how can I use induction to prove b^n - a^n is divisible by (b-a) when n is a natural number?

Question

b>a>0

Answer 1

Hipothesis

For n=1 : P (1)= b-a is divisible by

(b-a)--> (b-a)/b-a)=1 true.

For n=k (b^k-a^k)/(b-a) it is true -->

(b^k-a^k)/(b-a)=m-->

(b^k-a^k)=m*(b-a)--->

b^k=a^k+m ( b-a).



thesis

For n=k+1

[b^(k+1)-a^(k+1)]/(b-a)

Dimostrazione:

[b^(k+1)-a^(k+1)]=

=[b*b^k-a*a^k)=

=[b*(a^k+m (b-a)]- a*a^k]=

=[ b*a^k+bm (b-a) -a*a^k]=

=[(b*a^k-a*a^k)+bm (b-a)]=

= [a^k (b-a)+bm (b-a)]=

=(b-a) ( a^k+bm)

-->[ a^(k+1)-b^(k+1) ] is multiply of

( b-a) --> is c divisible for (b-a).

Answer 2

Assume that b^n - a^n is divisible by (b - a)

For n = 1: (b -a) is divisible by (b - a) i.e. itself

For n = 2: (b^2 -a^2) is divisible by (b - a) as (b^2 -a^2) = (b-a)(b+a)



Let us check if this holds good for n = n+1:

b^(n+1) - a^(n+1)

= b*b^n - a*a^n

= (b-a+a)*b^n - a*a^n

= (b-a)*b^n +a*b^n - a*a^n

= (b-a)*b^n + a(b^n - a^n)

In the above expression (b-a)*b^n is divisible by (b-a) and

a(b^n - a^n) is divisible by (b-a)



So by induction it is proven that b^n - a^n is divisible by (b - a)

Answer 3

for n =1 it is true so let it be true for n = k



P(k) = (b^k - a^k)



so b^k - a^k = (b-a)(P(k)) ----------------------(1)



so it should be true for n = k+1



P(k+1) = b^(k+1) - a^(k+1)



note the identity b^(k+1) - a^(k+1) = (1/2)*[(b^k + a^k)(b -a) + (b^k - a^k)*(b+a)]



so P(k+1) = (1/2)*[(b^k + a^k)(b -a) + (b^k - a^k)*(b+a)]



P(k+1) = (1/2)*[(b^k + a^k)(b -a) + (b-a)*P(k)*(b+a)] --------------------[ From equation 1]



take b - a common.



P(k+1) = b-a*(1/2)*[b^k + a^k + P(k)*(b+a)]



so P(k+1) is divisible by b-a



Hence it is true for any n belongs to natural number.

Answer 4

Let | be "divides".



n=1, b^1-a^1=b-a (b-a)|(b-a) true

assume true for n => ((b-a)|(b^n-a^n)



For n+1, b^(n+1)-a^(n+1)=

b*b^n - a*a^n=

b*b^n - a*b^n + a*b^n - a*a^n=

b^n(b-a) +a(b^n-a^n)

(b-a) | both terms. The second by assumption. Therefore (b-a) | the desired form for n+1

QED