Find the first six terms of recursive sequence.
{eq}a_1 = 7, \; a_{n + 1} = \frac{a_n}{n} {/eq}
Question:
Find the first six terms of recursive sequence.
{eq}a_1 = 7, \; a_{n + 1} = \frac{a_n}{n} {/eq}
Recurrence Relation:
Basically, it is a relation which defines a sequence recursively.
Consider a sequence {eq}\displaystyle f(n) {/eq} in which we define a recurrence relation in the following form
{eq}\displaystyle f\left( {n + 1} \right) = f\left( n \right) + f\left( {n - 1} \right), {/eq} with {eq}\displaystyle f\left( 0 \right) = \text{constant}. {/eq}
Answer and Explanation: 1
Given, {eq}\displaystyle a_1 = 7, \; a_{n + 1} = \frac{a_n}{n}. {/eq}
Pitting {eq}\displaystyle n=1, 2, 3, 4, 5, 6 {/eq} in the above recurrence relation,
{eq}\displaystyle \eqalign{ & {a_2} = {a_{1 + 1}} = \frac{{{a_1}}}{1} = \frac{7}{1} = 7 \cr & {a_3} = {a_{2 + 1}} = \frac{{{a_2}}}{2} = \frac{7}{2} \cr & {a_4} = {a_{3 + 1}} = \frac{{{a_3}}}{3} = \frac{1}{3} \cdot \frac{7}{2} = \frac{7}{6} \cr & {a_5} = {a_{4 + 1}} = \frac{{{a_4}}}{4} = \frac{1}{4} \cdot \frac{7}{6} = \frac{7}{{24}} \cr & {a_6} = {a_{5 + 1}} = \frac{{{a_5}}}{5} = \frac{1}{5} \cdot \frac{7}{{24}} = \frac{7}{{120}} \cr & {a_7} = {a_{6 + 1}} = \frac{{{a_6}}}{6} = \frac{1}{6} \cdot \frac{7}{{120}} = \frac{7}{{720}}. \cr} {/eq}
Learn more about this topic:
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Chapter 9 / Lesson 11Learn what a relation is in math and three different ways to represent mathematical relations. Examples are provided to support understanding.