MATHS

Asked by YeeunFirst we make some nth term that almost fits what we have. Almost, because we have to compare our nth term and the question’s nth term to fill in the blanks. So since it’s a quadratic sequence, we have to find the second difference, halve it and stick it in front of the n^2. Our first differences are: 6 (+4) 10 (+8) 18 (+12) 30 Our second differences are: 4 (+4) 8 (+4) 12 So our second difference is 4. Then we halve it and stick the number in front of n^2. This gives us 2n^2. This is our nth term that almost fits. The sequence with nth term 2n^2 has terms: 2, 8, 18, 32. To get from our template sequence to the sequence in the question, we have to: (+4), (+2), (+0) (+-2) These differences form a subsequence: 4, 2, 0, -2, which is linear and has the common difference -2. It has an nth term of -2n + 6. So if we add this onto our template, we should reach the quadratic sequence in the question. This gives us an nth term rule of 2n^2 - 2n + 6. To double check, the terms of this sequence are: 2(1) - 2 + 6, 2(4) - 4 + 6, 2(9) - 6 + 6, 2(16) - 8 + 6 = 6, 10, 18, 30 which is the same as the sequence in the question. A little extra but it’s good peace of mind. Hope that makes sense ~ Sohini

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