A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk = `(d_(k - 1))/"k"` for all natural numbers, k ≥ 2. Show that dn = `2/(n!)` for all n ∈ N.

Given that: d1 = 2 and dk = `(d_(k - 1))/k` Let P(n): dn = `2/(n!)` Step 1: P(1) : d1 = `2/(1!)` = 2 which is true for P(1). Step 2: P(k) : dk = `2/(k!)`. Let it be true for P(k). Step 3: Given that: dk = `(d_(k - 1))/k` &there4; dk+1 = `(d_(k + 1 - 1))/(k + 1) = d_k/(k + 1)` ⇒ dk+1 = `1/(k + 1) . d_k = 1/(k + 1) . 2/(k!)` ⇒ dk+1 = `2/((k + 1)!)` Which is true for P(k + 1). Hence, P(k + 1) is true whenever P(k) is true.

A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk = kdk-1k for all natural numbers, k ≥ 2. Show that dn = 2n! for all n ∈ N.

Question

A sequence d₁, d₂, d₃ ... is defined by letting d₁ = 2 and d_k = `(d_(k - 1))/"k"` for all natural numbers, k ≥ 2. Show that d_n = `2/(n!)` for all n ∈ N.

Theorem

Solution

Given that: d₁ = 2 and d_k = `(d_(k - 1))/k`

Let P(n): d_n = `2/(n!)`

Step 1: P(1) : d₁ = `2/(1!)` = 2 which is true for P(1).

Step 2: P(k) : d_k = `2/(k!)`. Let it be true for P(k).

Step 3: Given that: d_k = `(d_(k - 1))/k`

∴ d_k+1 = `(d_(k + 1 - 1))/(k + 1) = d_k/(k + 1)`

⇒ d_k+1 = `1/(k + 1) . d_k = 1/(k + 1) . 2/(k!)`

⇒ d_k+1 = `2/((k + 1)!)` Which is true for P(k + 1).

Hence, P(k + 1) is true whenever P(k) is true.

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Principle of Mathematical Induction

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Q 19 Q 18 Q 20

Chapter 4: Principle of Mathematical Induction - Exercise [Page 71]

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NCERT Exemplar Mathematics Exemplar [English] Class 11

Chapter 4 Principle of Mathematical Induction
Exercise | Q 19 | Page 71

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