求和:1+2+3+ … +(n-2)的和

2025-03-16 23:26:47
推荐回答(5个)
回答1:

(1+n) * n/2
适用于等差数列 :
(首项+末项)*项数/2=数列和
例题:1+2+3+4+5……+99+100
1就是首项,100就是末项,一共有100个项数
1+2+3+...+100
=(1+100)*100/2
=101*100/2
=10100/2
=5050
另外:末项=首项+(项数-1)*公差
项数=(末项-首项)/公差+1
首项=末项-(项数-1)*公差

回答2:

(1+n) * n/2
适用于等差数列 :
(首项+末项)*项数/2=数列和
例题:1+2+3+4+5……+99+100
1就是首项,100就是末项,一共有100个项数
1+2+3+...+100
=(1+100)*100/2
=101*100/2
=10100/2
=5050
另外:末项=首项+(项数-1)*公差
项数=(末项-首项)/公差+1
首项=末项-(项数-1)*公差

回答3:

因为是连续自然数的和,一共有n-2个数,首项是1,末项是n-2。所以和就是(n-2)×(1+n-2)÷2=(n-2)(n-1)/2

回答4:

该等差数列前n-2之和
=(1+n)×n÷2-n-(n-1)
=(n+n²)╱2-2n+1
=n²╱2-3n╱2+1
=(n-1)×(n-2)╱2

回答5:

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