# Induction practice for beginners

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 $n \geq 4$. Note by Sharky Kesa
5 years, 5 months ago

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.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

PROOF BY INDUCTION

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 -

$\left(1 - \dfrac {2}{4}\right)\left(1 - \dfrac {2}{5}\right)\ldots\left(1 - \dfrac{2}{k}\right) = \dfrac{6}{k(k-1)}$

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 $\forall n \geq 4$.

- 5 years, 5 months ago

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

- 5 years, 5 months ago

Haha :), thanks !

- 5 years, 5 months ago

No need to thank me, Genius :)

- 5 years, 5 months ago

Lol good joke ! By the way, you are of Class 10 too ?

- 5 years, 5 months ago

No brother, I am class 9 :) Btw, That was not a joke. :P

- 5 years, 5 months ago

So just entered class 9 right ? Nice, you are really talented.. You plan to give RMO ?

- 5 years, 5 months ago

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!

- 5 years, 5 months ago

No idea, I too am gonna write the RMO for the first time...

- 5 years, 5 months ago

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

- 5 years, 5 months ago

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

- 5 years, 5 months ago

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

- 5 years, 5 months ago