\What is the sum of the following series?
$$2.12 + 1.0012 + 0.000012 + 0.00000012 + 0.0000000012 +....\infty$$

We can rewrite the series as 
$$2+0.12+1+0.0012+0.000012+\ ...$$ and so on
=$$2+1+0.12+0.0012+0.000012+0.00000012+.....$$
=$$3+\frac{12}{100}+\frac{12}{100^2}+\frac{12}{100^3}+\frac{12}{100^4}+.....$$
=$$3+12\left(\frac{1}{100}+\frac{1}{100^2}+\frac{1}{100^3}+.....\right)$$
Using the formula for the sum of an infinite series we get, 
$$3+\frac{12\left(\frac{1}{100}\right)}{\left(1-\frac{1}{100}\right)}$$
=$$3+\frac{12}{99}$$
$$\frac{309}{99}=\frac{103}{33}$$

And hence, Option A is the correct answer.