MATH SOLVE

10 months ago

Q:
# If a(sub n)=24, which recursive formula could represent the sequence below? 24, 88, 664, 8408

Accepted Solution

A:

The answer would be: [tex] a_{n} [/tex]= ([tex] a_{n-2} [/tex])^2 + [tex] a_{n-1} [/tex]

To answer the question let:

[tex] a_{n-3} [/tex]=24

[tex] a_{n-2} [/tex]=88

[tex] a_{n-1} [/tex]= 664

[tex] a_{n} [/tex]= 8408

To solve this question, you can put the number into the equation. Third option will have minus result. So you can try the first, second and fourth option.

Second option

[tex] a_{n} [/tex]= 3[tex] a_{n-1}+16 [/tex]

[tex] a_{n} [/tex]= 3(664)+16= 2008

fourth option

[tex] a_{n} [/tex]= 2[tex] a_{n-2} [/tex] +7 [tex] a_{n-1} [/tex]

[tex] a_{n} [/tex]= 2(88)+7 (664)Β = 4824

first option

[tex] a_{n} [/tex]= ([tex] a_{n-2} [/tex])^2 + [tex] a_{n-1} [/tex]

[tex] a_{n} [/tex]=(88)^2+ 664= 8408 (true)

To answer the question let:

[tex] a_{n-3} [/tex]=24

[tex] a_{n-2} [/tex]=88

[tex] a_{n-1} [/tex]= 664

[tex] a_{n} [/tex]= 8408

To solve this question, you can put the number into the equation. Third option will have minus result. So you can try the first, second and fourth option.

Second option

[tex] a_{n} [/tex]= 3[tex] a_{n-1}+16 [/tex]

[tex] a_{n} [/tex]= 3(664)+16= 2008

fourth option

[tex] a_{n} [/tex]= 2[tex] a_{n-2} [/tex] +7 [tex] a_{n-1} [/tex]

[tex] a_{n} [/tex]= 2(88)+7 (664)Β = 4824

first option

[tex] a_{n} [/tex]= ([tex] a_{n-2} [/tex])^2 + [tex] a_{n-1} [/tex]

[tex] a_{n} [/tex]=(88)^2+ 664= 8408 (true)