(2n)^2/(2n-1)(2n+1)=(2n)^2/[(2n)²-1]=1+1/[(2n)²-1]=1+1/(2n-1)(2n+1)=1+1/2[1/(2n-1)-1/(2n+1)]∴Sn=2^2/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)=n+1/2[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]=n+1/2[1-1/...∴Sn=2^2/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)
=n+1/2[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]為啥1/2前面的東西加起來是n,能詳細(xì)解答下嗎，后面過程都懂了1/2[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=1/2×2n/(2n+1)
=n/(2n+1)

1. 這一步我是懂的，我是問2^2/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)=n+1/2[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]中的n咋加的
         

(2n)^2/(2n-1)(2n+1)
=(2n)^2/[(2n)²-1]
=1+1/[(2n)²-1]
=1+1/(2n-1)(2n+1)
=1+1/2[1/(2n-1)-1/(2n+1)]
所以每一項(xiàng)都有一個(gè)1
n個(gè)1相加當(dāng)然等于n啰