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

If P=n(n212)(n222)(n232)...(n2r2),n>r,nϵ N, then P is divisible by

A
(2r+2)!
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(2r1)!
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(2r+1)!
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct options are
B (2r1)!
D (2r+1)!
P=n(n21)(n222)...(n2r2)
P=n(n1)(n+1)(n2)(n+2)...(nr)(n+r)
Rearranging the terms we get,
P=(nr)(nr+1)...(n+r1)(n+r)
This is a product of 2r+1 consecutive terms. Hence it must be divisible by (2r+1)!
Hence, options B and C are correct.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Polynomial and Its General Form
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon