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Use Mathematical Induction to prove: a+b is a factor of a^2n+1 + b^2n+1

 Jan 5, 2015

Best Answer 

 #2
avatar+105411 
+10

First, let us show that it is true for n=1

So....a^(2(1) + 1) + b^(2(1) + 1)  =  a^3 + b^3  =  (a +b)(a^2 -ab + b^2)...and (a + b) is a factor

Now, let us assume it is true for  k=1, that is, a^(2(k) + 1) + b^(2(k) + 1) is true

Now, let's prove that it is true for k + 1

So we have

a^(2(k+1)+ 1) + b^(2(k+1) + 1) =

a^(2k + 3) + b^(2k + 3) =

a^3*a^(2k)  + b^3*b^(2k)......and using   a^3  =  (a +b)(a^2 -ab + b^2) - b^3 , we can write

[(a+b)(a^2 -ab + b^2) - b^3]a^(2k)  + b^3*b^(2k)  =

(a+b)(a^2 -ab + b^2)a^(2k) - b^3[ a^(2k) -  b^(2k)]

Notice that the last term is a difference of two even powers.........and it can be shown that the difference of two even powers can be factored with (a + b) as a factor thusly...

(a^(2n) - b^(2n)) = (a + b)[a^(2n-1) - a^(2n-2)b + a^(2n-3)b^2 - a^(2n-4)b^3 +.....+ ab^(2n-2) - b^(2n-1) ]

Thus, (a+b) is a factor of both terms, so (a +b) is a factor of a^(2k + 3) + b^(2k + 3)

 

 Jan 5, 2015
 #1
avatar+106046 
0

Could  the person who posted this please insert brackets to clarify the meaning?

 Jan 5, 2015
 #2
avatar+105411 
+10
Best Answer

First, let us show that it is true for n=1

So....a^(2(1) + 1) + b^(2(1) + 1)  =  a^3 + b^3  =  (a +b)(a^2 -ab + b^2)...and (a + b) is a factor

Now, let us assume it is true for  k=1, that is, a^(2(k) + 1) + b^(2(k) + 1) is true

Now, let's prove that it is true for k + 1

So we have

a^(2(k+1)+ 1) + b^(2(k+1) + 1) =

a^(2k + 3) + b^(2k + 3) =

a^3*a^(2k)  + b^3*b^(2k)......and using   a^3  =  (a +b)(a^2 -ab + b^2) - b^3 , we can write

[(a+b)(a^2 -ab + b^2) - b^3]a^(2k)  + b^3*b^(2k)  =

(a+b)(a^2 -ab + b^2)a^(2k) - b^3[ a^(2k) -  b^(2k)]

Notice that the last term is a difference of two even powers.........and it can be shown that the difference of two even powers can be factored with (a + b) as a factor thusly...

(a^(2n) - b^(2n)) = (a + b)[a^(2n-1) - a^(2n-2)b + a^(2n-3)b^2 - a^(2n-4)b^3 +.....+ ab^(2n-2) - b^(2n-1) ]

Thus, (a+b) is a factor of both terms, so (a +b) is a factor of a^(2k + 3) + b^(2k + 3)

 

CPhill Jan 5, 2015
 #3
avatar+106046 
0

This is one I want to look at properly.  If only there were  infinite hours in each day :)

THEN we could all procrastinate forever and it would not make one iota of difference :)

 Jan 5, 2015

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