The closed formula for this particular sequence is an = n² + 2.
Take note that the odd numbers 3, 5, 7, 9, and 11 are separate consecutive terms. This shows that the first n odd numbers can be added to the initial term, az, to get the nth term. Hence, the following is how we may represent the nth term a = az + 1 + 3 + 5 + ... + (2n-3) (2n-3). We may utilize the formula for the sum of an arithmetic series to make the sum of odd integers simpler that is 1 + 3 + 5 + ... + (2n-3) = n².
As a result, we get a = az + n^2 - 1. In conclusion, the equation for the series (an)n21, where a1 = az and an is the result of adding the first n odd numbers to az, is as a = az + n^2 - 1. We have the following for the given series where a1 = az = 3.
So, the closed formula for this particular sequence is an = n² + 2.
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Your question is incomplete. The complete question is:
Find the closed formula for the sequence (an)n21. Assume the first term given is az. an = 3, 6, 11, 18, 27... Hint: Think about the perfect squares.