Question

初一数学求解法，谢谢高手啊

1+3+5···+2n-3+2n-1

Answer 1

1+3+5…+2n-3+2n-1
解：原式 =(2×1-1)+(2×2-1)+（2×3-1)....+[2(n-1)-1]+(2n-1)
=2×1-1+2×2-1+2×3-1....+2(n-1)-1+2n-1
=2×1+2×2+2×3....+2(n-1)+2n-1×n
=2×(1+2+3...+n-1+n)-n
=2×[(1+n)×n÷2]-n
=2×(n+n的二次方）÷2-n
=n+n的二次方-n
=n的二次方

Answer 2

解：原式 =(2×1-1)+(2×2-1)+（2×3-1)....+[2(n-1)-1]+(2n-1)
=2×1-1+2×2-1+2×3-1....+2(n-1)-1+2n-1
=2×1+2×2+2×3....+2(n-1)+2n-1×n
=2×(1+2+3...+n-1+n)-n
=2×[(1+n)×n÷2]-n
=2×(n+n^2）÷2-n
=n+n^2-n
=n^2

Answer 3

(1+3+5···+2n-3+2n-1 +
2n-1+2n-3+...+5+3+1)/2
(2n)*n/2=n^2

或者利用数列的公式：
（1+2n-1）*n/2=n^2