Let x,y \epsilon r. use mathmatical induction to

Ответ:

Kin90ful

02.01.2022

Mathematics

Step-by-step explanation:

We will prove by mathematical induction that, for every natural n,

$(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}$

We will prove our base case (when n=1) to be true:

Base case:

$(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=(x-y)(x^{1}+y^{1})=x^2-y^2=x^{1+1}-y^{1+1}$

Inductive hypothesis:

Given a natural n,

$x^{n+1}-y^{n+1}=(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})$

Now, we will assume the inductive hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

Observe that, for y=0 the conclusion is clear. Then we will assume that $y\neq 0.$

$(x-y)(x^{n+1}+x^{n}y+...+xy^{n}+y^{n+1})=(x-y)y(\frac{x^{n+1}}{y}+x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+(x-y)y(x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+y(x^{n+1}-y^{n+1})=(x-y)x^{n+1}+y(x^{n+1}-y^{n+1})=x^{n+2}-yx^{n+1}+yx^{n+1}-y^{n+2}=x^{n+2}-y^{n+2}\\$

With this we have proved our statement to be true for n+1.

In conlusion, for every natural n,

$(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}$

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Let x,y \epsilon r. use mathmatical induction to prove the identity.
x^{n+1}-y^{n+1}=(x-y)(x^{n}+x^{n-1}y++xy^{n-1}+y^{n})

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Let x,y \epsilon r. use mathmatical induction to prove the identity. x^{n+1}-y^{n+1}=(x-y)(x^{n}+x^{n-1}y++xy^{n-1}+y^{n})

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Let x,y \epsilon r. use mathmatical induction to prove the identity.
x^{n+1}-y^{n+1}=(x-y)(x^{n}+x^{n-1}y++xy^{n-1}+y^{n})