The number of terms in the expansion of `(1+5 sqrt2x)^(19)+ (1-5 sqrt2x)^(19)` is

Options

A. 10
B. 7
C. 13
D. 4

Answer

To find the number of terms in the expansion of \( (1 + 5\sqrt{2}x)^{19} + (1 - 5\sqrt{2}x)^{19} \), 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 \( (1 + 5\sqrt{2}x)^{19} \) and \( (1 - 5\sqrt{2}x)^{19} \).

### Step 2: Expand Each Term
1. **Expansion of \( (1 + 5\sqrt{2}x)^{19} \)**:
\[ (1 + 5\sqrt{2}x)^{19} = \sum_{k=0}^{19} \binom{19}{k} (1)^{19-k} (5\sqrt{2}x)^k = \sum_{k=0}^{19} \binom{19}{k} (5\sqrt{2})^k x^k \]

2. **Expansion of \( (1 - 5\sqrt{2}x)^{19} \)**:
\[ (1 - 5\sqrt{2}x)^{19} = \sum_{k=0}^{19} \binom{19}{k} (1)^{19-k} (-5\sqrt{2}x)^k = \sum_{k=0}^{19} \binom{19}{k} (-5\sqrt{2})^k x^k \]

### Step 3: Combine the Expansions
Now, we add the two expansions:
\[ (1 + 5\sqrt{2}x)^{19} + (1 - 5\sqrt{2}x)^{19} = \sum_{k=0}^{19} \binom{19}{k} (5\sqrt{2})^k x^k + \sum_{k=0}^{19} \binom{19}{k} (-5\sqrt{2})^k x^k \]
This simplifies to:
\[ = \sum_{k=0}^{19} \binom{19}{k} ((5\sqrt{2})^k + (-5\sqrt{2})^k) x^k \]

### Step 4: Identify Non-Zero Terms
Notice that \( (5\sqrt{2})^k + (-5\sqrt{2})^k \) will be zero for odd \( k \) and non-zero for even \( k \). Therefore, only the even powers of \( k \) will contribute to the final sum.

### Step 5: Count the Number of Terms
The even values of \( k \) from 0 to 19 are:
- \( k = 0, 2, 4, 6, 8, 10, 12, 14, 16, 18 \)

This gives us a total of 10 terms (0 through 18, inclusive, with a step of 2).

### Final Answer
Thus, the number of terms in the expansion of \( (1 + 5\sqrt{2}x)^{19} + (1 - 5\sqrt{2}x)^{19} \) is **10**.

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