Solving the recurrence relation that contains summation of nth term $$T(n)=1+2\sum_{i=1}^{n-1}T(i) , \quad n > 1$$
$$T(1)=1$$
any hint or how to solve?

Using a spreadsheet, I note that $T(n)=3^{(n-1)}$  This is easily verified by induction.
$T(1)=1=3^0$.
Then if it is true up to $n$, $$T(n+1)=1+2\sum_{i=0}^{n-1}3^i=1+2\frac{3^n-1}{3-1}$$
