求和Sn=2²/1·3+4²/3·5+...+(2n)²/(2n-1)(2n+1)

题目

求和Sn=2²/1·3+4²/3·5+...+(2n)²/(2n-1)(2n+1)

答案

令an=(2n)²/(2n-1)(2n+1)
=1/[1-(1/2n)][1+(1/2n)]
=(1/2)*{[1-(1/2n)]+[1+(1/2n)]}/[1-(1/2n)][1+(1/2n)]
=(1/2)*{1/[1+(1/2n)] + 1/[1-(1/2n)]}
=(1/2)*[2n/(2n+1) + 2n/(2n-1)]
=(1/2)*{1-[1/(2n+1)]+1+[1/(2n-1)]}
=1 + (1/2)*[1/(2n-1) - 1/(2n+1)]
Sn
=a1+a2+...+an
=1*n + (1/2)*[(1/1-1/3)+(1/3 - 1/5)+(1/5 - 1/7)+.+1/(2n-1) - 1/(2n+1)]
=n + (1/2)*[1 - 1/(2n+1) ]
=n{1+[1/(2n+1)]}