Prove that x2n−1+y2n−1 is divisible by x+y for all n∈N.
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
=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)
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
=[[(x+y)λ−y2k−1].x2+y2k−1.y2] [ from equation(i)]
=(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.