代数恒等式证明1/(1*n) +1/(2*(n-1))+1/(3*(n-2))+.+1/(i*(n-i+1))+.+1/

证明思路：
1/n=(1/1+1/n)/(n+1)
1/[2(n-1)]=[1/2+1/(n-1)]/(n+1)
……………………
所以可证明：
左边=1/(1*n) +1/(2*(n-1))+1/(3*(n-2))+……+1/(i*(n-i+1))+.……+1/(n*1)
=1/(n+1)(1/1+1/n+1/2+1/(n-1)+1/3+1/(n-2)+……+1/i+1/(n-i+1)+……+1/n+1/1)
=1/(n+1)[(1/1+1/2+1/3+……+1/i+……+1/n)+(1/n+1/(n-1)+1/(n-2)+……+1/(n-i+1)+……+1/1)]
=2/(n+1)(1/1+1/2+1/3+……+1/i+……+1/n)
=右边