SOLUTION: A sequence is a function with domain the natural numbers 1,2,3,.... Using function notation, you can write the sequence an= a1+(n-1)d as an=a1+(n-1)d and the sequence an=a1r^n-1

Algebra ->  Sequences-and-series -> SOLUTION: A sequence is a function with domain the natural numbers 1,2,3,.... Using function notation, you can write the sequence an= a1+(n-1)d as an=a1+(n-1)d and the sequence an=a1r^n-1       Log On


   



Question 1031545: A sequence is a function with domain the natural numbers 1,2,3,.... Using function notation, you can write the sequence an= a1+(n-1)d as an=a1+(n-1)d and the sequence an=a1r^n-1 as a(n)=a1r^n-1
a. s a(n)=a1+(n-1)d a linear function ? Explain, if not how can you adjust its definition so that it is a linear function ? What is the slope?
b.what type of function does an=a1r^n-1 suggest? To what family of functions does this function belong? explain how its related to the parent function os that family. Draw that graph.
c. what type of function is suggested by the sum sequence sn=a1(1-r^n)/1-r ?By sn= n/2 (a1 + a(n))? explan each answer

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A sequence is a function with domain the natural numbers 1,2,3,.... Using function notation, you can write the sequence an= a1+(n-1)d as an=a1+(n-1)d and the sequence an=a1r^n-1 as a(n)=a1r^n-1
a. s a(n)=a1+(n-1)d a linear function ? Explain, if not how can you adjust its definition so that it is a linear function ? What is the slope?
a(n) = (n-1)d + a(1)
slope = (n-1)
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b.what type of function does an=a1r^n-1 suggest? To what family of functions does this function belong? explain how its related to the parent function in that family. Draw that graph.
a(n) = a(1)*r^(n-1)
Exponential
Because the exponent is n-1, the parent function is moved 1 unit to the right.
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c. what type of function is suggested by the sum sequence sn=a1(1-r^n)/1-r ?By sn= n/2 (a1 + a(n))? explan each answer
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s(n) = (n/2)*a(n) + a(1)/2
Linear with slope = (n/2) and intercept = a(1)/2
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Cheers,
Stan H.
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