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Marianna [84]
1 year ago
11

What is the nth term rule of the quadratic sequence below?

Mathematics
1 answer:
Vladimir [108]1 year ago
6 0

Answer:

3n² + 5n - 2

Step-by-step explanation:

<u>Given sequence</u>:

6, 20, 40, 66, 98, 136, ...

Calculate the <u>first differences</u> between the terms:

6 \underset{+14}{\longrightarrow} 20 \underset{+20}{\longrightarrow} 40 \underset{+26}{\longrightarrow} 66 \underset{+32}{\longrightarrow} 98 \underset{+38}{\longrightarrow} 136

As the first differences are not the same, calculate the <u>second differences:</u>

14 \underset{+6}{\longrightarrow} 20 \underset{+6}{\longrightarrow} 26 \underset{+6}{\longrightarrow} 32 \underset{+6}{\longrightarrow} 38

As the <u>second differences are the same</u>, the sequence is quadratic and will contain an n² term.

The <u>coefficient</u> of the n² term is <u>half of the second difference</u>.

Therefore, the n² term is:  3n²

Compare 3n² with the given sequence:

\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}

The second operations are different, therefore calculate the differences <em>between</em> the second operations:

3 \underset{+5}{\longrightarrow} 8 \underset{+5}{\longrightarrow} 13\underset{+5}{\longrightarrow} 18

As the differences are the same, we need to add 5n as the second operation:

\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}

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

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