B1. Let $(1+\sqrt{2})^{5}=a+b \sqrt{2}$, where $a$ and $b$ are positive integers. Determine the value of $a+b$.
Solution 1:
Note, if $(1+\sqrt{2})^{n}=a_{n}+b_{n} \sqrt{2}$, then
$$
\begin{aligned}
(1+\sqrt{2})^{n+1} & =(1+\sqrt{2})(1+\sqrt{2})^{n} \\
& =(1+\sqrt{2})\left(a_{n}+b_{n} \sqrt{2}\right) \\
& =a_{n}+b_{n} \sqrt{2}+a_{n} \sqrt{2}+2 b_{n} \\
& =\left(a_{n}+2 b_{n}\right)+\left(a_{n}+b_{n}\right) \sqrt{2}
\end{aligned}
$$

Thus, since $(1+\sqrt{2})^{1}=1+1 \times \sqrt{2}$, we can conclude that for any $n \in Z^{+}, n>1$
$$
(1+\sqrt{2})^{n}=a_{n}+b_{n} \sqrt{2}
$$
where $a_{1}=b_{1}=1$ and $a_{n}=a_{n-1}+2 b_{n-1}, b_{n}=a_{n-1}+b_{n-1}$. Thus we can easily calculate
\begin{tabular}{|c|c|c|}
\hline$n$ & $a_{n}$ & $b_{n}$ \\
\hline 1 & 1 & 1 \\
2 & 3 & 2 \\
3 & 7 & 5 \\
4 & 17 & 12 \\
5 & 41 & 29 \\
\hline
\end{tabular}

Hence $a+b=a_{5}+b_{5}=41+29=70$.
The answer is 70 .
Solution 2:
Expanding using the binomial theorem we get
$$
\begin{aligned}
(1+\sqrt{2})^{5} & = \\
& =\binom{5}{0}+\binom{5}{1}(\sqrt{2})+\binom{5}{2}(\sqrt{2})^{2}+\binom{5}{3}(\sqrt{2})^{3}+\binom{5}{4}(\sqrt{2})^{4}+\binom{5}{5}(\sqrt{2})^{5} \\
& =\binom{5}{0}+\binom{5}{1}(\sqrt{2})+2\binom{5}{2}+2 \sqrt{2}\binom{5}{3}+4\binom{5}{4}+4 \sqrt{2}\binom{5}{5} \\
& =\left(\binom{5}{0}+2\binom{5}{2}+4\binom{5}{4}\right)+\left(\binom{5}{1}+2\binom{5}{3}+4\binom{5}{5}\right) \sqrt{2} \\
& =a+b \sqrt{2}
\end{aligned}
$$

Hence
$$
\begin{aligned}
a+b & =\left(\binom{5}{0}+2\binom{5}{2}+4\binom{5}{4}\right)+\left(\binom{5}{1}+2\binom{5}{3}+4\binom{5}{5}\right) \\
& =\binom{5}{0}+\binom{5}{1}+2\binom{5}{2}+2\binom{5}{3}+4\binom{5}{4}+4\binom{5}{5} \\
& =1+5+2(10)+2(10)+4(5)+4(1) \\
& =70
\end{aligned}
$$

Solution 3:
Using direct calculation, we have:
$$
\begin{array}{l}
(1+\sqrt{2})^{2}=1+2 \sqrt{2}+2=3+2 \sqrt{2} \\
(1+\sqrt{2})^{4}=(3+2 \sqrt{2})^{2}=9+12 \sqrt{2}+8=17+12 \sqrt{2} \\
(1+\sqrt{2})^{5}=(17+12 \sqrt{2})(1+\sqrt{2})=17+24+(17+12) \sqrt{2}=41+29 \sqrt{2}
\end{array}
$$

The answer is $41+29=70$.