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# What is the fourth term of a sequence with the recursive rule f(n)=-2f(n-1); f(1)=-3.5

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What is the fourth term of a sequence with the recursive rule f(n)=-2f(n-1); f(1)=-3.5

Guest May 10, 2015

#1
+26470
+8

Each term beyond the first is -2 times the previous one, and there are n-1 terms beyond the first, so the n'th term is given by (-2)n-1 times the first term or:  f(n) = (-2)n-1*(-3.5)

The fourth term is then: f(4) = (-2)3*(-3.5)  = -8*(-3.5) = 28

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Alan  May 10, 2015
Sort:

#1
+26470
+8

Each term beyond the first is -2 times the previous one, and there are n-1 terms beyond the first, so the n'th term is given by (-2)n-1 times the first term or:  f(n) = (-2)n-1*(-3.5)

The fourth term is then: f(4) = (-2)3*(-3.5)  = -8*(-3.5) = 28

.

Alan  May 10, 2015

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