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प्रश्न
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.
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उत्तर
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`
∴ 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.
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