No. of term in `(1 + 5sqrt2x)^9 + (1 - 5sqrt2x)^9` if x > 0 is

Options

A. 3
B. 5
C. 4
D. 6

Answer

To find the number of terms in the expression \((1 + 5\sqrt{2}x)^9 + (1 - 5\sqrt{2}x)^9\), we can follow these steps:

### Step 1: Expand both expressions using the Binomial Theorem
Using the Binomial Theorem, we can expand both \((1 + 5\sqrt{2}x)^9\) and \((1 - 5\sqrt{2}x)^9\).

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

For \((1 + 5\sqrt{2}x)^9\):
\[ (1 + 5\sqrt{2}x)^9 = \sum_{k=0}^{9} \binom{9}{k} (1)^{9-k} (5\sqrt{2}x)^k = \sum_{k=0}^{9} \binom{9}{k} 5^k (\sqrt{2})^k x^k \]

For \((1 - 5\sqrt{2}x)^9\):
\[ (1 - 5\sqrt{2}x)^9 = \sum_{k=0}^{9} \binom{9}{k} (1)^{9-k} (-5\sqrt{2}x)^k = \sum_{k=0}^{9} \binom{9}{k} (-5)^k (\sqrt{2})^k x^k \]

### Step 2: Combine the two expansions
Now, we add the two expansions:
\[ (1 + 5\sqrt{2}x)^9 + (1 - 5\sqrt{2}x)^9 = \sum_{k=0}^{9} \binom{9}{k} 5^k (\sqrt{2})^k x^k + \sum_{k=0}^{9} \binom{9}{k} (-5)^k (\sqrt{2})^k x^k \]

### Step 3: Simplify the combined expression
Notice that when we combine the two expansions, the terms will cancel out for odd \(k\) due to the alternating signs:
- For even \(k\), the terms will add up.
- For odd \(k\), the terms will cancel out.

### Step 4: Identify the remaining terms
The remaining terms will be those where \(k\) is even. The even values of \(k\) from 0 to 9 are:
- \(k = 0\)
- \(k = 2\)
- \(k = 4\)
- \(k = 6\)
- \(k = 8\)

### Step 5: Count the number of terms
The even values of \(k\) are \(0, 2, 4, 6, 8\), which gives us a total of 5 terms.

### Final Answer
Thus, the number of terms in the expression \((1 + 5\sqrt{2}x)^9 + (1 - 5\sqrt{2}x)^9\) is **5**.

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