The given sequence is

\(\frac{1}{2}\) + \(\frac{3}{4}\) + \(\frac{7}{8}\) + \(\frac{15}{16}\) + ⋯ + n terms

We can write each individual as,

\(\frac{1}{2}\) = 1 − \(\frac{1}{2}\)

\(\frac{3}{4}\) = 1 − \(\frac{1}{4}\)

\(\frac{7}{8}\) = 1 − \(\frac{1}{8}\) and … till n terms

Now writing each term in its new form, we get,

\(\big(1-\frac{1}{2}\big)+\big(1-\frac{1}{4}\big)+\big(1-\frac{1}{8}\big)+...+\big(1-\frac{1}{2_n}\big)\)

= (1 + 1 + 1 … n terms) − \(\big(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\big)\)

= n − \(\big(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\big)\) … (i)

Now, \(\big(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\big)\) is a G.P with common ratio = \(\frac{1}{2}\) so,

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\) = \(\frac{\frac{1}{2} \big(1 − (\frac{1}{2})^n\big)}{1 − \frac{1}{2}}\)

= \(\bigg[1 −\big (\frac{1}{2}\big)^n \bigg]\)

Putting this value in eq.(i), we get

n − \(\big(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\big)\)

= n − \(\bigg[1 −\big (\frac{1}{2}\big)^n \bigg]\) = n − 1 + \(\big(\frac{1}{2}\big)^n\)