Skip to content
Exercise 4.4 · Q5

Q.Using the Mathematical induction, show that for any natural number n≥2n\ge 2,
[!FORMULA] (1−122)(1−132)(1−142)⋯(1−1n2)=n+12n.\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\cdots\left(1-\dfrac{1}{n^2}\right) = \dfrac{n+1}{2n}.

Puducherry TnboardTextbookSubjectiveImportance★★★★★
49% · 66/134 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Let P(n):(1−122)⋯(1−1n2)=n+12nP(n):\left(1-\dfrac1{2^2}\right)\cdots\left(1-\dfrac1{n^2}\right)=\dfrac{n+1}{2n}, n≥2n\ge2.

Step 1. Base case. P(2)P(2): LHS =1−14=34=1-\dfrac14=\dfrac34; RHS =34=\dfrac{3}{4}. True.

Step 2. Inductive hypothesis. Assume P(k):(1−122)⋯(1−1k2)=k+12kP(k):\left(1-\dfrac1{2^2}\right)\cdots\left(1-\dfrac1{k^2}\right)=\dfrac{k+1}{2k}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.