11/18/2023 0 Comments Arithmetic recursive sequence formula![]() ![]() So "f" of one is going toīe 12 minus seven times one minus one, that's a zero. And let's see if this actually works out. Whatever term we're on, that term minus one time. So it looks like whatever term we're on, we're subtracting seven "n" minus one, we're subtracting seven Going to subtract seven? So for the first term, we And what are we going to subtract seven? How many times are we But then, we are going to subtract seven. Of it as we're starting at, The first term is going to be 12. What this needs to be so that if you input "n" here, it gives you the appropriate "f" of "n." So let's think about it a little bit. Way, see if you can define this function of "n." If youĬan define it explicitly. What have we done? We've subtracted seven. The second term of this sequence is five. And so one way to think about it is this function "f" is defining a sequence where the first term ![]() Of "n" is negative two, when "n" is four, "f" "n" is 2, "f" of "n" is 5, and when "n" is three, "f" Maybe these having two levels of numbers to calculate the current number would imply that it would be some kind of quadratic function just as if I only had 1 level, it would be linear which is easier to calculate by hand.So we see here in this table, that for given "n's", when "n" is one, "f" of "n" is 12, when This gives us any number we want in the series. I do not know any good way to find out what the quadratic might be without doing a quadratic regression in the calculator, in the TI series, this is known as STAT, so plugging the original numbers in, I ended with the equation:į(x) = 17.5x^2 - 27.5x + 15. Then the second difference (60 - 25 = 35, 95-60 = 35, 130-95=35, 165-130 = 35) gives a second common difference, so we know that it is quadratic. = a ( 4 ) + 2 =a(4)+2 = a ( 4 ) + 2 equals, a, left parenthesis, 4, right parenthesis, plus, 2 = 9 =\goldD9 = 9 equals, start color #e07d10, 9, end color #e07d10Ī ( 5 ) a(5) a ( 5 ) a, left parenthesis, 5, right parenthesis = 7 + 2 =\blueD 7+2 = 7 + 2 equals, start color #11accd, 7, end color #11accd, plus, 2 = a ( 3 ) + 2 =a(3)+2 = a ( 3 ) + 2 equals, a, left parenthesis, 3, right parenthesis, plus, 2 = 7 =\blueD 7 = 7 equals, start color #11accd, 7, end color #11accdĪ ( 4 ) a(4) a ( 4 ) a, left parenthesis, 4, right parenthesis = 5 + 2 =\purpleC5+2 = 5 + 2 equals, start color #aa87ff, 5, end color #aa87ff, plus, 2 = a ( 2 ) + 2 =a(2)+2 = a ( 2 ) + 2 equals, a, left parenthesis, 2, right parenthesis, plus, 2 = 5 =\purpleC5 = 5 equals, start color #aa87ff, 5, end color #aa87ffĪ ( 3 ) a(3) a ( 3 ) a, left parenthesis, 3, right parenthesis = a ( 1 ) + 2 =a(1)+2 = a ( 1 ) + 2 equals, a, left parenthesis, 1, right parenthesis, plus, 2 ![]() = 3 =\greenE 3 = 3 equals, start color #0d923f, 3, end color #0d923fĪ ( 2 ) a(2) a ( 2 ) a, left parenthesis, 2, right parenthesis = a ( n − 1 ) + 2 =a(n\!-\!\!1)+2 = a ( n − 1 ) + 2 equals, a, left parenthesis, n, minus, 1, right parenthesis, plus, 2Ī ( 1 ) a(1) a ( 1 ) a, left parenthesis, 1, right parenthesis A ( n ) a(n) a ( n ) a, left parenthesis, n, right parenthesis ![]()
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