• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Induction Question (1 Viewer)

HoldingOn

Active Member
Joined
Dec 18, 2016
Messages
318
Location
The Cosmos
Gender
Male
HSC
2018
It's pretty straightforward but I can't seem to get it out. Thanks in advance.

Prove by induction that for all integers n≥1: (n+1)(n+2)...(2n-1)2n= 2^n [1x3x...x(2n-1)]
 

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
It's pretty straightforward but I can't seem to get it out. Thanks in advance.

Prove by induction that for all integers n≥1: (n+1)(n+2)...(2n-1)2n= 2^n [1x3x...x(2n-1)]
For n=1: LHS (1 factor) = (1+1) = 2 = RHS

Assume true for n = k >= 1

i.e. (k+1)(k+2) . . . (2k) = 2k(1 x 3 x . . . . (2k-1)]

For n = k+1:

LHS = (k+1+1)(k+1+2)(k+1+3) . . . (2k+1)(2k+2)

= (k+2)(k+3)(k+4) . . . (2k)(2K+1) x 2 x (k+1)

= 2 x {(k+1)(k+2) . . . (2k)} x (2[k+1]-1)

= 2 x {2k (1 x 3 x . . . (2k-1)] } x (2[k+1]-1)

= 2k+1 (1 x 3 x 5 x . . . x (2[k+1] - 1)

.: if formula is true for n = k then it is also true n = k+1

.: by the principle of mathematical induction . . .
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top