1 + 3 = (1 + 3) * 2 / 2; 1 + 3 + 5 = (1 + 5) * 3; 1 + 3 + 5 + 7 = (1 + 7) * 4 / 2 ', (2n-1) is equal to? a n*n b (n+1)*(n+1) c (1+n)*n/2 d (n+1)*n/2

1 + 3 = (1 + 3) * 2 / 2; 1 + 3 + 5 = (1 + 5) * 3; 1 + 3 + 5 + 7 = (1 + 7) * 4 / 2 ', (2n-1) is equal to? a n*n b (n+1)*(n+1) c (1+n)*n/2 d (n+1)*n/2

The first three show the law
Sum = (first item + last item) * number of items / 2
Number of items = (last item + 1) / 2
So 1 + 3 + 5 + 7 + 9 + '+ (2n-1)
=（1+2n-1)*(2n/2)/2
=2n*n/2
=n*n
Choose a

(1 + 2 + 3 +... + n) / N ^ 2 n tends to infinite limit
The solution to this limit is very simple, that is, add the molecules to get 1 / 2n (n + 1) / N ^ 2, and finally equal to 1 / 2
But if I separate each item into 1 / N ^ 2 2 / N ^ 2. N / N ^ 2, I find that each item tends to 0. And should be 0. Where is the low error? I've been depressed all night. I'm looking for solutions from experts. I don't have much points. I just want to ask for solutions from good people. Thank you. Wait

The addition of infinitely many zeros is an infinitive
It is equivalent to 0 / 0 or ∞ / ∞ type indeterminate, so the case of adding infinite numbers can not be divided into the case of finding the limit of each term and then adding