Let `(1 + x)^(36) = a_(0) +a_(1) x + a_(2) x^(2) +...+ a_(36)x^(36)`. Then
Statememt-1: `a_(0) +a_(3) +a_(6) +…+a_(36)= (2)/(3) (2^(35) +1)`
Statement-2: `a_(0) + a_(2) +a_(4) +…+ a_(36) = 2^(35)`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the two statements given regarding the coefficients of the expansion of \((1 + x)^{36}\).

### Step 1: Understanding the Coefficients
The coefficients \(a_k\) in the expansion of \((1 + x)^{36}\) can be expressed using the binomial theorem:
\[ (1 + x)^{36} = \sum_{k=0}^{36} a_k x^k = \sum_{k=0}^{36} \binom{36}{k} x^k \]
Thus, \(a_k = \binom{36}{k}\).

### Step 2: Analyzing Statement 1
Statement 1 claims:
\[ a_0 + a_3 + a_6 + \ldots + a_{36} = \frac{2}{3} (2^{35} + 1) \]

To find this sum, we can use the roots of unity filter. We will use the cube roots of unity, denoted as \(1, \omega, \omega^2\), where \(\omega = e^{2\pi i / 3}\).

1. **Calculate \(S_0\)**:
\[ S_0 = (1 + 1)^{36} = 2^{36} \]

2. **Calculate \(S_1\)**:
\[ S_1 = (1 + \omega)^{36} = (1 - \omega^2)^{36} = (-\omega^2)^{36} = \omega^{72} = 1 \]

3. **Calculate \(S_2\)**:
\[ S_2 = (1 + \omega^2)^{36} = (1 - \omega)^{36} = (-\omega)^{36} = \omega^{36} = 1 \]

Now, we can sum these results:
\[ S_0 + S_1 + S_2 = 2^{36} + 1 + 1 = 2^{36} + 2 \]

### Step 3: Using Roots of Unity
The sum of coefficients at indices that are multiples of 3 can be found using:
\[ \frac{S_0 + S_1 + S_2}{3} = \frac{2^{36} + 2}{3} \]

This simplifies to:
\[ \frac{2^{36} + 2}{3} = \frac{2(2^{35} + 1)}{3} \]

Thus, we have:
\[ a_0 + a_3 + a_6 + \ldots + a_{36} = \frac{2}{3}(2^{35} + 1) \]
This confirms that Statement 1 is true.

### Step 4: Analyzing Statement 2
Statement 2 claims:
\[ a_0 + a_2 + a_4 + \ldots + a_{36} = 2^{35} \]

To find this sum, we can use the same approach with \(x = 1\) and \(x = -1\):

1. **Calculate \(S_0\)** (as before):
\[ S_0 = 2^{36} \]

2. **Calculate \(S_{-1}\)**:
\[ S_{-1} = (1 - 1)^{36} = 0 \]

Adding these two results:
\[ S_0 + S_{-1} = 2^{36} + 0 = 2^{36} \]

Now, we can express the sums of even and odd indexed coefficients:
\[ S_0 = a_0 + a_2 + a_4 + \ldots + a_{36} + (a_1 + a_3 + a_5 + \ldots + a_{35}) \]
\[ S_{-1} = a_0 - a_1 + a_2 - a_3 + \ldots + a_{36} = 0 \]

Adding these two equations gives:
\[ 2(a_0 + a_2 + a_4 + \ldots + a_{36}) = 2^{36} \]
Thus,
\[ a_0 + a_2 + a_4 + \ldots + a_{36} = 2^{35} \]
This confirms that Statement 2 is also true.

### Conclusion
Both statements are true, but Statement 2 is not the correct explanation for Statement 1. Thus, the final answer is that both statements are true, but Statement 2 does not explain Statement 1.