Answer:
6
Explanation:
We can prove that every number after the second will be a six by induction.
Base case. Since
, so the third term is a six.
Inductive hypothesis. Fix the number of terms to be n and assume that
![(1)/(n) \sum\limits_(i=1)^(n)t_i = 6](https://img.qammunity.org/2020/formulas/mathematics/high-school/o6321yfahah3xu6h7x30n14czef2a5h7df.png)
Inductive step. We will now show that
.
Notice that
![$\begin{array}{lll}(1)/(n+1) \sum\limits_(i=1)^(n+1)t_i & = (n)/(n(n+1)) \sum\limits_(i=1)^(n+1)t_i & \\& = (t_(n+1))/(n+1) + (n)/(n(n+1)) \sum\limits_(i=1)^(n)t_i &\\& = (t_(n+1))/(n+1) + (6n)/(n+1) & \text{(by the IH)}\\& = (6)/(n+1) + (6n)/(n+1) & \text{by de\\finition}\\& = (6(n+1))/(n+1) & \\& = 6 & \end{array} \square$](https://img.qammunity.org/2020/formulas/mathematics/high-school/hn19s53gp61wftfocskz6cvdyekv6v3su1.png)