We have two responses for you

You must have seen that the first differences were 2,3,4,5,6 so a reasonable expectation is that the next difference and the next term are 7 and 29 respectively. That 7th term = 29 is therefore the first term of your progression, 2, plus the sum of the first 6 differences 2+3+4+5+6+7. The nth term will be the first term of your progression, 2, plus the sum of the first n-1 differences = 2+3+4+... + n. Thus in general you need to know the sum 2 + 2+3+4+... + n = 1 + 1 + 2+3+4+... + n. I suspect that you've seen the sum 1+2+3+4+... + n before.

Penny

Hi Simon.

Take a look at the sequence 1, 3, 6, 10, 15, 21, ... This is exactly one lower than your sequence, so if I can come up with a formula for the nth term in my sequence, yours is the same + 1.

The 1st element is 1. The 2nd is 1+2, the 3rd is 1+2+3, the 4th is 1+2+3+4, etc.

There is a well-known formula for the sum of the first n integers. That would represent the nth term for me. Can you find this formula (or derive it, even better!) and add one to it?

Stephen La Rocque.