2an+1<\/sub> – an<\/sub> = 3, a0<\/sub>=12<\/p>\n then generated the first few terms:<\/p>\n The general solution is a formula that, when n is inputted, yields an<\/sub>.<\/p>\n Example: Find the general solution to 2an+1<\/sub> – an<\/sub> = 3, a0<\/sub>=12<\/strong><\/p>\n Solution:<\/p>\n Grimaldi reveals that for an equation of the form<\/p>\n an+1<\/sub> + ran<\/sub> = k<\/p>\n the solution will take the form<\/p>\n an<\/sub> = Arn<\/sup> + B<\/p>\n where A, B are constants that need to be solved.<\/p>\n To bring our equation to that form, we divide both sides by two:<\/p>\n an+1<\/sub> – 0.5an<\/sub> = 1.5<\/p>\n The general solution will have the form<\/p>\n an<\/sub> = A(0.5)n<\/sup> + B<\/p>\n In our case, a0<\/sub> = 12:<\/p>\n a0<\/sub> = 12 = A(0.5)0<\/sup> + B<\/p>\n so that<\/p>\n 12 = A + B<\/p>\n Solving for B we get<\/p>\n 12 – A = B<\/p>\n Then, for a1<\/sub> = 7.5, we get<\/p>\n 7.5 = A(0.5)1<\/sup> + B<\/p>\n which means<\/p>\n 7.5 = A(0.5) + B<\/p>\n Recalling 12 – A = B, we can rewrite:<\/p>\n 7.5 = A(0.5) + 12 – A<\/p>\n which gives<\/p>\n 7.5-12 = -0.5A<\/p>\n then<\/p>\n -4.5 = -0.5A<\/p>\n Next, dividing both sides by -0.5, we arrive at<\/p>\n 9 = A<\/p>\n Then<\/p>\n 12 – A = B<\/p>\n becomes<\/p>\n 12 – 9 = B<\/p>\n so that<\/p>\n B = 3<\/p>\n Apparently, the general solution to 2an+1<\/sub> – an<\/sub> = 3, a0<\/sub>=12, is an<\/sub> = 9(0.5)n<\/sup> + 3<\/p>\n We seek to confirm the solution: From the table, a4<\/sub> = 3.5625. Plugging 4 in for n in our general solution gives<\/p>\n a4<\/sub> = 9(0.5)4<\/sup> + 3 = 0.5625 + 3 = 3.5625<\/p>\n HTH:<\/p>\n Source:<\/p>\n Grimaldi, Ralph P. Discrete and Combinatorial Mathematics<\/u>. Don Mills: Addison-![](\"\/..\/images\/recurtab_mar4_16.png\")
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Wesley, 1994.<\/p>\n