To find the coefficient of $x^4$ in the expansion of $(x + \frac{1}{3})^3 (x + \frac{2}{x^2})^{10}$, we will carefully analyze each term's contribution to the coefficient of $x^4$.

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Step 1: General Term of $(x + \frac{1}{3})^3$

Using the binomial expansion formula:

$(x + \frac{1}{3})^3 = \sum_{r=0}^{3} \binom{3}{r} x^{3-r} \left(\frac{1}{3}\right)^r$

The general term is:

$T_r^{(1)} = \binom{3}{r} x^{3-r} \left(\frac{1}{3}\right)^r$

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Step 2: General Term of $(x + \frac{2}{x^2})^{10}$

Using the binomial expansion formula:

$(x + \frac{2}{x^2})^{10} = \sum_{s=0}^{10} \binom{10}{s} x^{10-s} \left(\frac{2}{x^2}\right)^s$

The general term is:

$T_s^{(2)} = \binom{10}{s} x^{10-s} \left(\frac{2}{x^2}\right)^s = \binom{10}{s} x^{10-s - 2s} 2^s = \binom{10}{s} x^{10-3s} 2^s$

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Step 3: Combined Expansion

Now, the product of these expansions is:

$(x + \frac{1}{3})^3 (x + \frac{2}{x^2})^{10}$

The general term of the product is obtained by multiplying terms $T_r^{(1)}$ and $T_s^{(2)}$:

$T_{r,s} = \binom{3}{r} x^{3-r} \left(\frac{1}{3}\right)^r \cdot \binom{10}{s} x^{10-3s} 2^s$

Combine the powers of $x$:

$T_{r,s} = \binom{3}{r} \binom{10}{s} \left(\frac{1}{3}\right)^r 2^s x^{(3-r) + (10-3s)}$

Simplify the power of $x$:

$T_{r,s} = \binom{3}{r} \binom{10}{s} \left(\frac{1}{3}\right)^r 2^s x^{13 - r - 3s}$

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Step 4: Find Coefficient of $x^4$

We need the total power of $x$ to be $4$:

$13 - r - 3s = 4$

Solve for $r + 3s$:

$r + 3s = 9$

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Step 5: Solve for Valid Combinations of $r$ and $s$

The possible values of $r$ and $s$ are integers that satisfy $r \leq 3$ (from $\binom{3}{r}$) and $s \leq 10$ (from $\binom{10}{s}$):

Case 1: $s = 2$

$r + 3(2) = 9 \implies r = 3$

Case 2: $s = 3$

$r + 3(3) = 9 \implies r = 0$

The valid pairs are:

$(r, s) = (3, 2) \quad \text{and} \quad (r, s) = (0, 3)$

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Step 6: Compute Coefficients

For each pair, calculate the contribution to the coefficient of $x^4$:

Pair $(r, s) = (3, 2)$

$T_{3,2} = \binom{3}{3} \binom{10}{2} \left(\frac{1}{3}\right)^3 2^2$

$T_{3,2} = 1 \cdot 45 \cdot \frac{1}{27} \cdot 4 = \frac{45 \cdot 4}{27} = \frac{180}{27} = \frac{20}{3}$

Pair $(r, s) = (0, 3)$

$T_{0,3} = \binom{3}{0} \binom{10}{3} \left(\frac{1}{3}\right)^0 2^3$

$T_{0,3} = 1 \cdot 120 \cdot 1 \cdot 8 = 960$

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Step 7: Total Coefficient

Add the contributions from both cases:

$\text{Coefficient of } x^4 = \frac{20}{3} + 960 = \frac{20 + 2880}{3} = \frac{2900}{3}$

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Final Answer:

The coefficient of $x^4$ is:

$\boxed{\frac{2900}{3}}$

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Do you want more details on this solution or have specific questions?

Related Questions:

1. How is the binomial expansion formula applied in general?
2. Why does the power $r + 3s = 9$ arise in this problem?
3. Can this method be generalized for other powers of $x$?
4. What happens if one of the terms had a negative exponent?
5. How can we verify the solution using computational tools?

Tip:

Always double-check your indices and constraints when summing terms in multi-variable expansions!