The question is : Prove by mathematical induction that :
$ a^{2n-1} + b^{2n+1} $
is divisible by $a+b$
I've done a lot of stuff but can't put them down in tex properly. Thanks.
The question is : Prove by mathematical induction that :
$ a^{2n-1} + b^{2n+1} $
is divisible by $a+b$
I've done a lot of stuff but can't put them down in tex properly. Thanks.
Write it as
$a^{2n+1}+b^{2n+1}=(a^2+b^2)(a^{2n-1}+b^{2n-1})-a^2b^2(a^{2n-3}+b^{2n-3})$ and then use induction on $n$.
A mild variant of the answer by dmm: $a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}+b^{2n})-ab(a^{2n-1}+b^{2n-1}).$
Then directly from $a+b$ divides $a^{2k-1}+b^{2k-1}$ we can conclude that $a+b$ divides $a^{2k+1}+b^{2k+1}$. This may feel more comfortably familiar.
Added: In answer to a comment, the OP has indicated that $a^{2n-1}+b^{2n+1}$, which looked like an obvious typo, is not. If so, the conjectured result is false. Let $a=2$, $b=3$, $n=1$.
By induction $\rm\:(-b)^{2n-1}\! = - b^{2n-1}$ so $\rm\:mod\ a\!+\!b\!:\ a\equiv-b\:\Rightarrow\:a^{2n-1}\!\equiv(-b)^{2n-1}\!\equiv {-}b^{2n-1}$
Hint:
$\forall\,n\in\Bbb N\,\,\,,\,\,a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+....-ab^{2n-1}+b^{2n})$
Formally, the above still requires a little proof by induction.