等差数列求和公式推导

等差数列求和公式推导 等差数列求和公式推导等差数列求和公式推导过程二级等差数列求和公式及推导

去百度文库，查看完整内容> 内容来自用户:xc3678658119 二级等差数列求和公式就是后一项减前一项是等差数列,怎样求原数列的和?二级等差数列求和公式就是后一项减前一项是等差数列,怎样求原数列的和?a2-a1=ka3-a2=k+da4-a3=k+2d……an-a(n-1)=k+(n-2)d相加an-a1=(n-1)k+[1+2+……+(n-2)]d=(n-1)k+(n-2)(n-1)d/2所以an=a1+(n-1)k+(n-2)(n-1)d/2二阶等差数列怎样求和a1=1an-a(n-1)=2n-1Sn=?a1 = 1a2 - a1 = 2*2 -1a3 - a2 = 2*3 -1a4 - a3 = 2*4 -1……an - a(n-1) = 2*n - 1以上等式相加后,得到通项公式an = 1 + 2(2+3+4+……+n) - 1-1-1-……-1=2(1+2+3+……+n) - n=n(n+1) - n=n^2检验:a2 - a1 = 4 - 1 = 2*2 - 1a3 - a2 = 9 - 4 = 2*3 - 1a4 - a3 = 16 -9 = 2*4 - 1成立下面求Sn = 1^2 + 2^2 + 3^2 +……+ n^2(n+1)^3 - n^3 = (n^3 + 3n^2 + 3n + 1) - n^3 = 3*n^2 + 3n + 1利用上面这个式子有：2^3 - 1^3 = 3*1^2 + 3*1 + 13^3 - 2^3 = 3*2^2 + 3*2 + 14^3 - 3^3 = 3*3^2 + 3*3 + 15^3 - 4^3 = 3*4^2 + 3*4 + 1……(n+1)^3 - n^3 = 3*n^2 + 3n + 1把上述各等式左右分别相加得到：(n+1)^3 - 1^3 = 3*(1^2+2^2+3^2+……+n^2) + 3*(1+2+3+……+n) + n*1n^3 + 3n^2 + 3n + 1 - 1 = 3*(1^2+2^2+3^2+……+n^2) + 3*n(n+1)/2 + n整理1^2 + 2^2 + 3^2 +……+ n^2 = n(n+1)(2n+1