  Question

# Prove that x2n−1+y2n−1 is divisible by x+y for all n∈N.

Solution

## We shall prove the given statement by the principle of mathematical induction.Let the given statement be  P(n) : x2n−14+y2n−1 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., x2k−1+y2k−1 is divisible by (x+y) x2k−1+y2k+1=(x+y)λ      ⇒x2k−1=(x+y)λ−y2k+1 .......(1) Step III: We shall prove the given statement true for n=k+1i.e., It is sufficient to prove thatx2k+1+y2k+1=(x+y)μ  for some μ Now, Consider the R.H.S of above equation=x2k+1+y2k+1=x2k−1.x2+y2k−1.y2=[[(x+y)λ−y2k−1].x2+y2k−1.y2] [ from equation(i)] =(x+y)λx2−y2k−1.x2+y2k−1.y2 =(x+y)λx2−y2k−1(x2−y2) =(x+y)λx2−y2k−1(x+y)(x−y) =(x+y)[λx2−y2k−1(x−y)] =(x+y)μ    where μ=λx2−y2k−1(x−y)]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.  Suggest corrections  Similar questions
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