Question 1176668
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A formal algebraic solution would probably be very messy....<br>
But since the sequence is all integers, we can find the sequence we are looking for simply by investigation.<br>
Second term 2: 1, 2, 4, 6, 9, 12, 16, 20, 25, 30;  sum 55
Second term 3: 1, 3, 9, 15, 25, 35, 49, 63, 81, 109;  sum 190
Second term 4: 1, 4, 16, 28, 49, 70, 100, 130, 169, 208;  sum 377
Second term 5: 1, 5, 25, 45, 81, 117, 169, 221, 289, 357; sum 646<br>
That is the sequence we want; continue the sequence until the numbers exceed 1000.<br>
1, 5, 25, 45, 81, 117, 169, 221, 289, 357, 441, 525, 625, 725, 841, 957, 1089<br>
The largest number in the sequence less than 1000 is 957.<br>