一道数学题 急
2^(n+1)a(n+1)=1/2((n+1)^2-5(n+1)-n^2+5n)=n-2;
2^na(n)=n-3,
a(n)=n/2^n-3/2^n,
n=1,a(1）=1/2-3/2=-1；
或，代入2a(1)=(1-5)/2,a(1)=-1,
a(n+1)a(n-1)=(n-2)(n-4)/(2^n)^2≠(a(n))^2,
a(n1)=n/2^n,
s(n1)*(1-1/2)=1/2+1/4+1/8+…+1/2^n-n/2^(n+1）
s(n1)=2-(n+2)/2^n;
a(n2)=1/2^n,
s(n2)=1-1/2^n,
s(n)=2-(n+2)/2^n-3+3/2^n=-1...全部
2^(n+1)a(n+1)=1/2((n+1)^2-5(n+1)-n^2+5n)=n-2;
2^na(n)=n-3,
a(n)=n/2^n-3/2^n,
n=1,a(1）=1/2-3/2=-1；
或，代入2a(1)=(1-5)/2,a(1)=-1,
a(n+1)a(n-1)=(n-2)(n-4)/(2^n)^2≠(a(n))^2,
a(n1)=n/2^n,
s(n1)*(1-1/2)=1/2+1/4+1/8+…+1/2^n-n/2^(n+1）
s(n1)=2-(n+2)/2^n;
a(n2)=1/2^n,
s(n2)=1-1/2^n,
s(n)=2-(n+2)/2^n-3+3/2^n=-1-(n-1)/2^n。
收起
