Question

$\frac{3}{2}+\frac{9}{4}+\frac{25}{8}+\frac{65}{16}+...$ to n terms.

Answer 1

$\frac{3}{2}+\frac{9}{4}+\frac{25}{8}+\frac{65}{16}+....+n$ terms

$=(1+\frac{1}{2})+(2+\frac{1}{4})+(3+\frac{1}{8})+.......+n$ terms

$=(1+2+3+.....n)+(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{2^n})$

$=\frac{n(n+1)}{2}+\frac{\frac{1}{2}(1-{\left(\frac{1}{2}\right)}^n)}{1-\frac{1}{2}}$

$=\frac{n(n+1)}{2}+1-{\left(\frac{1}{2}\right)}^n$