Answer:
<em>The n-th term for the sequence will be:
</em>
Step-by-step explanation:
Given sequence is: 7, 14, 23, 34, 47. 62, 79, ........
The n-th term of a quadratic sequence is: ![t_{n}=an^2 +bn+c](https://tex.z-dn.net/?f=t_%7Bn%7D%3Dan%5E2%20%2Bbn%2Bc)
For
....
![t_{1}=a(1)^2+b(1)+c\\ \\ a+b+c=7 .............................(1)](https://tex.z-dn.net/?f=t_%7B1%7D%3Da%281%29%5E2%2Bb%281%29%2Bc%5C%5C%20%5C%5C%20a%2Bb%2Bc%3D7%20.............................%281%29)
For
....
![t_{2}=a(2)^2+b(2)+c\\ \\ 4a+2b+c=14 .............................(2)](https://tex.z-dn.net/?f=t_%7B2%7D%3Da%282%29%5E2%2Bb%282%29%2Bc%5C%5C%20%5C%5C%204a%2B2b%2Bc%3D14%20.............................%282%29)
For
....
![t_{3}=a(3)^2+b(3)+c\\ \\ 9a+3b+c=23 .............................(3)](https://tex.z-dn.net/?f=t_%7B3%7D%3Da%283%29%5E2%2Bb%283%29%2Bc%5C%5C%20%5C%5C%209a%2B3b%2Bc%3D23%20.............................%283%29)
Subtracting equation (1) from equation (2), we will get......
![3a+b=7..........................(4)](https://tex.z-dn.net/?f=3a%2Bb%3D7..........................%284%29)
Subtracting equation (2) from equation (3), we will get.......
![5a+b=9..........................(5)](https://tex.z-dn.net/?f=5a%2Bb%3D9..........................%285%29)
Now, subtracting equation (4) from equation (5)...........
![2a=2\\ \\ a=\frac{2}{2}=1](https://tex.z-dn.net/?f=2a%3D2%5C%5C%20%5C%5C%20a%3D%5Cfrac%7B2%7D%7B2%7D%3D1)
Plugging this
into equation (4), we will get....
![3(1)+b=7\\ \\ 3+b=7\\ \\ b=7-3=4](https://tex.z-dn.net/?f=3%281%29%2Bb%3D7%5C%5C%20%5C%5C%203%2Bb%3D7%5C%5C%20%5C%5C%20b%3D7-3%3D4)
Now, plugging
and
into equation (1) .........
![1+4+c=7\\ \\ 5+c=7\\ \\ c=7-5=2](https://tex.z-dn.net/?f=1%2B4%2Bc%3D7%5C%5C%20%5C%5C%205%2Bc%3D7%5C%5C%20%5C%5C%20c%3D7-5%3D2)
Thus, the n-th term for the sequence will be: ![n^2+4n+2](https://tex.z-dn.net/?f=n%5E2%2B4n%2B2)