2. Find the largest term in the expansion of $(1+\sqrt{2})^{50}$.
Solution. According to the binomial formula, the $(n+1)$-th term of the expansion of $(1+\sqrt{2})^{50}$ is

$$
T_{n+1}=\binom{50}{n} \sqrt{2^{n}}
$$

Thus, we have $T_{1}=1, T_{2}=50 \sqrt{2}, T_{3}=255$, which means that $T_{1}51(2-\sqrt{2})
$$

The smallest natural number greater than $51(2-\sqrt{2})$ is 30, which means that

$$
T_{30}>T_{31}>T_{32}>\ldots>T_{50} \text { and } T_{1}<T_{2}<T_{3}<\ldots<T_{30}
$$

According to this, the largest term in the expansion of $(1+\sqrt{2})^{50}$ is the term

$$
T_{30}=\binom{50}{29} \sqrt{2^{29}}
$$