The question in the image asks to:

"Find the term containing $x^4$ in the expansion of $(x + \frac{2}{x^2})^8$."

Solution

This problem involves finding a specific term in the binomial expansion of $(x + \frac{2}{x^2})^8$.

The general term in the binomial expansion of $(a + b)^n$ is given by: $T_{k+1} = \binom{n}{k} a^{n-k} b^k$ For this problem:

• $n = 8$
• $a = x$
• $b = \frac{2}{x^2}$

The general term, $T_{k+1}$, in this expansion is: $T_{k+1} = \binom{8}{k} \cdot x^{8-k} \cdot \left(\frac{2}{x^2}\right)^k$ Simplifying $\left(\frac{2}{x^2}\right)^k$, we get: $T_{k+1} = \binom{8}{k} \cdot x^{8-k} \cdot \frac{2^k}{x^{2k}} = \binom{8}{k} \cdot 2^k \cdot x^{8 - k - 2k}$ This simplifies to: $T_{k+1} = \binom{8}{k} \cdot 2^k \cdot x^{8 - 3k}$ We need the exponent of $x$ to be 4, so we set $8 - 3k = 4$: $8 - 3k = 4 \implies 3k = 4 \implies k = \frac{4}{3}$

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