$解:因为1×3=(2×1-1)(2×1+1)=4×1^2-1,$
$3×5=(2×2-1)(2×2+1)=4×2^2-1,$
$5×7=(2×3-1)(2×3+1)=4×3^2-1, $
$···$
$(2n-1)(2n+1)=4n^2-1，$
$所以1×3+3×5+5×7+···+(2n-1)(2n+1)$
$=4×1^2-1+4×2^2-1+4×3^2-1+···+4×n^2-1$
$=4×(1^2+2^2+3^2+···+n^2)-n$
$=4×\frac {1}{6}n(n+1)(2n+1)-n$
$=\frac {2}{3}n(n+1)(2n+1)-n.$
