Question

Решить задачу (1курс+)

Answer 1

$x_{n+1}=\frac{1}{2}\left ( x_n+x_{n-1} \right )\Rightarrow x_{n+1}-x_n=-\frac{1}{2}\left ( x_n-x_{n-1} \right )=\ldots=\left ( -\frac{1}{2} \right )^{n-1}\left ( x_2-x_1 \right )\\x_n=x_1+\left ( x_2-x_1 \right )+\ldots+\left ( x_n-x_{n-1} \right )=a+(b-a)\left ( 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots \right )\longrightarrow \\\longrightarrow a+(b-a)\cdot \frac{1}{1+1/2}=a+\frac{2}{3}(b-a)=\frac{2}{3}b+\frac{1}{3}a$