Let $$f\left( 1 \right) = 1$$   and $$f\left( n \right) = 2\sum\limits_{r = 1}^{n - 1} {f\left( r \right).} $$    Then $$\sum\limits_{n = 1}^m {f\left( n \right)} $$   is equal to :

Solution :
$$\eqalign{ & f\left( 1 \right) = 1 \cr & f\left( 2 \right) = 2\sum\limits_{x = 1}^1 {f\left( x \right)} \cr & = 2f\left( 1 \right) = 2 \times 1 = 2 \cr & f\left( 3 \right) = 2\sum\limits_{x = 1}^2 {f\left( x \right)} \cr & = 2f\left( 1 \right) + 2f\left( 2 \right) = 2 \times 1 + 2 \times 2 = 6 \cr & f\left( 4 \right) = 2\sum\limits_{x = 1}^3 {f\left( x \right)} \cr & = 2\left[ {f\left( 1 \right) + f\left( 2 \right) + f\left( 3 \right)} \right] \cr & = 2\left[ {1 + 2 + 6} \right] = 18 \cr & {\text{The sequence is}} = 1 + 2 + 6 + 18 + ..... \cr & \sum\limits_{n = 1}^m {f\left( n \right)} = 1 + 2\left( {1 + 3 + 9 + .....} \right) \cr & \sum\limits_{n = 1}^m {f\left( n \right)} = 1 + 2\frac{{\left( 1 \right)\left( {{3^{m - 1}} - 1} \right)}}{2} \cr & \sum\limits_{n = 1}^m {f\left( n \right)} = {3^{m - 1}} \cr} $$