Question

前n项和为sn=n2-n／2 则数列

Answer 1

等下,计算中……

Tn =[ 2^(n+1) / (n+1) ] - 2

设Tn的项为bn;

则bn = (n-1)2^n / n*(n+1)

=(2^n/n) - (1/n)(2^(n+1)/(n+1))

Tn = (2^1 / 1) +……+(2^n-1)/(n-1) - (1/(n-1))(2^n/n) + (2^n/n) - 1/n (2^(n+1)/(n+1))

= (2^1 / 1) +……+(2^n-1)/(n-1) - [(1/(n-1))(2^n/n)+(2^n/n) ] - 1/n (2^(n+1)/(n+1))

∵ [(1/(n-1))(2^n/n)+(2^n/n) ]= 2*b(n-1)

∴ =2+2*b1+……+2b(n-1) - 1/n(2^(n+1) / (n+1))

2Tn = 2b1+…………+2bn

下面的式子减上面的式子得：Tn = 2bn + ( 1/n )(2^(n+1)/(n+1)) - 2

=( 2^(n+1) / (n+1) ) -2

很辛苦的说……望采纳啊,亲!