Question

已知数列{an}的前n项和为Sn,a1=1/2，且Sn=n^2An-n(n-1)

已知数列{an}的前n项和为Sn,a1=1/2，且Sn=n^2An-n(n-1) (1)写出sn与sn-1的递推关系 并求sn关于n的表达式 (2)设f(x)=(sn/n)
x^(n+1).bn=f(x)‘’(p),p∈r 求数列<bn>的前n项和tn

Answer 1

s(n) = n^2a(n) - n(n-1),
s(n+1) = (n+1)^2a(n+1) - (n+1)n = (n+1)^2[s(n+1)-s(n)] - (n+1)n,
n(n+2)s(n+1) = (n+1)^2s(n) + (n+1)n
s(n+1) = (n+1)[(n+1)s(n)+n]/[n(n+2)],

(n+2)s(n+1)/(n+1) = (n+1)s(n)/n + 1,
{(n+1)s(n)/n}是首项为2s(1)/1 = 2a(1)=1,公差为1的等差数列。

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

f(x) = [s(n)/n]x^(n+1) = nx^(n+1)/(n+1).
f'(x) = nx^n,
f''(x) = n^2x^(n-1),

b(n) = n^2p^(n-1),

t(n) = b(1) + b(2) + b(3) + ... + b(n-1) + b(n) = 1 + 2^2p + 3^2p^2 + ... + (n-1)^2p^(n-2) + n^2p^(n-1),

p=1时，b(n) = n^2,
t(n) = n(n+1)(2n+1)/6.

p不为1时，
pt(n) = p + 2^2p^2 + 3^2p^3 + ... + (n-1)^2p^(n-1) + n^2p^n,
(1-p)t(n) = 1 + (2^2-1)p + (3^2-2^2)p^2 + ... + [n^2-(n-1)^2]p^(n-1) - n^2p^n

= 1 + (2*1+1)p + (2*2+1)p^2 + ... + [2(n-1)+1]p^(n-1) - n^2p^n
= 1 + p + p^2 + ... + p^(n-1) + 2[1*p + 2*p^2 + 3*p^3 + ... + (n-2)p^(n-2) + (n-1)p^(n-1)] - n^2p^n
= [1 - p^n]/(1-p) + 2c(n) - n^2p^n.

c(n) = 1*p + 2*p^2 + 3*p^3 + ... + (n-2)p^(n-2) + (n-1)p^(n-1),
pc(n) = 1*p^2 + 2*p^3 + ... + (n-2)p^(n-1)+ (n-1)p^n,
(1-p)c(n) = p + p^2 + p^3 + ... + p^(n-1) - (n-1)p^n
= 1 + p + p^2 + ... + p^(n-1) - 1 - (n-1)p^n
= [1-p^n]/(1-p) - 1 - (n-1)p^n.

c(n) = [1-p^n]/(1-p)^2 - 1/(1-p) - (n-1)p^n/(1-p).

(1-p)t(n) = [1-p^n]/(1-p) + 2c(n) - n^2p^n
= [1-p^n]/(1-p) + 2[1-p^n]/(1-p)^2 - 2/(1-p) - 2(n-1)p^n/(1-p) - n^2p^n,

t(n) = [1-p^n]/(1-p)^2 + 2[1-p^n]/(1-p)^3 - 2/(1-p)^2 - 2(n-1)p^n/(1-p)^2 - n^2p^n/(1-p).