Question

a1=1,2an+1=(1+1/n)^2*an 证明{an/n^2}为等比数列 求{an}通项公式 令bn=（an+1）

Answer 1

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

2^(n+1)a(n+1)=(n+1)^2[2^na(n)]/n^2

2^(n+1)a(n+1)/(n+1)^2=2^na(n)/n^2=...=2a(1)/1=2

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

b(n)=a(n+1)=(n+1)^2/2^n=n(n+1)/2^n + (n+1)/2^n=c(n)+d(n)

c(n)=n(n+1)/2^n,d(n)=(n+1)/2^n

D(n)=d(1)+d(2)+...+d(n)=2/2 + 3/2^2 + 4/2^3 + ...+ (n-1+1)/2^(n-1) + (n+1)/2^n

2D(n)=2 + 3/2 + 4/2^2 +...+ (n-1+1)/2^(n-2) + (n+1)/2^(n-1)

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

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

=3-(n+3)/2^n

C(n)=c(1)+c(2)+...+c(n)=1*2/2 + 2*3/2^2 + 3*4/2^3 + ...+(n-1)n/2^(n-1) + n(n+1)/2^n

2C(n)=1*2 + 2*3/2 + 3*4/2^2 + ...+ (n-1)n/2^(n-2) + n(n+1)/2^(n-1)

C(n)=2C(n)-C(n)=1*2 + 2(3-1)/2 + 3(4-2)/2^2 + ...+ n(n+1-n+1)/2^(n-1) - n(n+1)/2^n

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

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

S(n)=C(n)+D(n)=2+2[D(n)-(n+1)/2^n] - n(n+1)/2^n + D(n)

=2-n(n+1)/2^n - 2(n+1)/2^n + 3D(n)

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

=11-(n^2+6n+11)/2^n