求和：Sn=2^/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)
问题描述：
求和：Sn=2^/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)
答
=(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)]
所以每一项都有一个1
n个1相加当然等于n啰
(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)]为啥1/2前面的东西加起来是n,能详细解答下吗，后面过程都懂了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)
这一步我是懂的，我是问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)²-1]
=1+1/[(2n)²-1]
=1+1/(2n-1)(2n+1)
=1+1/2[1/(2n-1)-1/(2n+1)]
所以每一项都有一个1
n个1相加当然等于n啰