#a^(n+1) - b^(n+1) = (a-b)(a^n+a^(n-1)b+....+ab^(n-1)+b^n)# How I do get this result using induction???

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If you would have some book in pdf for indicate for me about this subject, I would be very happy.

I have this result for now:
#a^(n+1)-b^(n+1) = a^na - b^nb = a^na - a^nb + a^nb - b^nb = a^n(a-b) + (a^n - b^n)b#

I have this, but I want to have to get the other equality that is :

#a^(n+1) - b^(n+1) = (a-b)(a^n+a^(n-1)b+....+ab^(n-1)+b^n)#

But I don't how I can do it this.

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Answer 1 · 2018-04-22T12:41:19

See below.

Let's call this relation we have here #p(n)#:

#p(n) : a^(n+1)-b^(n+1) = (a-b)(a^n+a^(n-1)b + ... + ab^(n-1)+b^n)#

Let's take the base case to be #p(1)#.

#p(1) : a^2 - b^2 = (a-b)(a+b)#

#nu[p(1)]=1#

Note: #nu["statement"]# is the truth value of that mathematical statement.

Let's assume that our equality is true for some integer #x#.

#p(x): a^(x+1) - b^(x+1) = (a-b)(a^x+a^(x-1)b+...+ab^(x-1)+b^x)#

The sum in the second pair of brackets is a bit messy. Let's use Sigma notation to clean it up, which I hope you're familiar with.

#p(x): a^(x+1)-b^(x+1) = (a-b)color(blue)(sum_(k=0)^x a^(x-k)b^k)#

Since we assumed it to be true, we have

#nu[p(x)] = 1#

Let's try to see if the case is true for #x+1#.

#p(x+1): a^(x+2) - b^(x+2) = #
#=(a-b)(a^(x+1)+a^(x)b+a^(x-1)b^2+...+ab^x+b^(x+1))#

The result you got is very useful right now:

#a^(x+2) - b^(x+2) = a^(x+1)(a-b)+(a^(x+1)-b^(x+1))b#

But we know that #a^(x+1)-b^(x+1)=(a-b)color(blue)(sum_(k=0)^x a^(x-k)b^k)#, so we have

#a^(x+2) - b^(x+2) = a^(x+1)(a-b) + (a-b)color(blue)(sum_(k=0)^x a^(x-k)b^k)*b#

Let's factor out the #(a-b)#.

#a^(x+2) - b^(x+2) = (a-b)(a^(x+1)+sum_(k=0)^x a^(x-k)b^(k+color(red)1))#

#b# is raised to the #k+1#-th power because we had our sum multiplied by #b#, making each power of it (of #b#) one degree more.

We should write out the sum we have in brackets:

#a^(x+2) - b^(x+2) = (a-b)(a^(x+1)+a^xb + a^(x-1)b^2+...+ab^x+b^(x+1))#

This is exactly what #p(x+1)# is stating! Basically, if #p(x)# is true, then so is #p(x+1)#. In math terms:

#:. nu[p(x)] =1 => nu[p(x+1)]=1#

And since the base case is true, we have proven #p(n)# is true for all integers #n# bigger than or equal to #1#.

In other words,

#a^(n+1)-b^(n+1) = (a-b)(a^n+a^(n-1)b + ... + ab^(n-1)+b^n),#
#forall n in ZZ, n>=1#