作业帮等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,最前面的d/2和(a1-d/2)n听说能等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,里面有个性质,最前面的d/2和(a1-d/2)n
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![等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,最前面的d/2和(a1-d/2)n听说能等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,里面有个性质,最前面的d/2和(a1-d/2)n](/uploads/image/z/693045-45-5.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E6%B1%82%E5%92%8C%E5%85%AC%E5%BC%8FSn%3D%28d%2F2%29%2An%5E2%2B%28a1-d%2F2%29n%E6%98%AF%E7%94%B1+Sn%3DnA1%2B%5Bn%28n-1%29d%5D%2F2%E5%8C%96%E7%AE%80%E8%80%8C%E6%9D%A5%2C%E6%9C%80%E5%89%8D%E9%9D%A2%E7%9A%84d%2F2%E5%92%8C%28a1-d%2F2%29n%E5%90%AC%E8%AF%B4%E8%83%BD%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E6%B1%82%E5%92%8C%E5%85%AC%E5%BC%8FSn%3D%28d%2F2%29%2An%5E2%2B%28a1-d%2F2%29n%E6%98%AF%E7%94%B1+Sn%3DnA1%2B%5Bn%28n-1%29d%5D%2F2%E5%8C%96%E7%AE%80%E8%80%8C%E6%9D%A5%2C%E9%87%8C%E9%9D%A2%E6%9C%89%E4%B8%AA%E6%80%A7%E8%B4%A8%2C%E6%9C%80%E5%89%8D%E9%9D%A2%E7%9A%84d%2F2%E5%92%8C%28a1-d%2F2%29n)
等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,最前面的d/2和(a1-d/2)n听说能等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,里面有个性质,最前面的d/2和(a1-d/2)n
等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,最前面的d/2和(a1-d/2)n听说能
等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,里面有个性质,最前面的d/2和(a1-d/2)n好像能直接求出什么的,不用过分麻烦,不过没记忘记礼物,有谁知道,速回.
等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,最前面的d/2和(a1-d/2)n听说能等差数列求和公式Sn=(d/2)*n^2+(a1-d/2)n是由 Sn=nA1+[n(n-1)d]/2化简而来,里面有个性质,最前面的d/2和(a1-d/2)n
通项公式：
An=A1+(n-1)d
An=Am+(n-m)d
等差数列的前n项和：
Sn=[n(A1+An)]/2; Sn=nA1+[n(n-1)d]/2
等差数列求和公式:等差数列的和=(首数+尾数)*项数/2;
项数的公式:等差数列的项数=[(尾数-首数)/公差]+1.
化简得(n-1)an-1-(n-2)an=a1,这对于任一N均成立
当n取n-1时式子变为,(n-3)an-1-(n-2)an-2=a1=(n-2)an-(n-1)an-1
得
2(n-2)an-1=(n-2)*(an+an-2)
当n大于2时得2an-1=an+an-2 显然证得它是等差数列
和=（首项+末项）×项数÷2
项数=（末项-首项）÷公差+1
首项=2和÷项数-末项
末项=2和÷项数-首项
末项=首项+（项数-1）×公差
性质：
若 m、n、p、q∈N
①若m+n=p+q,则am+an=ap+aq
②若m+n=2q,则am+an=2aq
注意：上述公式中an表示等差数列的第n项.
求和公式
Sn=(a1+an)n/2
Sn=n(2a1+(n-1)d)/2; d=公差
Sn=An2+Bn; A=d/2,B=a1-(d/2)