What is the sum of the first #n# terms of the series #1/2+3/4+5/8+...# ?

1 Answer
Feb 15, 2018

Answer:

#S_n=3-(2n+3)/2^n#

Explanation:

Let #S_n=1/2+3/4+5/8+.................."n terms"#

What is #n^(th)# term of the series? Well numerators are in arithmatic sequence and numerator of #n^(th)# term is #2n-1# and denominator is in geometric series with #n^(th)# term being #2^n#. Hence #n^(th)# term of given series is #(2n-1)/2^n# and series is

#S_n=1/2+3/4+5/8+7/16+..................+(2n-1)/2^n# and then

#1/2S_n=1/4+3/8+5/16+..................+(2n-1)/2^(n+1)#

Subtracting latter from former, we get

#S_n-1/2S_n=1/2+2/4+2/8+....+2/2^n-(2n-1)/2^(n+1)#

or #1/2S_n=1/2+2/4[1+1/2+....+1/2^(n-2)]-(2n-1)/2^(n+1)#

= #1/2+1/2(1-1/2^(n-1))/(1-1/2)-(2n-1)/2^(n+1)#

= #1/2+1-1/2^(n-1)-(2n-1)/2^(n+1)#

= #3/2-(4+2n-1)/2^(n+1)#

= #3/2-(2n+3)/2^(n+1)#

and #S_n=3-(2n+3)/2^n#