Answer:
3n² + 5n - 2
Step-by-step explanation:
Given sequence:
6, 20, 40, 66, 98, 136, ...
Calculate the first differences between the terms:
[tex]6 \underset{+14}{\longrightarrow} 20 \underset{+20}{\longrightarrow} 40 \underset{+26}{\longrightarrow} 66 \underset{+32}{\longrightarrow} 98 \underset{+38}{\longrightarrow} 136[/tex]
As the first differences are not the same, calculate the second differences:
[tex]14 \underset{+6}{\longrightarrow} 20 \underset{+6}{\longrightarrow} 26 \underset{+6}{\longrightarrow} 32 \underset{+6}{\longrightarrow} 38[/tex]
As the second differences are the same, the sequence is quadratic and will contain an n² term.
The coefficient of the n² term is half of the second difference.
Therefore, the n² term is: 3n²
Compare 3n² with the given sequence:
[tex]\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2 & 3 & 12 & 27 & 48 \\\cline{1-5} \sf operation & +3&+8 & +13 & +18 \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}[/tex]
The second operations are different, therefore calculate the differences between the second operations:
[tex]3 \underset{+5}{\longrightarrow} 8 \underset{+5}{\longrightarrow} 13\underset{+5}{\longrightarrow} 18[/tex]
As the differences are the same, we need to add 5n as the second operation:
[tex]\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2 +5n & 8&22 & 42 & 68\\\cline{1-5}\sf operation & -2 &-2 &-2 & -2 \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}[/tex]
Finally, we can clearly see that the operation to get from 3n² + 5n to the given sequence is to subtract 2.
Therefore, the nth term of the quadratic sequence is:
3n² + 5n - 2
