C-13-Q127mediumsingle_mcqThe sum of the first 444 terms of 1+13+19+127+⋯1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \cdots1+31+91+271+⋯ is —a4027\dfrac{40}{27}2740b139\dfrac{13}{9}913c12181\dfrac{121}{81}81121d127\dfrac{1}{27}271ব্যাখ্যাThis is geometric with a=1, r=13a=1,\ r=\tfrac13a=1, r=31, so S4=1−(1/3)41−1/3=80/812/3=4027S_4=\dfrac{1-(1/3)^4}{1-1/3}=\dfrac{80/81}{2/3}=\dfrac{40}{27}S4=1−1/31−(1/3)4=2/380/81=2740.