Question

Prove by induction
${{x}^{n}}-{{y}^{n}}$ is divisible by (x+y) when n is even.

Answer 1

HINT: Mathematical induction is a mathematical proof technique. A proof by induction consists of two cases. The first, the base case (or basis), proves the statement for n = 0 without assuming any knowledge of other cases. The second case, the induction step, proves that if the statement holds for any given case n = k, then it must also hold for the next case n = k + 1. These two steps establish that the statement holds for every natural number n.[3] The base case does not necessarily begin with n = 0, but often with n = 1, and possibly with any fixed natural number n = N, establishing the truth of the statement for all natural numbers n ≥ N.

Complete step-by-step answer:
As mentioned in the question, we have to prove the statement through mathematical induction.
As mentioned in the hint,
Let P(n) $={{x}^{n}}-{{y}^{n}}$ is divisible by (x+y)
Case 1:-
For n=2(being the smallest even number), we get
P(2)

& ={{x}^{2}}-{{y}^{2}} \\\ & =(x+y)(x-y) \\\ \end{aligned}$$ Hence, P(n) is true for n=2. Case 2:- For n=2k(an even number) Let P(k) ‘ $$={{x}^{2k}}-{{y}^{2k}}$$ is divisible by x+y ‘ be true. Hence, we can write $$\begin{aligned} & {{x}^{2k}}-{{y}^{2k}}=n(x+y) \\\ & {{x}^{2k}}=\ n(x+y)+\ {{y}^{2k}}\ \ \ ...(a) \\\ \end{aligned}$$ Case 3:- Now, we have to prove P(k+1) to be true. Hence, P(k+1) $$\begin{aligned} & ={{x}^{2(k+1)}}-{{y}^{2(k+1)}} \\\ & ={{x}^{2k+2}}-{{y}^{2k+2}} \\\ & ={{x}^{2k}}\cdot {{x}^{2}}-{{y}^{2k}}.{{y}^{2}} \\\ \end{aligned}$$ Now, using equation (a), we get $$\begin{aligned} & ={{x}^{2k}}\cdot {{x}^{2}}-{{y}^{2k}}.{{y}^{2}} \\\ & =\left( n(x+y)+\ {{y}^{2k}} \right)\cdot {{x}^{2}}-{{y}^{2k}}.{{y}^{2}} \\\ & =n(x+y)\cdot {{x}^{2}}+{{y}^{2k}}({{x}^{2}}-{{y}^{2}}) \\\ & =n(x+y)\cdot {{x}^{2}}+{{y}^{2k}}(x-y)(x+y) \\\ & =(x+y)\left( n{{x}^{2}}+{{y}^{2k}}(x-y) \right)\ \ \ \ \ ...(b) \\\ \end{aligned}$$ As equation (b) is divisible by (x+y), therefore, P(k+1) is true . Hence, the statement is proved by use of mathematical induction. $$$$ NOTE: A student can make an error if they don’t know about the process of mathematical induction which is as follows A proof by induction consists of two cases. The first, the base case (or basis), proves the statement for n = 0 without assuming any knowledge of other cases. The second case, the induction step, proves that if the statement holds for any given case n = k, then it must also hold for the next case n = k + 1. These two steps establish that the statement holds for every natural number n.[3] The base case does not necessarily begin with n = 0, but often with n = 1, and possibly with any fixed natural number n = N, establishing the truth of the statement for all natural numbers n ≥ N.