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photoshop1234 [79]
3 years ago
6

Identify the recursive formula for the following sequence. 8, 16, 32, 64, ...

Mathematics
1 answer:
Dominik [7]3 years ago
7 0

Answer:

The  recursive formula for the given sequence

                 t_{n}=  (2)^{n+2}

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Given sequence is 8 , 16, 32, 64 ,.....

First term is   a = 8

common ratio   r = \frac{16}{8} =2

                       r = \frac{32}{16} =2

                      r = \frac{64}{32} =2   and so on..

Given sequence of the common ratio 'r' is equal

The n^{th} term of the sequence

                   t_{n} = a r^{n-1}

                   t_{n}= 8 (2)^{n-1}

                   t_{n}=  (2)^{(3+n-1)}

                  t_{n}=  (2)^{n+2}

This is  recursive formula for the given sequence

                 t_{n}=  (2)^{n+2}

<u><em>Verification</em></u>:-

           t_{n}=  (2)^{n+2}

put n=1   ⇒ t₁ = 8

put n=2   ⇒ t₂ = 16

put n=3   ⇒ t₃ = 32

put n=4   ⇒ t₄= 64    

and so on..

The sequence  8 ,16 , 32 , 64 ....

                 

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<h3>What is simplification?</h3>

Simplification in mathematics to solve the given condition on its operators.


\frac{10000}{(1+\frac{0.115}{4})^2}+68\frac{1-\frac{1}{(1+\frac{0.115}{4})^2}}{\frac{0.115}{4}}

= \frac{10000}{1.028^2} +68*\frac{1-\frac{1}{1.028^2} }{\frac{0.115}{4} } \\9451+68*\frac{1-0.94}{\frac{0.115}{4} } \\\\9451+68*\frac{0.054}{\frac{0.115}{4} } \\\\\\9451+3.72{\frac{4}{0.115} } \\\\\\9451+129\\

= 9580


The required solution is given as 9580.

Learn more about simplification here:
brainly.com/question/9218183

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