Question 172790
Take note that the first term is *[Tex \LARGE a_{1}], the second term is *[Tex \LARGE a_{3}], etc.


So the fourth term is *[Tex \LARGE a_{4}] where {{{n=4}}}



So when *[Tex \LARGE a_{n}=a_{4}], then *[Tex \LARGE a_{n-1}=a_{3}] (the previous term), *[Tex \LARGE a_{n-2}=a_{2}] (the next previous term), and *[Tex \LARGE a_{n-3}=a_{1}] (the next previous term)






*[Tex \LARGE a_{n}=a_{n-1}+\left(\frac{1}{2}\right)a_{n-2}+\left(\frac{1}{3}\right)a_{n-3}] Start with the given equation



*[Tex \LARGE a_{4}=a_{3}+\left(\frac{1}{2}\right)a_{2}+\left(\frac{1}{3}\right)a_{1}] Plug in *[Tex \LARGE a_{n}=a_{4}], *[Tex \LARGE a_{n-1}=a_{3}], *[Tex \LARGE a_{n-2}=a_{2}], and *[Tex \LARGE a_{n-3}=a_{1}] (see above)




*[Tex \LARGE a_{4}=1+\left(\frac{1}{2}\right)(3)+\left(\frac{1}{3}\right)1] Plug in *[Tex \LARGE a_{1}=1], *[Tex \LARGE a_{2}=3], and *[Tex \LARGE a_{3}=1] (these are given) 



*[Tex \LARGE a_{4}=1+\frac{3}{2}+\frac{1}{3}] Multiply



*[Tex \LARGE a_{4}=\frac{17}{6}] Combine the fractions.




So the 4th term is *[Tex \LARGE a_{4}=\frac{17}{6}]