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?

Take a look at the first few terms:
$$T(1)=1, T(2)=3, T(3)=9$$
and conclude that it is a geometric series. So, $$\sum_{i=1}^{n-1}=a(r^m-1)/(r-1)=\frac{3^{n-1}-1}{2}$$
Now, $$\boxed{T(n)}=1+2\sum_{i=1}^{n-1}=1+3^{n-1}-1 = \boxed{3^{n-1}}$$
