用裂项法来解1／x^2-1＋1／x^2+4x+3＋1／x^2+8x+15＋......＋1／x^2+4nx+4n^2-1

显示全部楼层 · 2012-11-29 19:35:24

1/(x2-1)+1/(x2+4x+3)+1/(x2+8x+15)+…+1/(x2+4nx+4n2-1)=1/(x-1)(x+1) + 1/(x+1)(x+3) + 1/(x+3)(x+5) + ... + 1/(x+2n-1)(x+2n+1)=1/2 * [1/(x-1) - 1/(x+1) + 1/(x+1) - 1/(x+3) + 1/(x+3) - 1/(x+5) + ... + 1/(x+2n-1) - 1/(x+2n+1)]中间的各式相加为零=1/2 * [1/(x-1) - 1/(x+2n+1)]=1/2 * [(x+2n+1-x+1)/(x-1)(x+2n+1)]...

千问 · 2012-11-29 19:35:24

1/x^2-1+1/x^2+4x+3+1/x^2+8x+15+…+1/x^2+4nx+4n^2-1 =a*1/x^2+bx+c 代入n=1,2,3 求a,b,c;再归纳与n的关系。...