Write the first three terms of each sequence:

Solution (a) The $n$th term of this sequence is $b_{n}=\dfrac{1}{3n-2}.$ The first three terms are $$b_{1}=\dfrac{1}{3\cdot 1-2}=1\qquad b_{2}=\dfrac{1}{3\cdot 2-2}=\dfrac{1}{4}\qquad b_{3}=\dfrac{1}{3\cdot 3-2}=\dfrac{1}{7}$$

(b) The $n$th term of this sequence is $c_{n}=\dfrac{2n-1}{n^{3}}$. Then $$c_{1}=\dfrac{2(1)-1}{1^{3}}=1\qquad c_{2}=\dfrac{2(2) -1}{2^{3}}=\dfrac{3}{8}\qquad c_{3}=\dfrac{2 (3) -1}{3^{3}}=\dfrac{5}{27}$$