3. Let $M_{1}, M_{2}, M_{3}, \ldots, M_{n}$ be collinear points in this order, such that:

$$
M_{1} M_{2}=9 ; M_{2} M_{3}=17, M_{3} M_{4}=33, M_{4} M_{5}=65, \ldots, M_{n-1} M_{n}
$$

a) Find the length of the segment $\left[M_{8} M_{9}\right]$.

b) Find $n \in \mathbb{N}$ if $M_{1} M_{n}=8194$.

c) Find the length of the segment $M_{2} A$ where $A$ is the midpoint of the segment $\left[M_{7} M_{9}\right]$.
Solution. a) $M_{n} M_{n+1}=2^{n+2}+1 ; M_{8} M_{9}=2^{10}+1=1025$.

b)

$$
\begin{aligned}
& M_{1} M_{n}=M_{1} M_{2}+M_{2} M_{3}+M_{3} M_{4}+\ldots+M_{n-1} M_{n}= \\
& =\left(2^{3}+1\right)+\left(2^{4}+1\right)+\ldots+\left(2^{n+1}+1\right)=2^{n+2}-2^{3}+n-1=8194 \\
& M_{1} M_{10}=2^{11+2}-8+11-1=2^{13}+2=8194 \Rightarrow n=11
\end{aligned}
$$

c) $M_{2} A=M_{2} M_{7}+M_{7} M_{9}: 2=\left(2^{4}+1\right)+\left(2^{5}+1\right)+\left(2^{6}+1\right)+\left(2^{7}+1\right)+\left(2^{8}+1\right)+\left[\left(2^{9}+1\right)+\left(2^{10}+1\right)\right]: 2=$ $=\left(2^{9}-2^{4}\right)+5+2^{8}+2^{9}+1=1270$.