凍 凍鴛鴦 Member Joined Nov 10, 2004 Messages 53 Gender Undisclosed HSC 2006 Oct 9, 2005 #1 Hi again, Can someone please help with this? Prove by induction that: (1+x)^n >= nx+1 n and x are positive integers Thanks in advance!
Hi again, Can someone please help with this? Prove by induction that: (1+x)^n >= nx+1 n and x are positive integers Thanks in advance!
gman03 Active Member Joined Feb 7, 2004 Messages 1,283 Gender Male HSC 2003 Oct 9, 2005 #2 Induction over n true for n = 1 Assume true for n = k that is, (1+x)^k >= kx+1 Then for n = k + 1 LHS = (1+x)^(k+1) = (1+x) * (1+x)^k >= (1+x) * (kx+1) (by induction) = kx^2 + (k+1)x + 1 > 0 + (k+1)x + 1 (k is positive) = (k+1)x + 1 So it is true for n = k + 1 Hence by induction the statement is true for positive n, x
Induction over n true for n = 1 Assume true for n = k that is, (1+x)^k >= kx+1 Then for n = k + 1 LHS = (1+x)^(k+1) = (1+x) * (1+x)^k >= (1+x) * (kx+1) (by induction) = kx^2 + (k+1)x + 1 > 0 + (k+1)x + 1 (k is positive) = (k+1)x + 1 So it is true for n = k + 1 Hence by induction the statement is true for positive n, x
凍 凍鴛鴦 Member Joined Nov 10, 2004 Messages 53 Gender Undisclosed HSC 2006 Oct 9, 2005 #3 Cheers! Thanks heaps =D
davidw89 Member Joined Feb 7, 2006 Messages 211 Gender Male HSC 2007 Dec 14, 2006 #4 gman03 said: Induction over n > 0 + (k+1)x + 1 (k is positive) = (k+1)x + 1 Click to expand... Why did u replace x^2 with 0?
gman03 said: Induction over n > 0 + (k+1)x + 1 (k is positive) = (k+1)x + 1 Click to expand... Why did u replace x^2 with 0?
Affinity Active Member Joined Jun 9, 2003 Messages 2,046 Location Oslo Gender Undisclosed HSC 2003 Dec 14, 2006 #5 Coz it suits him
J jyu Member Joined Nov 14, 2005 Messages 622 Gender Male HSC 2006 Dec 14, 2006 #6 davidw89 said: Why did u replace x^2 with 0? Click to expand... Because x^2 > 0 for positive x.