Using induction, prove that

\[\left(1 - \dfrac {2}{4}\right)\left(1 - \dfrac {2}{5}\right)\ldots\left(1 - \dfrac{2}{n}\right) = \dfrac{6}{n(n-1)}\]

for \(n \geq 4\).

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TopNewestPROOF BY INDUCTIONLet \( T(n) \) be the proposition that \( \forall n \geq 4 \), we have \[\left(1 - \dfrac {2}{4}\right)\left(1 - \dfrac {2}{5}\right)\ldots\left(1 - \dfrac{2}{n}\right) = \dfrac{6}{n(n-1)}\]

Base Case:- \( T(4) \) is true because \( (1 - \dfrac{2}{4}) = \dfrac{6}{4(4-1)} \)Inductive Step:- Let \( T(k) \) be true for some \( k \geq 4 \), that is -\[\left(1 - \dfrac {2}{4}\right)\left(1 - \dfrac {2}{5}\right)\ldots\left(1 - \dfrac{2}{k}\right) = \dfrac{6}{k(k-1)}\]

Multiplying both sides of the above equation by \( (1 - \dfrac{2}{k+1}) \), we get that

LHS of \( T(k+1) = ( \dfrac{6}{k(k-1)}) \times (1 - \dfrac{2}{k+1}) = \dfrac{6}{k(k+1)} = \) RHS of \( T(k+1) \).

Hence, as \( T(k) \) true \( \Rightarrow T(k+1) \) true, our induction step is now complete.

Therefore, by First Principle of Mathematical Induction, we now conclude that \( T(n) \) is true \( \forall n \geq 4 \).

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Yeah. I learnt Induction today! What a coincidence @Sharky Kesa. And, Flawless proof Karthik Venkata . I solved it using the same way. And also Got it!

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Haha :), thanks !

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