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# How to find the recursive formula by hand for a quadratic sequence? Topic: How to write a recursive sequence
July 15, 2019 / By Jaclyn
Question: I know that this sequence (3,12,25,42,63) is quadratic, but how do I find the recursive formula for it? Help! ## Best Answers: How to find the recursive formula by hand for a quadratic sequence? Emelia | 8 days ago
I've two examples you might find useful. a) -1, 4, 11, 20, 31, 44... Differences between each term and the last are: +5 (from -1 to 4) +7 (from 4 to 11) +9 (from 11 to 20) +11 (from 20 to 31) +13 (from 31 to 44) Can you see the trend yet? If you look at the differences of the differences, you'll see: +2 (from +5 to 7) +2 (from +7 to +9) +2 (from +9 to +11) +2 (from +11 to +13) So you can see that the second differences are all 2. The first 'first difference' was +5, so the mth first difference has the form: 5 + 2(m-1) Check: m=1, first difference = 5 + 2 (1-1) = 5 ok m=2, first difference = 5 + 2 (2-1) = 7 ok m=3, first difference = 5 + 2 (3-1) = 9 ok etc... Now for the nth term of the original series, we need to look at the first term: -1 and the sum of the first (n-1) 'first differences'. The first (n-1) 'first differences' will consist of: 5+7+9+...(5 + 2((n-3)-1))+((5 + 2((n-2)-1))+(5 + 2((n-1)-1)) = 5+7+9+...(5 + 2(n-4))+((5 + 2(n-3))+(5 + 2(n-2)) There's another trick which lets us work out this sum: If you reverse the terms, you get: (5 + 2(n-2))+(5 + 2(n-3))+(5+2(n-4))+...+9+7+5 Add this to the last expression, gives: [5+5+2(n-2)] + [7+5+2(n-3)]+ [9+5+2(n-4)]+ ... + [9+5+2(n-4)]+[7+5+2(n-3)]+ [5+5+2((n-2)] If you rearrange this a bit, you get: [10+2(n-2)] + [10+2(n-2)]+ [10+2(n-2)]+ ... + [10+2(n-2)]+[10+2(n-2)]+ [10+2((n-2)] which is n-1 lots of [10+2(n-2)] and we need to remember this will be twice the value we're looking for (because we reversed the expression and added it to itself). So the expression we want is: (1/2)(n-1)(10+2(n-2)) which simplifies to: (1/2)(n-1)(10+2n-4) = (1/2)(n-1)(6+2n) = (n-1)(3+n) and that, in turn, multiplies up to give: 3n+n^2-3-n =n^2+2n-3 So the value of the nth term we've calculated is: -1+(n^2+2n-3) Let's check that: n=1, term = -1+(1+2-3) = -1 ok. n=2, term = -1+(4+4-3) = -1+5 = 4 ok n=3, term = -1+(9+6-3) = -1+12 = 11 ok n=4, term = -1+(16+8-3) = -1+21 = 20 ok n=5, term = -1+(25+10-3) = -1+32 = 31 ok n=6, term = -1+(36+12-3) = -1+45 = 44 ok ====================== 1) 7,10,15,22,31,42 7....10....15.....22.....31.....42 ....+3....+5....+7....+9....+11 .........[first difference with its first term=3] .........+2.....+2....+2...+2............ difference] write the mth term of the 1st difference, 3+2(m-1) testing to see if the formula is sounded. m=1, 3+2(1-1)=3 m=2, 3+2(2-1)=5 m=3, 3+2(3-2)=7...... ........ The sum of the first difference consists of S= 3+5+7+.....+[3+2((n-3)-1)] +[3+2((n-2)-1)] +[3+2((n-1)-1)] simplify it a lil. S=3+5+7+.....+[3+2((n-4))] +[3+2((n-3))] +[3+2((n-2))] write the expression in reverse S= [3+2(n-2)]+[3+2(n-3)]+[3+2(n-4)+......+7... 2S= [3+3+2(n-2)] +[3+5+2(n-3)]+[3+7+2(n-4)]........ + [3+7+2(n-4)] + [3+5+2(n-3)]+[3+3+2(n-2)] rewrite it in term of (n-2) you have (n-1) lots of (3+3+2(n-2)) 2S= (n-1)(6+2(n-2)) S= (n-1)(6+2(n-2))/2 simplify S=n²-1 the formula for the nth term of the original series is 7+(n²-1) testing n=1, 7+(1²-1)=7 n=2, 7+(2²-1)=10 n=3, 7+(3²-1)=15 n=4,7+(4²-1)=22 n=5, 7+(5²-1)=31 n=6, 7+(6²-1)= 42 =======================
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We found more questions related to the topic: How to write a recursive sequence Originally Answered: Find a closed formula for the sequence?
The first step is realizing it is a geometric sequence where each term is equal to the prior term multiplied by -3. This is the common ratio. The general formula for a geometric sequence is: a(n) = a₁ · rⁿ⁻¹ a₁ : first term r : common ratio n : number of the term you want In your case, the first term is 7 and the common ratio is -3. So the formula for this sequence is: a(n) = 7(-3)ⁿ⁻¹ If you can't type superscripts, write it as follows: a(n) = 7(-3)^(n-1) Update: One slight modification. Your questions asks you to use an index of *0* for the first term (rather than 1). So we don't need to use (n-1), we can just use n. c(n) = 7*-3^n, where n = 0 is the index of the first term. Originally Answered: Find the roots of quadratic formula?
I'm not intent on doing your homework for you, so I will do a similar, but different problem instead. x^2 - 4x + k = 0 Given that one of the roots is half the of the other, find the value of k. Let's refer to the roots as 'a'. This would mean that the roots are a and 2a. First, write down a, b and c: a = 1 b = -4 c = k Then, find the sum of roots and product of roots (SOR and POR respectively): The SOR is a + 2a = 3a SOR = -b/a = -(-4)/1 = 4 3a = 4 a = 4/3 The POR is a*2a = 2a^2 POR = c/a = k/1 2a^2 = k Because we know the value of a, simply substitute a into the last equation: Before: 2a^2 = k After: 2(4/3)^2 = k Which gives us: 32/9 Originally Answered: Find the roots of quadratic formula?
use quadratic formula a=1 b=-6 c=k x1=[-b-sqrt(b^2-4ac)]/(2a) x1=[6 -sqrt(36-4k)]/2 x2=[-b+sqrt(b^2-4ac)]/(2a) x2=[6 +sqrt(36-4k)]/2 those are the roots and the problem says one is 3 times bigger than other which we write as x2=3*x1 [6 +sqrt(36-4k)]/2 = 3* [6 -sqrt(36-4k)]/2 [6 +sqrt(36-4k)] = 3* [6 -sqrt(36-4k)] 6 +sqrt(36-4k) = 18 -3sqrt(36-4k) 3sqrt(36-4k)+sqrt(36-4k) = 18+6 4sqrt(36-4k)=24 sqrt(36-4k)=24/6 sqrt(36-4k)=4 square it 36-4k=16 4k=36-16 4k=20 k=20/4 k=5

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