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Ilya [14]
3 years ago
5

Let x,y \epsilon R. Use mathmatical induction to prove the identity.

Mathematics
1 answer:
mariarad [96]3 years ago
8 0

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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Step-by-step explanation:

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