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 \(n \geq 4\).

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 \(n \geq 4\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## 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 \( \forall n \geq 4 \). – Karthik Venkata · 1 year, 9 months ago

Log in to reply

@Sharky Kesa. And, Flawless proof Karthik Venkata . I solved it using the same way. And also Got it!

Yeah. I learnt Induction today! What a coincidenceHappy Dance– Mehul Arora · 1 year, 9 months agoLog in to reply

– Karthik Venkata · 1 year, 9 months ago

Haha :), thanks !Log in to reply

– Mehul Arora · 1 year, 9 months ago

No need to thank me, Genius :)Log in to reply

– Karthik Venkata · 1 year, 9 months ago

Lol good joke ! By the way, you are of Class 10 too ?Log in to reply

– Mehul Arora · 1 year, 9 months ago

No brother, I am class 9 :) Btw, That was not a joke. :PLog in to reply

– Karthik Venkata · 1 year, 9 months ago

So just entered class 9 right ? Nice, you are really talented.. You plan to give RMO ?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! – Mehul Arora · 1 year, 9 months ago

Log in to reply

– Karthik Venkata · 1 year, 9 months ago

No idea, I too am gonna write the RMO for the first time...Log in to reply

– Mehul Arora · 1 year, 9 months ago

Okay! Let's compete xD xD ALthough I am sure you will win :P xDLog in to reply

– Karthik Venkata · 1 year, 9 months ago

Haha, hope we meet in person at the INMO Camp next year :P !Log in to reply

– Mehul Arora · 1 year, 9 months ago

Yeah, Sure! I sure want to Meet a genius in person xDLog in to reply