二级等差数列求和公式及推导.doc

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+2dan-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 -1an - 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=n2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 = 12 + 22 + 32 + + n2(n+1)3 - n3 = (n3 + 3n2 + 3n + 1) - n3 = 3*n2 + 3n + 1 利用上面这个式子有： 23 - 13 = 3*12 + 3*1 + 1 33 - 23 = 3*22 + 3*2 + 1 43 - 33 = 3*32 + 3*3 + 1 53 - 43 = 3*42 + 3*4 + 1 (n+1)3 - n3 = 3*n2 + 3n + 1 把上述各等式左右分别相加 得到： (n+1)3 - 13 = 3*(12+22+32+n2) + 3*(1+2+3+n) + n*1 n3 + 3n2 + 3n + 1 - 1 = 3*(12+22+32+n2) + 3*n(n+1)/2 + n 整理12 + 22 + 32 + + n2 = n(n+1)(2n+1)/6Sn = n(n+1)(2n+1)/6