数列的问题2

求和{Sn=(2^2/1.3)+4^2/3.5+…(2n)^2/[(2n-1)(2n+1)]

热心网友

(2n)^2/[(2n-1)(2n+1) = {[2n)^2 -1]+1}/[(2n-1)(2n+1)]= [(2n-1)(2n+1)+1]/[(2n-1)(2n+1)]= 1 + [1/(2n-1) - 1/(2n+1)]/2==== Sn=(2^2/1.3)+4^2/3.5+…(2n)^2/[(2n-1)(2n+1)]= n + [ 1/1 - 1/(2n+1)]/2= 2n(n+1)/(2n+1)