$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$$ is approximately equal to ___________ .

Expression = $$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$$

= $$(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6})-(\frac{1}{8})$$

The first term above is a geometric progression with first term, $$a=1$$ and common ratio, $$r=\frac{1}{2}$$

Number of terms, $$n=7$$

Sum of $$n$$ terms of G.P. (when $$r<1$$) = $$\frac{a(1-r^n)}{1-r}$$

= $$[\frac{1(1-\frac{1}{2}^7)}{1-\frac{1}{2}}]-(\frac{1}{8})$$

= $$[2\times(1-\frac{1}{128})]-(\frac{1}{8})$$

= $$\frac{127}{64}-\frac{8}{64}$$

= $$\frac{119}{64}=1.85\approx2$$

=> Ans - (B)