Question

Here are the first four terms of a quadratic sequence

Answer 1

Answer:

2n² + n

Step-by-step explanation:

Given sequence:

3, 10, 21, 36, ...

Work out the first differences between terms:

$3 \underset{+7}{\longrightarrow} 10 \underset{+11}{\longrightarrow} 21 \underset{+15}{\longrightarrow} 36$

As the first differences are not the same, work out the second differences.

$7 \underset{+4}{\longrightarrow} 11 \underset{+4}{\longrightarrow} 15$

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

The coefficient of n² is always half of the second difference. Therefore, as the second difference is 4, the coefficient of n² is 2.

To work out the nth term of the sequence, write out the numbers in the sequence 2n² and compare with the given sequence:

$\begin{array}{|l|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4 \\\cline{1-5} 2n^2 & 2 & 8 & 18 & 32\\\cline{1-5} \sf Operation & +1 & +2& +3& +4\\\cline{1-5} \sf Sequence & 3 & 10 & 21 & 36\\\cline{1-5}\end{array}$

Therefore, we need to add n to 2n² to match the sequence.

Therefore, the nth term of this sequence is 2n² + n.

Learn more about sequences here:

brainly.com/question/27953040

brainly.com/question/27775450