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Question

Prove that x2n1+y2n1 is divisible by x+y for all nN


Solution

We shall prove the given statement by the principle of mathematical induction.
Let the given statement be  P(n) : x2n14+y2n1 is divisible by x+y

Step I: For n=1

=x2(1)1+y2(1)1

=x+y is divisible by itself  x+y

P(n) is true for n=1

StepII: Assume P(n) is true for n=k,

i.e., x2k1+y2k1 is divisible by (x+y)

x2k1+y2k+1=(x+y)λ   
  x2k1=(x+y)λy2k+1 .......(1)

Step III: We shall prove the given statement true for n=k+1
i.e., It is sufficient to prove that
x2k+1+y2k+1=(x+y)μ  for some μ

Now, Consider the R.H.S of above equation
=x2k+1+y2k+1
=x2k1.x2+y2k1.y2
=[[(x+y)λy2k1].x2+y2k1.y2] [ from equation(i)]

=(x+y)λx2y2k1.x2+y2k1.y2

=(x+y)λx2y2k1(x2y2)

=(x+y)λx2y2k1(x+y)(xy)

=(x+y)[λx2y2k1(xy)]

=(x+y)μ    where μ=λx2y2k1(xy)]

Thus, x2k+1+y2k+1 is divisible by x+y

P(n) is true for n=k+1

Hence, P(n) is true for all n  N by principle of mathematical induction.

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