n^2 - n is even for any n >= 1 So far i have the base case: n=1 1^2-1 = 0 and 2^2-2=2 which are both even. then prove for n = k so k^2 - 2 for k>=1 then n= (k+1)^2 - (k+1).... having problems with further steps.
Follow Math Help Forum on Facebook and Google+
[LaTeX ERROR: Convert failed] .
Why do you have to use induction? If is odd and , then where and which is clearly even. If is even and , then where and which is clearly even. Therefore is even for any .
Originally Posted by zaj091000 n^2 - n is even for any n >= 1 So far i have the base case: n=1 1^2-1 = 0 and 2^2-2=2 which are both even. then prove for n = k so k^2 - 2 for k>=1 then n= (k+1)^2 - (k+1).... having problems with further steps. is even for You've completed the base case... Now you must show whether or not... causes P(k) is even P(k+1) is even if P(k) is true Proof Since 2k is even... If P(k) is true, then P(k+1) is definately true.
View Tag Cloud