Question

prove that Xn - an is divisible by X+a only when X is even​

Answer 1

Dear Student,

★ Answer:

It is Answered.

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★ Correction :

➠ Ques : Prove that $X^{n}$ - $a^{n}$  is divisible by  X + a only when X is even​

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★ Step-by-step explanation:

➼ The statement to be proved is:

P ( n ) :  $X^{n} - a^{n}$  is divisible by X + a when X is even​

➼ Step 1: Verify that the statement is true for the smallest value of X ,

here X =2

⇒ P ( 2 ) : $X^{2} - a^{2}$ is divisible by X + a

⇒ P (2) : (X + a) (X−a ) is divisible by X + a , which is true.

Therefore P(2) is true.

➼ Step 2 :  Assume that the statement is true for k

⇒ Let us assume that P (k) :  $X^{k} - a^{k}$  is divisible by X + a  where k is even.

➼ Step 3: Verify that the statement is true for the next possible integer, here for X =k+2

⇒ $X ^{k+2} - a^{k+2}$  =  $X^{k+2} - X^{2} a^{k} + X^{2} a^{k} - a^{k+2}$

= $X^{2} ( X^{k} -a^{k} ) + a^{k} (X^{2} - a^{2} )$

➥ Since,

⇒ $( X^{k} - a^k} )$  and $(X^{2} - a^{2} )$  are both divisible by (X + a ), the complete equality is divisible by X + a .

➥ Therefore,  

⇒ P ( k + 2)  :  $X ^{k+2} - a^{k+2}$ is divisible by X + a  where k+2 is even.

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Answer :

Therefore by principle of mathematical induction, P(n) is true.

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