Question

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

Answer 1

Answer: B. 0.25, 3.75, 5.2

Step-by-step explanation:

Since, the given sequence, $a_n= n^2 -a_{n-1}$

And, $a_5=14\frac{1}{4}$

On substituting n=5 in the given recursive formula,

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

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

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  $a_3=5.25$.

Answer 2

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