数列an的前n项和为Sn且an+Sn= -2n-1若数列bn满足b1=1且b(n+1)=bn+nan求bn的通项公式 急

题目

数列an的前n项和为Sn且an+Sn= -2n-1若数列bn满足b1=1且b(n+1)=bn+nan求bn的通项公式 急

答案

an+Sn= -2n-1 a1=-3/2 a1+1=-1/2
an-1+sn-1=-2n+1
2an-an-1=-2
2an+2=an-1+1-1
an+1=1/2(an-1+1)-1/2
1/2(an-1+1)=1/4(an-2+1)-1/4
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1/2^(n-2)(a2+1)=1/2^(n-1)(a1+1)-1/2^(n-1)
an+1=1/2^(n-1)(-1/2)-(1-1/2^(n-1))=-1/2^n-1+2/2^n
an=-2+1/2^n
b(n+1)=bn+nan=bn+n/2^n-2n
bn=b(n-1)+(n-1)/2^(n-1)-2(n-1)
b2=b1+1/2-2
2+...+2(n-1)=2(n-1+1)(n-1)/2=n(n-1)
s=1/2+2/2^2+3/2^3+..+(n-1)/2^(n-1)
s/2=1/2^2+2/2^2+3/2^3+...+(n-2)/2^(n-1)+(n-1)/2^n
s/2=1/2+1/2^2+1/2^3+...+1/2^(n-1)-(n-1)/2^n=1-1/2^(n-1))-(n-1)/2^n=1-(n+1)/2^n
s=2-(n+1)/2^(n-1)
bn=b1-n(n-1)+2-(n+1)/2^(n-1)=3-n(n-1)-(n+1)/2^(n-1)