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Mathematical Induction — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsMathematical Induction5 Marks
Question
A sequence $x_1, x_2, x_3, ...$ is defined by letting $x_1 = 2$ and $\text{x}_{\text{k}}=\frac{\text{x}_{\text{k}}-1}{\text{n}}$ for all natural numbers $k, \text{k}\geq2.$ Show that $\text{x}_{\text{n}}=\frac{2}{\text{n}!}$ for all $\text{n}\in\text{N}.$
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Answer

Let $p(n)$ be the statement given by $P(n): \text{x}_{\text{n}}=\frac{2}{\text{n}!}$ for all $\text{n}\in\text{N}.$
Step 1:
$p(2): \text{x}_{\text{n}}=\frac{2}{2!}=1$
Given that $\text{x}_{\text{k}}=\frac{\text{x}_{\text{k}}-1}{\text{n}}$ for all natural numbers $\text{k}\geq2$
$\text{x}_{2}=\frac{\text{x}_1}{2}=\frac{2}{2}=1$
$\therefore p(2)$ is true.
Step 2:
Let $p(m)$ is true. Then,
$\text{x}_{\text{m}}=\frac{2}{\text{m}!} \ ...(1)$
We have to prove that $p(m + 1)$ is true.
$\text{x}_{\text{m}+1}=\frac{\text{x}_{\text{m}+1-1}}{\text{m+1}}$
$\text{x}_{\text{m}+1}=\frac{\text{x}_{\text{m}}}{\text{m+1}}$
$\text{x}_{\text{m}+1}=\frac{{\frac{2}{\text{m}!}}}{\text{m+1}} \ ...[\text{From(1)}]$
$\text{x}_{\text{m}+1}=\frac{2}{\text{m}!(\text{m+1})}$
$\text{x}_{\text{m}+1}=\frac{2}{(\text{m+1})!}$
$\Rightarrow p(m + 1)$ is true.
Hence by the principle of mathematical induction, the given results is true for all $\text{n}\in\text{N}.$

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