notice that when we add the two expansions, terms where \(k\) is even will cancel out because they will have the same coefficient but opposite signs. Terms where \(k\) is odd will double because they will have the same sign.

### Step 6: Count the Remaining Terms
The odd values of \(k\) from \(0\) to \(17\) are \(1, 3, 5, 7, 9, 11, 13, 15, 17\). This gives us:
- The odd integers from \(1\) to \(17\) are \(9\) in total.

### Conclusion
Thus, the number of terms in the expansion of \((a + bx)^{17} - (a - bx)^{17}\) is \(9\).