A geometric sequence is defined recursively by #a_n = 5a_(n-1)#, the first term of the sequence is 0.45. What is the explicit formula for the nth?

2 Answers
Shwetank Mauria
Mar 5, 2016

#0.45xx5^(n-1)#

Explanation:

In a geometric series #{a,ar,ar^2,ar^3,.....}#, #a# is first term and ratio is #r#, #n^(th)# term is given by #ar^(n-1)#. Note #r=a_m/a_(m-1)# or #a_m=ra_(m-1)#

As #a_n=5a_(n-1)#, #r=5# and first term #a=0.45#

Hence, #n^(th)# term is given by #0.45xx5^(n-1)#.

Bio
Mar 5, 2016

#a_n = 5^((n-1)) (0.45)#

Explanation:

The recursive definition #a_n = 5 a_{n-1}# means that

#a_2 = 5 a_1#

#a_color(red)(3) = 5 a_2#

#= 5 * (5 a_1)#

#=5^2 a_1#

#= 5^((color(red)(3)-1)) a_1#

#a_color(red)(4) = 5 a_3#

#= 5 * (5^2 a_1)#

#= 5^3 a_1#

#= 5^((color(red)(4)-1)) a_1#

#...#

We can guess that

#a_color(red)(n) = 5^((color(red)(n)-1)) a_1#

and show that this is true by induction.

For #n = 1#,

#a_1 = 5^((1-1)) a_1 = 5^0 a_1 = a_1#.

For all positive integers #n# greater than 1,

#a_n = 5^((n-1)) a_1#

#= 5 * 5^((n-2)) a_1#

#= 5 * (5^(((n-1)-1)) a_1)#

#= 5 a_{n-1}#.

Since it is given that the first term (#a_1#) is 0.45, the explicit formula for the #n^{"th"}# term is

#a_n = 5^((n-1)) (0.45)#.

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