Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
659 views
in Algebra by (26.9k points)
closed by

Find the sum of first n terms of the series

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

1 Answer

+1 vote
by (30.0k points)
selected by
 
Best answer

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 + \(\big(\frac{1}{2}\big)^n\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...