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
- A \(a+b=c+d\)
- B \(a-b=c-d\)
- C \(a+b=c+d \cdot d^{2}+b^{2}=c^{2}+d^{2}\)
- D \(a-b=c-d \cdot d^{2}-b^{2}=c^{2}-d^{2}\)
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 \(2 a=2 c \Rightarrow a=c\) \(\Rightarrow \quad b=d\)…
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