The equation for nth term of this Quadratic Sequence is [tex]n^2[/tex] + 3n + 4
Quadratic Sequence is a sequence, having constant second difference between consecutive numbers.
1st term : 8 + 6 = 14
2nd : 14 + 8 = 22
3rd : 22 + 10 = 32
4th : 32 + 12 = 44
5th : 44 + 14 = 58
6th term : 58 + 16 = 74
As, in this case, the second difference is 8 - 6 = 2 , 10 - 8 = 2 , 12 - 10 = 2 , 14 - 12 = 2 , 16 - 14 = 2
According to the sequence, the 0th item would be = 8 - (6 - 2) = 8 - 4 = 4
The quadratic form is : y = [tex]an^2 + bn + c [/tex] , where c = 4
As y = 8 at n = 1 . So, 8 = a (1) + b (1) + 4 . Hence, 8 = a + b + 4
Similarly for 2nd term, 14 = 4a + 2b + 4. Hence, 7 = 2a + b + 2 (ii)
Solving above two equations, 4 = a + b & 5 = 2a + b :
We get, 4 = a + 5 - 2a {By substitution}
4 = 5 - a ; a = 5 - 4 = 1
Putting the value of 'a = 1' in 'a + b = 4'
b = 4 -1 = 3
Hence, the equation is [tex]n^2[/tex] + 3n + 4
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