Question

$x^{\frac{1}{2}}.x^{\frac{1}{4}}.x^{\frac{1}{8}}.x^{\frac{1}{16}}.....$ to $\infty$

• A. $0$
• B. $1$
• C. $x$
• D. $\infty$

Answer 1

The correct option is C $x$

$x^{1/2}.x^{1/4}......\infty =x^{(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16})}$

$=x(1/2+1/2^2+1/2^3+1/2^4+.......\infty )$

$=x^{\frac{1}{2}\left(\frac{1-\left(1/2\right)\infty }{1-1/2}\right)}$

$=x^1$

$=x$