17an+1<\/sub> – 5an<\/sub> = 11<\/p>\n The above is a first degree, nonhomogeneous recurrence relation. First degree means that the next term only depends on the current one. Nonhomogeneous means that there is a value in the equation that is independent from the sequence terms; in the case above, it’s 11. A homogeneous recurrence relation looks like<\/p>\n 10an+1<\/sub> – 7an<\/sub> = 0<\/p>\n Example:<\/p>\n Find the first few terms of the recurrence relation<\/p>\n 2an+1<\/sub> – an<\/sub> = 3, a0<\/sub> = 12<\/p>\n <\/strong> Note the sequence starts at the zeroth term: a0<\/sub> = 12. With n=0, the relation becomes<\/p>\n 2a1<\/sub> – a0<\/sub> = 3<\/p>\n which leads to<\/p>\n 2a1<\/sub> – 12 = 3<\/p>\n 2a1<\/sub> = 15<\/p>\n a1<\/sub> = 7.5<\/p>\n Setting n=1, we have<\/p>\n 2a2<\/sub> – a1<\/sub> = 3<\/p>\n Then, putting in 7.5 for a1<\/sub>, we have<\/p>\n 2a2<\/sub> – 7.5 = 3<\/p>\n which yields a2<\/sub> = 5.25<\/p>\n Continuing the same way, we come up with<\/p>\n and so on.<\/p>\n Arriving at, for instance, the 20th term using the method above would be laborious. The general solution<\/em> to a recurrence relation means a formula that can get the nth term, an<\/sub>, from just inputting the specific n. For example, if you need the 20th term, you can just substitute 20 for n in the general solution and calculate a20<\/sub>.<\/p>\n Next post I’ll show how to find the general solution to the recurrence relation of today’s example.<\/p>\n Source:<\/p>\n Grimaldi, Ralph P. Discrete and Combinatorial Mathematics<\/u>. Don Mills: Addison-
\nSolution:<\/p>\n![](\"\/..\/images\/recurtab_mar4_16.png\")
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Wesley, 1994.<\/p>\n