An algebra problem by prashant goyal

$Given\quad that\\ \qquad \qquad 1+{ x }+{ x }^{ 2 }+{ x }^{ 3 }+\quad ...............\quad =f(x)\quad ............\quad (i)\\ Putting\quad \boxed { x=\frac { 1 }{ 2 } } \quad in\quad (i)\quad \quad \quad \quad \quad \because \quad x=\frac { 1 }{ 2 } \\ we\quad have\quad \\ \qquad \qquad f(x)=\quad 1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\quad .......................\\ Multiplying\quad both\quad sides\quad by\quad "2"\\ we\quad get\\ \qquad \qquad 2f(x)=\quad 2+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\quad .......................\\ or\quad \quad \quad \quad 2f(x)=\quad 2+f(x)\\ \qquad \qquad \boxed { \because f(x)=\quad 1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\quad ....................... } \\ or\quad 2f(x)-f(x)=2\\ or\quad f(x)=\boxed { 2 } \\ \qquad$

Clearly $f(x)$ is a GP with ratio $x$ (if $|x|<1$ ) then: $f(x)=\frac{1}{1-x}$ $f\left(\frac{1}{2}\right)=\frac{1}{1-\frac{1}{2}}$ $f\left(\frac{1}{2}\right)=2$

$f\left( x \right) = 1 + x + x^2 + x^3 + ... = \sum_{n=0}^\infty{x^n}$ So $f(x)$ is a geometric progression and its value is given by: $\sum_{n=0}^\infty{x^n}=\cfrac{1}{1-x}$ Since $x=\frac{1}{2}$ , therefore: $f(\frac{1}{2}) = \cfrac{1}{1-\frac{1}{2}} = \boxed{2}$

Just use the sum of infinite geometric sequence formula for the x's. Then add 1 to the result.