If the sum to infinity of the series, $$1 + 4x + 7{x^2} + 10{x^3} + .....,\,{\text{is}}\frac{{35}}{{16}},$$       where $$\left| x \right| < 1,$$  then $$x$$ equals to

Solution :
$$\eqalign{ & S = 1 + 4x + 7{x^2} + 10{x^3} + ..... \cr & x.S = x + 4{x^2} + 7{x^3} + ..... \cr} $$
Subtract
$$\eqalign{ & S\left( {1 - x} \right) = 1 + 3x + 3{x^2} + 3{x^3} + ..... \cr & S\left( {1 - x} \right) = 1 + 3x\left( {\frac{1}{{1 - x}}} \right) \cr & \because \left| x \right| < 1 \cr & S = \frac{{1 + 2x}}{{{{\left( {1 - x} \right)}^2}}} \cr & {\text{Given : }}\frac{{1 + 2x}}{{{{\left( {1 - x} \right)}^2}}} = \frac{{35}}{{16}} \cr & \Rightarrow 16 + 32x = 35 + 35{x^2} - 70x \cr & \Rightarrow x = \frac{1}{5},\frac{{19}}{7} \cr & {\text{But, }}\left| x \right| < 1 \cr & \therefore x = \frac{1}{5} \cr} $$