If 11+103+1005+⋯n terms=10n+1+xn2+y9, then the value of x−y is
A
−19
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
19
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B19 Let Sn=11+103+1005+⋯to n terms
Now, ⇒Sn=(10+1)+(100+3)+(1000+5)+⋯{10n+(2n−1)}⇒Sn=(10+102+103+⋯+10n)+(1+3+5+⋯+(2n−1))⇒Sn=[10×(10n−1)10−1+n2(1+2n−1)]⇒Sn=109(10n−1)+n2⇒Sn=10n+1+9n2−109⇒x=9,y=−10∴x−y=19