Add the last number onto the end. So the sum of all numbers between 1 and 101 is 50 pairs of 101 plus one extra 101 and the end. It’d end up being 5050 + 101 or 51x101 or 5151
Nope, because what you’re doing is copying the entire sequence, reversing it, and pairing up each element left to right. There’s no way to have any leftovers because the original sequence and the new reversed sequence have the same number of elements.
A perhaps less intuitive way of thinking of it is you start with a sequence of 1 up to N, which contains exactly N elements. The sequence from 1 to N and its reverse together contain 2N elements, which is by definition an even number, regardless of whether N is even or odd. Because it’s even we can break it into pairs without leftovers.
Sorry if this is stupid but how to deal with sums to odd numbers ? Won’t you have a number left over after pairing all the others?
Add the last number onto the end. So the sum of all numbers between 1 and 101 is 50 pairs of 101 plus one extra 101 and the end. It’d end up being 5050 + 101 or 51x101 or 5151
Nope, because what you’re doing is copying the entire sequence, reversing it, and pairing up each element left to right. There’s no way to have any leftovers because the original sequence and the new reversed sequence have the same number of elements.
A perhaps less intuitive way of thinking of it is you start with a sequence of 1 up to N, which contains exactly N elements. The sequence from 1 to N and its reverse together contain 2N elements, which is by definition an even number, regardless of whether N is even or odd. Because it’s even we can break it into pairs without leftovers.