The correct option is A 32
Let S=∞∑i=12i−12i+1
⇒S=14+38+516+732+…
⇒4S=(1+32+54+78+…)
Now clearly, the series is an AGP with a=1, d=2, r=12.
Hence, using the formula for the sum of infinite terms of AGP, we get
S=14⎛⎜
⎜
⎜
⎜
⎜⎝11−12+2⋅12(1−12)2⎞⎟
⎟
⎟
⎟
⎟⎠
=14(2+4)=32