Question

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

Answer 1

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)]

=(n+1)/[(x-1)(x+2n+1)]