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WBJEE · Maths · Mathematical Induction

Let \(a, b, c\) and \(d\) be any four real numbers. Then. \(a^{n}+b^{n}=c^{n}+d^{n}\) holds for any natural number \(n\). if

  1. A \(a+b=c+d\)
  2. B \(a-b=c-d\)
  3. C \(a+b=c+d \cdot d^{2}+b^{2}=c^{2}+d^{2}\)
  4. D \(a-b=c-d \cdot d^{2}-b^{2}=c^{2}-d^{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(a-b=c-d \cdot d^{2}-b^{2}=c^{2}-d^{2}\)

Step-by-step Solution

Detailed explanation

From option (d), we have \(\begin{aligned} a-b &=c-d \\ a^{2}-b^{2} &=c^{2}-d^{2} \end{aligned}\) Consider, \(a^{2}-b^{2}=c^{2}-d^{2}\) \(\Rightarrow(a+b)(a-b)=(c-d)(c+d)\) \(\Rightarrow \quad a+b=c+d\) On adding Eqs we get \(2 a=2 c \Rightarrow a=c\) \(\Rightarrow \quad b=d\)…