The nth term of the sequence is 2n^2 +n+3.
Let's try to find a pattern in the given sequence:
6, 13, 24, 39, 58
The difference between consecutive terms might give us a clue:
13 - 6 = 7
24 - 13 = 11
39 - 24 = 15
58 - 39 = 19
Observing the differences, we can see that they are increasing by 4 each time: 7, 11, 15, 19. This is a second-order difference. When you have a second-order difference, the sequence might be a quadratic one.
Let's find a formula using the quadratic equation for the nth term of a sequence, which is an^2 +bn+c.
First, let's find the common differences between consecutive terms of the differences:
11 - 7 = 4
15 - 11 = 4
19 - 15 = 4
So, the second-order difference is constant at 4.
Now, let's proceed to find the nth term. The second-order difference suggests a quadratic equation.
The formula for the nth term of a quadratic sequence is
an^2 +bn+c.
To find the values of a, b, and c, we can use the terms of the sequence:
For the first term (n = 1):
a(1)^2 +b(1)+c=6
a+b+c=6 ---- Equation 1
For the second term (n = 2):
a(2)^2 +b(2)+c=13
4a+2b+c=13 ---- Equation 2
For the third term (n = 3):
a(3)^2 +b(3)+c=24
9a+3b+c=24 ---- Equation 3
Now, let's solve this system of equations to find the values of a, b, and c.
Subtracting Equation 1 from Equation 2, we get:
3a+b=7 ---- Equation 4
Subtracting Equation 1 from Equation 3, we get:
8a+2b=18 ---- Equation 5
Now, let's solve Equations 4 and 5 simultaneously to find the values of a and b.
From Equation 4:
b=7−3a ---- Equation 6
Substituting Equation 6 into Equation 5:
8a+2(7−3a)=18
8a+14−6a=18
2a=4
a=2
Substitute the value of a into Equation 6 to find b:
b=7−3(2)
b=7−6
b=1
Now that we have found a = 2 and b = 1, let's substitute these values into Equation 1 to find c:
a+b+c=6
2+1+c=6
c=6−3
c=3
So, the quadratic equation for the nth term is 2n^2 +n+3.
Let's check it for the 5th term (n = 5):
2(5)^2+5+3=2(25)+5+3=50+5+3=58
The equation 2n^2 +n+3 holds true for the sequence. Therefore, the nth term of the sequence is 2n^2 +n+3.