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3. We can name sequence A to be 1+2+3+…n.
We can arrange 2A as (1+n) + (2+n-1) + (3+n-2)….(n+1), and it simplifies to (1+n) + (1+n) + (1+n) + (1+n)…(1+n), a total of n times. So, 2A = n(n+1). To get A, we can divide by 2 to get A = n(n+1)/2. I am not sure how to prove the sum of squares, though.