Question

If r is a fixed positive integer, then if we use induction we can find that the expression (r+1)(r+2)(r+3)....(r+n) is always divisible by

A
n+1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
nn
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
n!
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
n2+1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C n!Let P(n): (r+1)(r+2)(r+3)....(r+n)=n!.k, Where k is an integer. When n=1,r+1=1!.(r+1) ∴P(1) is true. let P(m) be true,i.e., (r+1)(r+2)(r+3)....(r+m)=m!.k ....(1) Now, (r+1)(r+2)(r+3)....(r+m)(r+m+1) = r(r+1)(r+2)(r+3)....(r+m)+(m+1)(r+1)(r+2)(r+3)...(r+m) =(r+m)!(r−1)!+(m+1)(m!)k,using(1) =(m+1)!.(r+m)!(r−1)!(m+1)!+(m+1)!.k =(m+1)!.(r+mCr−1+k) =(m+1)!(integer+k) P(m+1) is true ⇒P(m+1)is true. P(n) is true for all nϵN.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Mathematical Induction
MATHEMATICS
Watch in App
Join BYJU'S Learning Program