qq -induction step for $1+3+5+...+(2k-1)=k^2$

Asked Sep 24 '17 at 21:16

Active Sep 24 '17 at 22:31

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I'm trying to prove that $ 1+3+5+...+(2n−1)=n^2$ where $ n>= 1$

but I'm a little confused how my induction step should look like for $S(k)= 1+3+5+...+(2k-1) = k^2$

should it be $S(k+1) = 1+3+5+...+(2k-1)+((2k+1)-1) = (k+1)^2$ or $S(k+1)=1+3+5+...+(2k-1)+(2k-1+1) = k^2+1$

$(k+1)^2 $ makes more sense, but it does not hold for $S(k+1)=k^2+((2k+1)-1)= (k+1)^2$ so therefore would $k^2 +1 $ is correct answer?

edited Sep 24 '17 at 22:31

kimi Tanaka

asked Sep 24 '17 at 21:16

Ward

2

$1+3+\dots+(2k-1)+(2(k+1)-1)$ should equal $(k+1)^2$. Note that if $f(x)=2x-1$ then $f(k+1)=2(k+1)-1$ and if $g(x)=x^2$ then $g(k+1)=(k+1)^2$ – JMoravitz Sep 24 '17 at 21:18
@JMoravitz awesome, thanks! now $S(k+1)=k^2+(2(k+1)-1)=(k+1)^2 $ holds. Is this enough to prove that $n^2$ is true or am I still missing something? – Ward Sep 24 '17 at 21:30
You are missing the base step: $S(1)=~?$ – Simply Beautiful Art Sep 24 '17 at 21:40
@Simply Beautiful Art so my base step would be $S(1)=1=1^2=1$? Is there any further verification I need to do with induction step? Thanks for reply :) – Ward Sep 24 '17 at 21:42
Nope, that's about it. – Simply Beautiful Art Sep 24 '17 at 21:42
@Simply Beautiful Art, cool. thank you! Always helps to verify with others! – Ward Sep 24 '17 at 21:43

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