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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.

A sequence a_1, a_2, a_3 ... is defined by letting a_1 = 3 \ and \ a_k = 7a_{k-1} for all natural numbers k ≥ 2. Show that a_n = 3.7^{n-1 }for all natural numbers.

Answers (1)

Given 

\begin{aligned} &a_{1}=3 \& a_{k}=7 a_{k-1}\\ &\text { Now, we'll substitute different values for k }\\ &\begin{aligned} a_{2} &=7 \times a_{1} \\ &=7 \times 3=21 \\ &=3.7^{2-1} \\ a_{3} &=7 \times a_{2} \\ &=7^{2} \times a_{1}=49 \times 3 \\ &=147 \end{aligned} \end{aligned}

=3.7^{3-1}

\begin{aligned} a_{n} &=7 a_{n-1} \\ &=3.7^{n-1} \end{aligned} \\

\\ \text { Thus, } a_{n+1}=7 a_{n+1-1} \\ =7 a_{n} \\ =7 \times 3.7^{n-1} \\ =3.7^{(m+1)-1}

Thus, an+1 is true if an is true

Hence, by mathematical induction,

For each natural no. n it is true that, a_n = 3.7^{n-1 }

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infoexpert21

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