Question

$\frac{2}{3}+\frac{5}{6}+\frac{11}{12}+\frac{23}{24}+...$ to n terms.

Answer 1

$\frac{2}{3}+\frac{5}{6}+\frac{11}{12}+\frac{23}{24}+..........$ to $n$ terms

$(1-\frac{1}{3})+(1-\frac{1}{6})+(1-\frac{1}{12})+(1-\frac{1}{24})+......$ to $n$ terms

$(1+1+......+tonterms)-(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+......tonterms)$

$n-\frac{1}{3}(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..........tonterms)$

$n-\frac{1}{3}.\frac{1(1-{\left(\frac{1}{2}\right)}^n)}{1-\frac{1}{2}}$

$=n-\frac{2}{3}.(1-\frac{1}{2})$

$\therefore \frac{2}{3}+\frac{5}{6}+\frac{11}{12}+.....$ to $n$ terms $=n-\frac{2}{3}(1-\frac{1}{2^n})$