# How do you write the first five terms of the sequence defined recursively a_1=81, a_(k+1)=1/3a_k, then how do you write the nth term of the sequence as a function of n?

Jun 13, 2017

Given: ${a}_{1} = 81$ and ${a}_{k + 1} = \frac{1}{3} {a}_{k}$

Starting with the first term:

${a}_{1} = 81 = {\left(\frac{1}{3}\right)}^{0} 81$

${a}_{2} = \left(\frac{1}{3}\right) 81 = {\left(\frac{1}{3}\right)}^{1} 81$

${a}_{3} = \left(\frac{1}{3}\right) {a}_{2} = \left(\frac{1}{3}\right) \left(\frac{1}{3}\right) 81 = {\left(\frac{1}{3}\right)}^{2} 81$

${a}_{4} = \left(\frac{1}{3}\right) {a}_{3} = \frac{1}{3} {\left(\frac{1}{3}\right)}^{2} 81 = {\left(\frac{1}{3}\right)}^{3} 81$

The equation for the nth term is:

${a}_{n} = {\left(\frac{1}{3}\right)}^{n - 1} 81$