已知Sn是数列{an}的前n项和,an>0,Sn=(an²+an)/2(1)求Sn(2)若数列{bn}满足b(n+1)=2^an+bn,求bn哦、漏了、还有b1=2.
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![已知Sn是数列{an}的前n项和,an>0,Sn=(an²+an)/2(1)求Sn(2)若数列{bn}满足b(n+1)=2^an+bn,求bn哦、漏了、还有b1=2.](/uploads/image/z/677122-34-2.jpg?t=%E5%B7%B2%E7%9F%A5Sn%E6%98%AF%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2Can%3E0%2CSn%3D%28an%26%23178%3B%2Ban%29%2F2%281%29%E6%B1%82Sn%EF%BC%882%EF%BC%89%E8%8B%A5%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3b%28n%2B1%29%3D2%5Ean%2Bbn%2C%E6%B1%82bn%E5%93%A6%E3%80%81%E6%BC%8F%E4%BA%86%E3%80%81%E8%BF%98%E6%9C%89b1%3D2.)
已知Sn是数列{an}的前n项和,an>0,Sn=(an²+an)/2(1)求Sn(2)若数列{bn}满足b(n+1)=2^an+bn,求bn哦、漏了、还有b1=2.
已知Sn是数列{an}的前n项和,an>0,Sn=(an²+an)/2
(1)求Sn
(2)若数列{bn}满足b(n+1)=2^an+bn,求bn
哦、漏了、还有b1=2.
已知Sn是数列{an}的前n项和,an>0,Sn=(an²+an)/2(1)求Sn(2)若数列{bn}满足b(n+1)=2^an+bn,求bn哦、漏了、还有b1=2.
1.
n=1时,S1=a1=(a1²+a1)/2,整理,得
a1²-a1=0
a1(a1-1)=0
a1=0(与已知不符,舍去)或a1=1
S1=a1=1
n≥2时,
Sn=(an²+an)/2 S(n-1)=[a(n-1)²+a(n-1)]/2
an=Sn-S(n-1)=(an²+an)/2-[a(n-1)²+a(n-1)]/2
整理,得
an²-a(n-1)²-an-a(n-1)=0
[an+a(n-1)][an-a(n-1)]-[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-1]=0
an>0,an+a(n-1)>0,要等式成立,只有an-a(n-1)-1=0
an-a(n-1)=1,为定值.
数列{an}是以1为首项,1为公差的等差数列.
an=n
Sn=(n²+n)/2
2.
b(n+1)=2^an +bn=2ⁿ+bn
b(n+1)-bn=2ⁿ
bn-b(n-1)=2^(n-1)
b(n-1)-b(n-2)=2^(n-2)
…………
b2-b1=2
累加
bn-b1=2+2²+...+2^(n-1)
题目缺少条件,估计原题有b1=1,那么:
bn=b1+2+2²+...+2^(n-1)=1+2+2²+...+2^(n-1)=1×(2ⁿ-1)/(2-1)=2ⁿ-1
(1)an=n
(2)bn=2𠆢n-1