Question 1137191: write the recursive formula for 5,15,30,50,...
please show work on how it's done
Found 3 solutions by rothauserc, MathTherapy, greenestamps:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Use the method of level of common difference
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5, 15, 30, 50
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Level 1
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10, 15, 20
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Level 2
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5, 5
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Level 2 has a common difference which tells us that we have an n^2 term
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We can write the following equations for the known terms
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1) a(1)^2 +b(1) +c = a +b +c = 5
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2) a(2)^2 +b(2) +c = 4a +2b +c = 15
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3) a(3)^2 +b(3) +c = 9a +3b +c = 30
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4) a(4)^2 +b(4) +c = 16a +4b +c = 50
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Since we have 3 unknowns(a, b, c), we can use three equations to form an augmented matrix
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1 1 1 5
4 2 1 15
9 3 1 30
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Use your favorite method to solve augmented matrices
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a = 2.5, b = 2.5, c = 0
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the formula for the nth term is
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x(n) = 2.5n^2 +2.5n
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x(1) is the first term of the sequence, then
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5) x(n) = x(n-1) +2a(n) +d
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x(1)=5, 2a = 2(2.5) = 5, use equation 5 to find d
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x(2) = 5 +5(2) +d = 15
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d = 0
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Therefore, the recursive function is
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x(1) = 5,
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x(n) = x(n-1) +5n
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Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
write the recursive formula for 5,15,30,50,...
please show work on how it's done
With n being the term number in the series, and the 1st term, or  , then each term or 
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
A recursive formula tells you how to get a term of a sequence from the preceding term (or, in more complicated examples, from some combination of preceding terms). So look for a pattern in how each number compares to the previous number:
1st term 5, second term 15; to get the 2nd term, you add 10 (= 2*5) to the 1st term.
2nd term 15, 3rd term 30; to get the 3rd term, you add 15 (= 3*5) to the 2nd term.
3rd term 30, 4th term 50; to get the 4th term, you add 20 (= 4*5) to the 3rd term.
A simple pattern should be recognizable: to get the n-th term (after the first), you add (n*5) to the (n-1)th term.
In algebraic notation:
t(n) = t(n-1)+5n
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