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A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers. - Mathematics

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प्रश्न

A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.

प्रमेय
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उत्तर

Given that: a1 = 3

a2 = 7a2 – 1 = 7.a1 = 7.3 = 21

a3 = 7.a3 – 1 = 7.a2 = 7.21 = 147

Let P(n): an = 3.7n – 1, ∀ n ∈ N

Step 1: P(2) : a2 = 3.72 – 1 = 21

⇒ 21 = 21 which is true for P(2).

Step 2: P(k): ak = 3.7k – 1.

Let it be true.

Step 3: ak = 7ak – 1   .......(Given)

Put k = k + 1

ak + 1 = 7ak = 7(3.7k – 1)

= 3.7k+1–1

= `3.7^(("k"+1) – 1`

Which is true for P(k + 1).

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

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अध्याय 4: Principle of Mathematical Induction - Exercise [पृष्ठ ७१]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 4 Principle of Mathematical Induction
Exercise | Q 17 | पृष्ठ ७१

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