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 $n \geq 4$ .

#Algebra #Sharky #Induction

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Comments

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 $\forall n \geq 4$ .

Venkata Karthik Bandaru - 6 years, 1 month 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