If $$a = \min \left\{ {{x^2} + 4x + 5,\,x\, \in \,R} \right\}$$      and $$b = \mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \,2\theta }}{{{\theta ^2}}},$$    then the value of $$\sum\limits_{r = 0}^n {{a^r} \cdot {b^{n - r}}} $$   is :

Solution :
$$\eqalign{ & {x^2} + 4x + 5 = {\left( {x + 2} \right)^2} + 1 \geqslant 1 \cr & {\text{So, }}a = 1 \cr & b = \mathop {\lim }\limits_{\theta \to 0} \frac{{2\,{{\sin }^2}\theta }}{{{\theta ^2}}} = 2 \cr & \sum\limits_{r = 0}^n {{a^r} \cdot {b^{n - r}}} = {b^n} + a{b^{n - 1}} + {a^2}{b^{n - 2}} + ..... + {a^n} \cr & = \frac{{{b^n}\left[ {1 - {{\left( {\frac{a}{b}} \right)}^{n + 1}}} \right]}}{{1 - \frac{a}{b}}} \cr & = \frac{{{2^n}\left[ {1 - {{\left( {\frac{1}{2}} \right)}^{n + 1}}} \right]}}{{1 - \frac{1}{2}}} \cr & = \frac{{{2^{n + 1}}\left( {{2^{n + 1}} - 1} \right)}}{{{2^{n + 1}}}} \cr & = \left( {{2^{n + 1}} - 1} \right) \cr} $$