Solution. Let's consider two consecutive terms of the expansion

$$
T_{k+1}=\binom{50}{k} \sqrt{2^{k}} x^{50-k}=A_{k+1} x^{50-k}, T_{k}=\binom{50}{k-1} \sqrt{2^{k-1}} x^{50-k+1}=A_{k} x^{50-k+1}
$$

Let's see for which values of $k$ the ratio between the coefficients $A_{k+1}$ and $A_{k}$ is less than 1. From $\frac{\binom{50}{k} \sqrt{2^{k}}}{\binom{50}{k-1} \sqrt{2^{k-1}}}51(2-\sqrt{2}) \approx 29.875$. Starting from the 30th term, the ratio $A_{k+1}: A_{k}$ is less than 1, so $A_{31}A_{29}>A_{28}>\ldots$. Therefore, the thirtieth term has the largest coefficient.
3B. Find the term with the largest coefficient in the expansion of the binomial

$$
(\sqrt{2}+x)^{50}
$$