！！在等差数列{an}中,已知a1=1,前n项和Sn满足条件S

Question

20.第二问不太一样 看着点！！！！！！！在等差数列{an}中,已知a1=1,前n项和Sn满足条件S2n/Sn=4,n=1，2，…
（1）求数列{an}的通项公式及Sn
（2）记bn=an×2^(an），求数列{bn}的前n项和Tn

Answer 1

a(n) = 1 + (n-1)d,
s(n) = n + n(n-1)d/2.
s(2n) = 2n + n(2n-1)d.

4 = s(2n)/s(n) = [2n + n(2n-1)d]/[n + n(n-1)d/2] = [2 + (2n-1)d]/[1 + (n-1)d/2],

4 + 2(n-1)d = 2 + (2n-1)d, d = 2.

a(n) = 1 + 2(n-1) = 2n-1.
s(n) = n + n(n-1) = n^2.

b(n) = a(n)2^[a(n)] = (2n-1)2^(2n-1) = (4n-2)2^(2n-2) = (4n-2)4^(n-1),

t(n) = b(1) + b(2) + b(3) + ... + b(n-1) + b(n)
= (4*1-2) + (4*2-2)4 + (4*3-2)4^2 + ... + [4(n-1)-2]4^(n-2) + [4n-2]4^(n-1),

4t(n) = (4*1-2)4 + (4*2-2)4^2 + ... + [4(n-1)-2]4^(n-1) + [4n-2]4^n,

3t(n) = 4t(n) - t(n) = -(4*1-2) - 4*4 - 4*4^2 - ... - 4*4^(n-1) + (4n-2)4^n
= (4n-2)4^n + 2 - 4(1+4+...+4^(n-1))
= (4n-2)4^n + 2 - 4[4^n - 1]/(4-1)
= (4n-2)4^n + 2 - (4/3)[4^n - 1]
= [(12n-10)/3]4^n + 10/3,

t(n) = [(12n-10)/9]4^n + 10/9

Answer 2

Sn =(a1 + an )*n/2
S2n =(a1 + a2n )*2n/2
S2n/Sn = (a1 + a2n)*2/(a1 + an )
an = a1 + (n-1)*d
a2n = a1 + (2n-1)*d
S2n/Sn = 4
带入得2*a1+(2n-1)*d = 2*(2*a1+(n-1)*d)
a1=1,解得d=2*a1=2