∵1/[(n-2)·n·(n+2)]=(1/8)·[1/(n-2)-2/n+1/(n+2)]
∴
1/(1·3·5)=(1/8)·(1/1-2/3+1/5)
1/(3·5·7)=(1/8)·(1/3-2/5+1/7)
1/(5·7·9)=(1/8)·(1/5-2/7+1/9)
……
1/(35·37·39)=(1/8)·(1/35-2/37+1/39)
1/(37·39·41)=(1/8)·(1/37-2/39+1/41)
1/(39·41·43)=(1/8)·(1/39-2/41+1/43)
∴1/(1·3·5)+1/(3·5·7)+1/(5·7·9)+…+1/(39·41·43)
=(1/8)·[(1/1-2/3+1/5)+(1/3-2/5+1/7)+(1/5-2/7+1/9)+…+(1/35-2/37+1/39)+(1/37-2/39+1/41)+(1/39-2/41+1/43)]
=(1/8)·(1/1-1/3-1/39+1/43)
=557/6708