Average Problem

For the general case of $k$ terms, $U_n = \frac{U_{n-k} + U_{n-k+1} + \dots + U_{n-1}}{k}$ .

Then $U_{n-k+1} + 2U_{n-k+2} + \dots kU_{n}$ $=$ $U_{n-k+1} + 2U_{n-k+2} + \dots + k \cdot \frac{U_{n-k} + U_{n-k+1} + \dots + U_{n-1}}{k}$ $=$ $U_{n-k} + 2U_{n-k+1} + \dots + kU_{n-1}$ .

Since $U_{n-k+1} + 2U_{n-k+2} + \dots + kU_{n}$ $=$ $U_{n-k} + 2U_{n-k+1} + \dots + kU_{n-1}$ , $n$ is independent, so $U_{n-k+1} + 2U_{n-k+2} + \dots + kU_{n}$ $=$ $U_{0} + 2U_{1} + \dots + kU_{k - 1}$ .

If $L = \lim_{n \to \infty} U_{n}$ , then $U_{1} + 2U_{2} + \dots + kU_{k}$ $=$ $L + 2L + \dots + kL$ $=$ $(1 + 2+ \dots + 3)L$ $=$ $\frac{k(k + 1)}{2}L$ . Therefore, $L = \frac{2}{k(k + 1)}(U_{0} + 2U_{1} + \dots + kU_{k - 1})$ , which means:

$\lim_{n \to \infty} U_{n} = \frac{2}{k(k + 1)}(U_{0} + 2U_{1} + \dots + kU_{k - 1})$

For the specific case when $k = 2$ and $U_0 = m$ and $U_1 = n$ , $\lim_{n \to \infty} U_{n}$ $=$ $\frac{2}{2(2 + 1)}(m + 2n)$ $=$ $\frac{m + 2n}{3}$ , which means $a = 1$ , $b = 2$ , and $c = 3$ , and $a + b + c$ $=$ $1 + 2 + 3$ $=$ $\boxed{6}$ .