(1)1+3+5+7+9+…+19=([1+19/2])2=100;
(2)1+3+5+7+9+…+(2n-1)+(2n+1)+(2n+3)=([1+2n+3/2])2=(n+2)2;
(3)101+103+…+197+199=([1+199/2])2-([1+99/2])2=10000-2500=7500.
故答案为:100;(n+2)2.
(1)1+3+5+7+9+…+19=([1+19/2])2=100;
(2)1+3+5+7+9+…+(2n-1)+(2n+1)+(2n+3)=([1+2n+3/2])2=(n+2)2;
(3)101+103+…+197+199=([1+199/2])2-([1+99/2])2=10000-2500=7500.
故答案为:100;(n+2)2.