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What are the first three terms of the sequence represented by the recursive formula
[tex]a_n=n^2-a-n_-_1[/tex] if [tex]a_5=[/tex] 14 1/4
A. 12.25, –10.25, 13.25
B. 0.25, 3.75, 5.25
C. –13.25, 17.25, –8.25
D. –4.25, –3.25, 7.25

Respuesta :

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[tex]a_n=n^2-a_{n-1}\to a_{n-1}=n^2-a_n[/tex]
[tex]a_5=14\dfrac{1}{4};\ n=5\\\\substitute\\\\a_4=4^2-14\dfrac{1}{4}=16-14\dfrac{1}{4}=1\dfrac{3}{4}=1.75[/tex]
[tex]a_3=3^2-1.75=9-1.75=7.25\\\\a_2=2^2-7.25=4-7.25=-3.25\\\\a_1=1^2-(-3.25)=1+3.25=4.25[/tex]
Answer: D. 4.25; -3.25; 7.25

Answer: B. 0.25, 3.75, 5.2

Step-by-step explanation:

Since, the given sequence, [tex]a_n= n^2 -a_{n-1}[/tex]

And, [tex]a_5=14\frac{1}{4}[/tex]

On substituting n=5 in the given recursive formula,

We get, [tex]a_5=5^2-a_{5-1}\implies a_5=25-a_4\implies 14\frac{1}{4}= 25-a_4\implies a_4=10\frac{3}{4}[/tex]

For, n=4 [tex]a_4=4^2-a_{4-1}\implies a_4=16-a_3\implies 10\frac{3}{4}= 16-a_3\implies a_3=5\frac{1}{4}\implies a_3=5.25[/tex]

For, n=3 [tex]a_3=3^2-a_{3-1}\implies a_3=9-a_2\implies 5.25= 9-a_2\implies a_2=3.75[/tex]

For, n=2 [tex]a_2=2^2-a_{2-1}\implies a_2=4-a_1\implies 3.75= 4-a_1\implies a_1=0.25[/tex]

Thus, First second and third terms are,

[tex]a_1=0.25[/tex], [tex]a_2=3.75[/tex] and  [tex]a_3=5.25[/tex].

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