Question

What are the first three terms of the sequence represented by the recursive formula


`a_n=n^2-a-n_-_1` if
`a_5=` 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

Answer 1

`a_n=n^2-a_(n-1)\to a_(n-1)=n^2-a_n`

`a_5=14(1)/(4);\ n=5\\\\substitute\\\\a_4=4^2-14(1)/(4)=16-14(1)/(4)=1(3)/(4)=1.75`

`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`
Answer: D. 4.25; -3.25; 7.25

Answer 2

Answer: B. 0.25, 3.75, 5.2

Explanation:

Since, the given sequence,
`a_n= n^2 -a_(n-1)`

And,
`a_5=14(1)/(4)`

On substituting n=5 in the given recursive formula,

We get,
`a_5=5^2-a_(5-1)\implies a_5=25-a_4\implies 14(1)/(4)= 25-a_4\implies a_4=10(3)/(4)`

For, n=4
`a_4=4^2-a_(4-1)\implies a_4=16-a_3\implies 10(3)/(4)= 16-a_3\implies a_3=5(1)/(4)\implies a_3=5.25`

For, n=3
`a_3=3^2-a_(3-1)\implies a_3=9-a_2\implies 5.25= 9-a_2\implies a_2=3.75`

For, n=2
`a_2=2^2-a_(2-1)\implies a_2=4-a_1\implies 3.75= 4-a_1\implies a_1=0.25`

Thus, First second and third terms are,

`a_1=0.25`,
`a_2=3.75` and
`a_3=5.25`.