Question

Here are the first five terms of a quadratic sequence.

5 11 21 35 53

Find, in terms of n , an expression for the nth term of this sequence

Answer 1

Answers

5, 11, 21, 35, 53, ... Find the nth term of this sequence.

By calculating the difference between each of the progressions, we see that the first difference is 6, the next is 10, then 14 and finally 18. It is easy to observe that the jump increases by 4 each time, and so we call this the second difference. Because the second difference is the same this tells us that the nth term will be quadratic and thus include a squared term. Halving the second difference will give us a value of 2 and tells us that the squared term is 2n^2. By putting this into the first term, we get 2(1)^2, which gives us 2. To reach 5 and satisfy the progression, we must add 3. In total, this gives us an nth term of 2n^2 + 3.

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

By calculating the difference between each of the progressions, we see that the first difference is 6, the next is 10, then 14 and finally 18. It is easy to observe that the jump increases by 4 each time.

Now, the squared term is 2 n^2. By putting this into the first term, we get 2 (1)^2, which gives us 2. To reach 5 and satisfy the progression, we must add 3. In total, this gives us an nth term of 2n^2 + 3.