Question

Prove by mathematical induction x^(2n)-y^(2n) is divisible by (x+y) for integers n>=1.

Answer 1

damn wtf. are you doing hsc 2014?

Answer 2

u have to prove that
\[\large (x+y)\mid(x^{2n}-y^{2n}) \]
for all integers \(n\geq1\)

Answer 3

yeah bro, 2 days to final, u up for that?

Answer 4

how helper_irwin kind sir?

Answer 5

(i) induction base: n=1
\[\large x^{2n}-y^{2n}=x^2-y^2=(x+y)(x-y) \]
which shows that \((x+y)\mid(x^2-y^2)\).

Answer 6

wait its that easy?, thought there's a need to prove n=k+1 something like that

Answer 7

no there is more steps, hes just typing give him time

Answer 8

thanks man

Answer 9

@SydneySider wait for him, next step will be it

Answer 10

thank you loser66 heyimtrash!

Answer 11

(ii) inductive step: let \(n>1\) and suppose that (inductive hypothesis)
\[\large (x+y)\mid(x^{2n}-y^{2n})\quad\text{i.e.}\quad
x^{2n}-y^{2n}=(x+y)\cdot\alpha. \]
we have to prove that (inductive thesis)
\[\large (x+y)\mid(x^{2(n+1)}-y^{2(n+1)})\quad\text{i.e.}\quad
x^{2(n+1)}-y^{2(n+1)}=(x+y)\cdot\beta. \]

Answer 12

thanks bro

Answer 13

what class u in btw?

Answer 14

let's see:
from the inductive hypothesis:
\[\large x^{2n}=y^{2n}+(x+y)\alpha \]
so
\[\large x^{2(n+1)}-y^{2(n+1)}=x^{2n}x^2-y^{2n}y^2 \]
\[\large =[y^{2n}+(x+y)\alpha]x^2-y^{2n}y^2 \]
\[\large =y^{2n}x^2+(x+y)\alpha x^2-y^{2n}y^2 \]
\[\large =y^{2n}(x^2-y^2)+(x+y)\alpha x^2 \]
\[\large =y^{2n}(x+y)(x-y)+(x+y)\alpha x^2 \]
\[\large =(x+y)[y^{2n}(x-y)+\alpha x^2] \]
which proves that \((x+y)\mid)(x^{2(n+1)}-y^{2(n+1)})\)

Answer 15

thank you so much good man

Answer 16

@helder_edwin thanks a lot. It's quite new to me ( I mean the way to prove) but it's perfect

Answer 17

i got stuck at the x^2nx^2 part :( damn. but very nice. thanks alot