已知等差数列{an}中,a2=5,其前n项和Sn=pn∧2-n,(n∈N*)求p的值及an,若bn=1/[(4n+1)*

(1) a2=S2-S1=(4p-2)-(p-1)=5,解得p=2
所以 a1=S1=p-1=1,d=a2-a1=4
an=a1+(n-1)d=4n-3
(2) bn=1/[(4n+1)(4n-3)]=(1/4)[1/(4n-3) -1/(4n+1)]
所以 Tn=b1+b2+...+bn
=(1/4)[1/1 -1/5 +1/5 -1/9+...+1/(4n-1) -1/(4n+3)]
=(1/4)[1-1/(4n+3)]
=(2n+1)/[2(4n+3)]
令 Tn>15/64
所以 (2n+1)/(4n+3)>15/32
64n+32>60n+45
n>13/4
从而n的最小值为4.

(1) a2=S2-S1=(4p-2)-(p-1)=5，解得p=2
所以 a1=S1=p-1=1，d=a2-a1=4
an=a1+(n-1)d=4n-3
(2) bn=1/[(4n+1)(4n-3)]=(1/4)[1/(4n-3) -1/(4n+1)]
所以 Tn=b1+b2+...+bn
=(1/4)[1/1 -1/5 +1/5 -1/9+...+1/(4...