wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let Tn=(n2+1)n! and Sn=T1+T2+T3++Tn. If T10S10=ab, where a and b are relatively prime natural numbers, then the value of (ba) is

A
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
9
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
14
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
13
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 9
Tn=(n2+1)n!
=[(n+1)n(n1)]n!
Tn=n(n+1)!(n1)n!

T1=1×2!0T2=2×3!1×2!T3=3×4!2×3!T4=4×5!3×4! Tn=n(n+1)!(n1)n!

Sn=T1+T2+T3++Tn
Sn=n(n+1)!
Now, T10S10=101×10!10×11!=101110
ba=110101=9

flag
Suggest Corrections
thumbs-up
11
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon