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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.