#number theory No.3
bleak willowBOT
umbral rock
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Sorry autocorrect
Try doing that for the general case
And looking at cancelations
knotty mural
the right side
but this proof would be without use of induction
i can try to write up one that uses induction
knotty mural
We want to prove
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1})$$
\emph{Case 1:} $x=y$
\
both sides equal 0 hence done
\
\emph{Case 2:} $x \neq y$. We will redefine our equation a bit
$$\frac{x^n-y^n}{x-y}= x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1}$$
Notice that for $n=1$ this is correct. Now the induction step $n \to n+1$:
$$\frac{x^{n+1}-y^{n+1}}{x-y}\overset{!}{=}$$
$$= x^{n}+x^{n-1}y+x^{n-2}y^2+\cdots +x^2y^{n-2}+xy^{n-1}+y^{n}$$
$$= x(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots +xy^{n-2}+y^{n-1})+y^{n}$$
By the induction hypothesus:
$$=x\left(\frac{x^n-y^n}{x-y}\right)+y^{n}$$
With some algebra we get again
$$=\frac{x^{n+1}-y^{n+1}}{x-y}$$
and hence we are done \qed
deft summitBOT
Homelama
fading bear
expanding the right side of the equation using the distributive property of multiplication over addition, we get:
fading bear
expanding the right side of the equation using the distributive property of mult...
as you can see, all the terms in the middle cancel out, leaving us with x^n−y^n on the right side of the equation, which is equal to the left side. this completes the proof.