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  • A sequence d_1, d_2, d_3 …. Is defined by letting d1 = 2 and for all natural numbers, k ≥ 2. Show that for all n ϵ N.

A sequence \mathrm{d}_{1}, \mathrm{d}_{2}, \mathrm{d}_{3} \ldots . . Is defined by letting

\mathrm{d}_{1}=2 and \mathrm{d}_{\mathrm{k}}=\frac{\mathrm{d}_{\mathrm{k}-1}}{\mathrm{k}} \mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n} !} numbers, k \geq 2.Show that \mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n} !}  for all n ϵ N.

Answers (1)

A sequence \mathrm{d}_{1}, \mathrm{d}_{2}, \mathrm{d}_{3} \ldots . . Is defined by letting

\mathrm{d}_{1}=2 \ and \ \mathrm{d}_{\mathrm{k}}=\frac{\mathrm{d}_{\mathrm{k}-1}}{\mathrm{k}}\mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n} !} \ numbers, k \geq 2

\mathrm{d}_{k}=\mathrm{d}_{k-1} / \mathrm{k}

Now, we'll substitute different values for k

\begin{aligned} \mathrm{d}_{2}=& \mathrm{d}_{1} / 2 \\ &=2 / 2=2 / 2 ! \\ \mathrm{d}_{3}=\mathrm{d}_{2} / 3 & \\ &=1 / 3=2 / 3 ! \end{aligned}

Now, let us consider,

\mathrm{d}_{m}=\mathrm{d}_{m-1} / \mathrm{m}=2 / \mathrm{m} !   to be true.

Thus, \\ d_{m+1}=d_{m+1-1} / m+1\\ \begin{array}{l} =\mathrm{d}_{\mathrm{m}} / \mathrm{m}+1 \\ =2 /(\mathrm{m}+1) \cdot \mathrm{m} ! \\ =2 /(\mathrm{m}+1) ! \end{array}

Thus, d_{m+1 }is true if d_{m } is true

Hence, by mathematical induction,

For each natural no. n it is true that, \mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n} !}

Posted by

infoexpert21

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