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Sagot :
Alright, let's solve the problem step-by-step in detail.
We are given the recursive formula:
[tex]\[ a_n = \left(a_{n-1}\right)^2 - a_{n-1} \][/tex]
and we know that:
[tex]\[ a_4 = 870 \][/tex]
Our task is to find the first two terms of this sequence.
1. Start by figuring out [tex]\( a_3 \)[/tex]:
[tex]\[ a_4 = \left(a_3\right)^2 - a_3 \][/tex]
[tex]\[ 870 = \left(a_3\right)^2 - a_3 \][/tex]
This is a quadratic equation in [tex]\( a_3 \)[/tex]:
[tex]\[ \left(a_3\right)^2 - a_3 - 870 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[ a_3 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -870 \)[/tex]:
[tex]\[ a_3 = \frac{1 \pm \sqrt{1 + 4 \cdot 870}}{2} \][/tex]
[tex]\[ a_3 = \frac{1 \pm \sqrt{3481}}{2} \][/tex]
[tex]\[ a_3 = \frac{1 \pm 59}{2} \][/tex]
[tex]\[ a_3 = 30 \text{ or } a_3 = -29 \][/tex]
Since we are working with whole numbers, we discard the negative value:
[tex]\[ a_3 = 30 \][/tex]
2. Next, find [tex]\( a_2 \)[/tex] using the same recursive formula:
[tex]\[ a_3 = \left(a_2\right)^2 - a_2 \][/tex]
[tex]\[ 30 = \left(a_2\right)^2 - a_2 \][/tex]
This is another quadratic equation in [tex]\( a_2 \)[/tex]:
[tex]\[ \left(a_2\right)^2 - a_2 - 30 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[ a_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -30 \)[/tex]:
[tex]\[ a_2 = \frac{1 \pm \sqrt{1 + 4 \cdot 30}}{2} \][/tex]
[tex]\[ a_2 = \frac{1 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ a_2 = \frac{1 \pm 11}{2} \][/tex]
[tex]\[ a_2 = 6 \text{ or } a_2 = -5 \][/tex]
Since we are working with whole numbers, we discard the negative value:
[tex]\[ a_2 = 6 \][/tex]
3. Finally, find [tex]\( a_1 \)[/tex]:
[tex]\[ a_2 = \left(a_1\right)^2 - a_1 \][/tex]
[tex]\[ 6 = \left(a_1\right)^2 - a_1 \][/tex]
This is one more quadratic equation in [tex]\( a_1 \)[/tex]:
[tex]\[ \left(a_1\right)^2 - a_1 - 6 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[ a_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -6 \]: \[ a_1 = \frac{1 \pm \sqrt{1 + 4 \cdot 6}}{2} \] \[ a_1 = \frac{1 \pm \sqrt{25}}{2} \] \[ a_1 = \frac{1 \pm 5}{2} \] \[ a_1 = 3 \text{ or } a_1 = -2 \] Since we are working with whole numbers, we discard the negative value: \[ a_1 = 3 \] Therefore, the first two terms of the sequence are: \[ a_1 = 3 \] \[ a_2 = 6 \] The correct set of answers is \( \boxed{3, 6} \)[/tex].
We are given the recursive formula:
[tex]\[ a_n = \left(a_{n-1}\right)^2 - a_{n-1} \][/tex]
and we know that:
[tex]\[ a_4 = 870 \][/tex]
Our task is to find the first two terms of this sequence.
1. Start by figuring out [tex]\( a_3 \)[/tex]:
[tex]\[ a_4 = \left(a_3\right)^2 - a_3 \][/tex]
[tex]\[ 870 = \left(a_3\right)^2 - a_3 \][/tex]
This is a quadratic equation in [tex]\( a_3 \)[/tex]:
[tex]\[ \left(a_3\right)^2 - a_3 - 870 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[ a_3 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -870 \)[/tex]:
[tex]\[ a_3 = \frac{1 \pm \sqrt{1 + 4 \cdot 870}}{2} \][/tex]
[tex]\[ a_3 = \frac{1 \pm \sqrt{3481}}{2} \][/tex]
[tex]\[ a_3 = \frac{1 \pm 59}{2} \][/tex]
[tex]\[ a_3 = 30 \text{ or } a_3 = -29 \][/tex]
Since we are working with whole numbers, we discard the negative value:
[tex]\[ a_3 = 30 \][/tex]
2. Next, find [tex]\( a_2 \)[/tex] using the same recursive formula:
[tex]\[ a_3 = \left(a_2\right)^2 - a_2 \][/tex]
[tex]\[ 30 = \left(a_2\right)^2 - a_2 \][/tex]
This is another quadratic equation in [tex]\( a_2 \)[/tex]:
[tex]\[ \left(a_2\right)^2 - a_2 - 30 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[ a_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -30 \)[/tex]:
[tex]\[ a_2 = \frac{1 \pm \sqrt{1 + 4 \cdot 30}}{2} \][/tex]
[tex]\[ a_2 = \frac{1 \pm \sqrt{121}}{2} \][/tex]
[tex]\[ a_2 = \frac{1 \pm 11}{2} \][/tex]
[tex]\[ a_2 = 6 \text{ or } a_2 = -5 \][/tex]
Since we are working with whole numbers, we discard the negative value:
[tex]\[ a_2 = 6 \][/tex]
3. Finally, find [tex]\( a_1 \)[/tex]:
[tex]\[ a_2 = \left(a_1\right)^2 - a_1 \][/tex]
[tex]\[ 6 = \left(a_1\right)^2 - a_1 \][/tex]
This is one more quadratic equation in [tex]\( a_1 \)[/tex]:
[tex]\[ \left(a_1\right)^2 - a_1 - 6 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula:
[tex]\[ a_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -6 \]: \[ a_1 = \frac{1 \pm \sqrt{1 + 4 \cdot 6}}{2} \] \[ a_1 = \frac{1 \pm \sqrt{25}}{2} \] \[ a_1 = \frac{1 \pm 5}{2} \] \[ a_1 = 3 \text{ or } a_1 = -2 \] Since we are working with whole numbers, we discard the negative value: \[ a_1 = 3 \] Therefore, the first two terms of the sequence are: \[ a_1 = 3 \] \[ a_2 = 6 \] The correct set of answers is \( \boxed{3, 6} \)[/tex].
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