The sum of the series S=1+2(1011)+3(1011)2+⋯upto∞ is equal to
A
121
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
120
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
111
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
110
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A121 Let 1011=x.
Then S=1+2x+3x2+⋯upto∞⋯(1) xS=x+2x2+⋯upto∞⋯(2)
Subtracting (1) and (2), S(1−x)=1+x+x2+⋯upto∞ ⇒S(1−x)=11−x[∵Common ratio=x<1] ⇒S=1(1−x)2
Hence, the required sum is, 1(1−1011)2=121