This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 $\forall n \geq 4$.

Log in to reply

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 DanceLog in to reply

Haha :), thanks !

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

@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!

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply