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
4,247
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