Consider the quadratic polynomial sequence an =2n2+1
. Calculate an
for n = 1, 2, 3, 4, 5(1 point)
Question 2
Using the results above, find the constant second difference.(1 point)
To calculate the values of an, we substitute the given values of n into the formula for an.
So, for n = 1,
a1 = 2(1)^2 + 1 = 2(1) + 1 = 2 + 1 = 3.
For n = 2,
a2 = 2(2)^2 + 1 = 2(4) + 1 = 8 + 1 = 9.
For n = 3,
a3 = 2(3)^2 + 1 = 2(9) + 1 = 18 + 1 = 19.
For n = 4,
a4 = 2(4)^2 + 1 = 2(16) + 1 = 32 + 1 = 33.
For n = 5,
a5 = 2(5)^2 + 1 = 2(25) + 1 = 50 + 1 = 51.
Therefore, an for n = 1, 2, 3, 4, 5 is 3, 9, 19, 33, 51.
To find the constant second difference, we need to find the difference between consecutive values of the first differences.
First differences: 9 - 3 = 6, 19 - 9 = 10, 33 - 19 = 14, 51 - 33 = 18.
Second differences: 10 - 6 = 4, 14 - 10 = 4, 18 - 14 = 4.
The constant second difference is 4.
To calculate the values of the quadratic polynomial sequence an = 2n^2 + 1 for n = 1, 2, 3, 4, 5, we substitute the values of n into the formula:
For n = 1:
a1 = 2(1)^2 + 1
a1 = 2 + 1
a1 = 3
For n = 2:
a2 = 2(2)^2 + 1
a2 = 2(4) + 1
a2 = 8 + 1
a2 = 9
For n = 3:
a3 = 2(3)^2 + 1
a3 = 2(9) + 1
a3 = 18 + 1
a3 = 19
For n = 4:
a4 = 2(4)^2 + 1
a4 = 2(16) + 1
a4 = 32 + 1
a4 = 33
For n = 5:
a5 = 2(5)^2 + 1
a5 = 2(25) + 1
a5 = 50 + 1
a5 = 51
Therefore, the values of the quadratic polynomial sequence for n = 1, 2, 3, 4, 5 are: 3, 9, 19, 33, 51.
To find the constant second difference, we take the difference between consecutive terms and then take the difference between those differences:
First difference:
9 - 3 = 6
19 - 9 = 10
33 - 19 = 14
51 - 33 = 18
Second difference:
10 - 6 = 4
14 - 10 = 4
18 - 14 = 4
The constant second difference is 4.
To calculate the values of an for n = 1, 2, 3, 4, and 5 in the quadratic polynomial sequence an = 2n^2 + 1, we can simply substitute the given values of n into the equation and solve for an.
For n = 1:
a1 = 2(1)^2 + 1
= 2(1) + 1
= 2 + 1
= 3
For n = 2:
a2 = 2(2)^2 + 1
= 2(4) + 1
= 8 + 1
= 9
For n = 3:
a3 = 2(3)^2 + 1
= 2(9) + 1
= 18 + 1
= 19
For n = 4:
a4 = 2(4)^2 + 1
= 2(16) + 1
= 32 + 1
= 33
For n = 5:
a5 = 2(5)^2 + 1
= 2(25) + 1
= 50 + 1
= 51
Hence, the values of an for n = 1, 2, 3, 4, and 5 are 3, 9, 19, 33, and 51 respectively.
Moving on to Question 2, to find the constant second difference, we need to calculate the difference between consecutive terms twice and observe whether the resulting values are the same.
The first differences between consecutive terms are:
d1 = a2 - a1 = 9 - 3 = 6
d2 = a3 - a2 = 19 - 9 = 10
d3 = a4 - a3 = 33 - 19 = 14
d4 = a5 - a4 = 51 - 33 = 18
The second differences between the consecutive first differences are:
d2 - d1 = 10 - 6 = 4
d3 - d2 = 14 - 10 = 4
d4 - d3 = 18 - 14 = 4
Since the resulting second differences are all equal to 4, we can conclude that the constant second difference for the given quadratic polynomial sequence is 4.