induction qs (1 Viewer)

xXnukerrrXx · Mar 20, 2012

if n is even prove n^2+2n is divisble by 8 thanks

Carrotsticks · Mar 20, 2012

n=2 etc etc.. (remember to use n=2 to begin with because we are testing even numbers).

Hypothesis:

$k^2+2k=8m \qquad $for some $ m \in \mathbb{Z} $ and $ k $ is even.$$

Inductive step (k --> k+2)

$\\ (k+2)^2+2(k+2) \\\\ = k^2 + 4k + 4 + 2k + 4 \\\\ = k^2 + 2k + 4k+8 \\\\ = 8m + 4k+8 \qquad $ but k is an even integer ie: 4k = 8j for some integer j$ \\\\ = 8(m+j+1) \qquad $, completing the inductive step.$$

Make a Difference – Donate to Bored of Studies!

induction qs (1 Viewer)

xXnukerrrXx

Member

Carrotsticks

Retired

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