①1+3=4; 1+3+5=9; 1+3+5+7=16; 1+3+5+7+9=25. ②从以上个题得出规律:有几个连续的奇数相加,和就是几乘几. 1+3+5+7+9+11, =6×6, =36; 1+3+5+7+9+11+13, =7×7, =49. ③1+3+5+7+…+99, =[(99-1)÷2+1]×[(99-1)÷2+1], =50×50, =2500; 101+103+105+…+199, =100×[(199-101)÷2+1]+(1+3+5+…+99), =100×50+[(199-101)÷2+1]×[(199-101)÷2+1], =5000+50×50, =7500. |