Respuesta :
Answer:
The nth term rule of the quadratic sequence is [tex]2n^{2} +5n-3[/tex].
Step-by-step explanation:
We are given the following quadratic sequence below;
4, 15, 30, 49, 72, 99, 130,...
As we know that the formula for the nth term of the quadratic sequence is given by = [tex]an^{2}+bn+c[/tex]
Firstly, we will find the difference between the term of the given sequence;
1st difference of the given sequence;
(2nd term - 1st term), (3rd term - 2nd term), (4th term - 3rd term), (5th term - 4th term), (6th term - 5th term), (7th term - 6th term)
= (15 - 4), (30 - 15), (49 - 30), (72 - 49), (99 - 72), (130 - 99),....
= (11, 15, 19, 23, 27, 31,.....)
Now, we will find the second difference of the given sequence, i.e;
= (15 - 11), (19 - 15), (23 - 19), (27 - 23), (31 - 27),....
= (4, 4, 4, 4, 4)
Since the differences are same now, so to find the value of a we have to divide the value of second difference by 2, i.e;
The value of a = [tex]\dfrac{4}{2}[/tex] = 2
SO, the first term of the nth term rule equation is [tex]an^{2}[/tex] = [tex]2n^{2}[/tex].
Now, in the term [tex]2n^{2}[/tex], put the value of n = 1, 2, 3, 4 and 5 and then form the sequence, i.e;
If n = 1, then [tex]2n^{2}[/tex] = [tex]2 \times (1)^{2}[/tex] = 2
If n = 2, then [tex]2n^{2}[/tex] = [tex]2 \times (2)^{2}[/tex] = 8
If n = 3, then [tex]2n^{2}[/tex] = [tex]2 \times (3)^{2}[/tex] = 18
If n = 4, then [tex]2n^{2}[/tex] = [tex]2 \times (4)^{2}[/tex] = 32
If n = 5, then [tex]2n^{2}[/tex] = [tex]2 \times (5)^{2}[/tex] = 50
SO, the sequence formed is (2, 8, 18, 32, 50).
Now, find the difference of this sequence and the original quadratic sequence, i.e;
= (4 - 2), (15 - 8), (30 - 18), (49 - 32), (72 - 50)
= (2, 7, 12, 17, 22)
Now, as we can see that the above sequence resembles the general form of (5n - 3) because:
If we put n = 1, then (5n - 3) = 5 - 3 = 2
If we put n = 2, then (5n - 3) = 10 - 3 = 7 and so on....
From this, we concluded that the value of b and c are 5 and (-3) respectively.
Hence, the nth term rule for the given quadratic sequence is [tex]2n^{2} +5n-3[/tex].
