Let `n in N, S_n=sum_(r=0)^(3n)^(3n)C_r` and `T_n=sum_(r=0)^n^(3n)C_(3r)`, then `|S_n-3T_n|` equals

Answer

To solve the problem, we need to find the value of \(|S_n - 3T_n|\), where:

- \(S_n = \sum_{r=0}^{3n} \binom{3n}{r}\)
- \(T_n = \sum_{r=0}^{n} \binom{3n}{3r}\)

### Step 1: Calculate \(S_n\)

Using the Binomial Theorem, we know that:

\[ (1 + x)^{3n} = \sum_{r=0}^{3n} \binom{3n}{r} x^r \]

Setting \(x = 1\):

\[ (1 + 1)^{3n} = 2^{3n} = \sum_{r=0}^{3n} \binom{3n}{r} \]

Thus,

\[ S_n = 2^{3n} \]

### Step 2: Calculate \(T_n\)

To find \(T_n\), we consider the sum of the coefficients where \(r\) is a multiple of 3. We can use the roots of unity filter.

Let \(\omega = e^{2\pi i / 3}\), which is a primitive cube root of unity. We can evaluate:

\[ T_n = \frac{1}{3} \left( S_n + (1 + \omega)^{3n} + (1 + \omega^2)^{3n} \right) \]

Calculating \(1 + \omega\) and \(1 + \omega^2\):

- \(1 + \omega = e^{\pi i / 3}\)
- \(1 + \omega^2 = e^{-\pi i / 3}\)

Thus,

\[ (1 + \omega)^{3n} = \left(e^{\pi i / 3}\right)^{3n} = e^{\pi i n} = (-1)^n \]

\[ (1 + \omega^2)^{3n} = \left(e^{-\pi i / 3}\right)^{3n} = e^{-\pi i n} = (-1)^n \]

Substituting back into the equation for \(T_n\):

\[ T_n = \frac{1}{3} \left( 2^{3n} + (-1)^n + (-1)^n \right) = \frac{1}{3} \left( 2^{3n} + 2(-1)^n \right) \]

Thus,

\[ T_n = \frac{2^{3n}}{3} + \frac{2(-1)^n}{3} \]

### Step 3: Calculate \(|S_n - 3T_n|\)

Now we substitute \(S_n\) and \(T_n\) into \(|S_n - 3T_n|\):

\[ 3T_n = 3 \left( \frac{2^{3n}}{3} + \frac{2(-1)^n}{3} \right) = 2^{3n} + 2(-1)^n \]

Thus,

\[ S_n - 3T_n = 2^{3n} - (2^{3n} + 2(-1)^n) = -2(-1)^n \]

Taking the absolute value:

\[ |S_n - 3T_n| = | -2(-1)^n | = 2 \]

### Final Answer

Therefore, the final result is:

\[ |S_n - 3T_n| = 2 \]