Let $T(n)$ 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 -

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! Happy Dance

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Yeah, I just entered. And No, I am not That Talented. There are People Much Smarter and intelligent Than me. Some of them Would include @Archit Boobna , @Rajdeep Dhingra and Many more.

I do plan to Give the RMO. Any Tips or Tricks? They Would be of great help!

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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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@Archit Boobna , @Rajdeep Dhingra and Many more.

Yeah, I just entered. And No, I am not That Talented. There are People Much Smarter and intelligent Than me. Some of them Would includeI do plan to Give the RMO. Any Tips or Tricks? They Would be of great help!

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