1+3+5+7+…+(2n-1)=( ) (n为正整数)
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1+3+5+7+…+(2n-1)=( ) (n为正整数)
1+3+5+7+…+(2n-1)=( ) (n为正整数)
1+3+5+7+…+(2n-1)=( ) (n为正整数)
1+3+5+7+…+(2n-1) = n²
等差数列求和公式
S= (首项 + 末项) ×项数 ÷ 2
= (1 + 2n -1) × n ÷ 2
= 2n× n ÷ 2
= n²
1+3+5+7+…+(2n-1)
=[1+(2n-1)]*n/2
=n^2
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