Question 12 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

Transcript

Question 12 By factoring the expression, check that 𝑛^3−𝑛 is always divisible by 6 for all natural numbers 𝑛. Give reasons. First, we factorise 𝑛^3−𝑛 Now, 𝒏^𝟑−𝒏=𝑛(𝑛^2−1) =𝑛(𝑛^2−1^2 ) =𝒏(𝒏−𝟏)(𝒏+𝟏) Let's arrange these in order from smallest to largest: (𝒏−𝟏)× 𝒏 × (𝒏+𝟏) Looking at the brackets, they represent three consecutive numbers (like 4, 5, 6 or 11, 12, 13). And, we know that In any three consecutive numbers, at least one of them MUST be an even number (divisible by 2) exactly one of them MUST be a multiple of 3 Now, If a number is multiplied by 2 , and then multiplied by 3 Then, Total product is guaranteed to be a multiple of 𝟐 × 𝟑=𝟔 Thus, our expression 𝑛^3−𝑛 is always divisible by 6 Hence proved