Question

高一数学，数列求和问题，，

Answer 1

答案：前 n 项和为 [ n(n-1) ]／[ 2(2n+1) ]

－－－－－－－－－－－

解析：

( 有问题欢迎追问 @_@ )

Answer 2

An = (n-1)(n+1)/[2*(n-1/2)*2*(n+1/2)]
= 1/4 * (n²-1)/(n² - 1/4)
= 1/4 * [(n²-1/4)-3/4]/(n² - 1/4)
= 1/4 [1 - (3/4)/(n² -1/4)]
= 1/4 - 3/16 * 1/[(n-1/2)(n+1/2)]
= 1/4 - 3/16 * [(n+1/2)-(n-1/2)]/[(n-1/2)(n+1/2)]
= 1/4 - 3/16 * [1/(n-1/2) - 1/(n+1/2)]
= 1/4 - 3/16 * [2/(2n-1) - 2/(2n+1)]
所以，前 n 项的和为：
S = n/4 - 3/16 * [(2/1 - 2/3) + (2/3 - 2/5) + (2/5 - 2/7) + …… + 2/(2n-1) - 2/(2n+1)]
= n/4 - 3/16 * [2/1 - 2/(2n+1)]
= n/4 - 3/16 * 2 * [1- 1/(2n+1)]
= n/4 - 3/8 * [(2n+1) - 1]/(2n+1)
= n/4 - 3/8 * (2n)/(2n+1)
= n/4 - 3/4 * n/(2n+1)
= n/4 * [1 - 3/(2n+1)]
= n/4 * [(2n+1) - 3]/(2n+1)
= n/4 * (2n-2)/(2n+1)
= n/2 * (n-1)/(2n+1)
= n(n-1)/(4n+2)

Answer 3

可以先分离系数，再裂项

所以可以求和得：n/4+(3/8)[1-1/(2n+1)]=n/4+3n/(8n+4)

=(n²+2n)/(4n+2)

上面是减3/4，顺应的都改为“减号”