If $$\alpha ,\beta $$  are roots of $$375{x^2} - 25x - 2 = 0$$     and $${s_n} = {\alpha ^n} + {\beta ^n}$$   then $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{s_r}} $$   is

Solution :
$$\eqalign{ & \sum\limits_{r = 1}^n {{s_r}} = \left( {\alpha + \beta } \right) + \left( {{\alpha ^2} + {\beta ^2}} \right) + ..... + \left( {{\alpha ^n} + {\beta ^n}} \right) \cr & \sum\limits_{r = 1}^n {{s_r}} = \left( {\alpha + {\alpha ^2} + ..... + {\alpha ^n}} \right) + \left( {\beta + {\beta ^2} + ..... + {\beta ^n}} \right) \cr & \therefore \,\,\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{s_r}} = \left( {\alpha + {\alpha ^2} + {\alpha ^3} + .....\,{\text{to }}\infty } \right) + \left( {\beta + {\beta ^2} + {\beta ^3} + .....\,{\text{to }}\infty } \right) \cr & \therefore \,\,\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{s_r}} = \frac{\alpha }{{1 - \alpha }} + \frac{\beta }{{1 - \beta }}\left( {{\text{clearly }}\left| \alpha \right| < 1,\left| \beta \right| < 1} \right). \cr} $$