



 
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 n1 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 wellknown 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.  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 